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Solutions Manual For

Engineering Mechanics Manoj Kumar Harbola IIT Kanpur

1

Chapter 1 1.1

Rotational speed of the earth is very small (about 7  10 5 radians per second). Its effect on particle motion over small distances is therefore negligible. This will not be true for intercontinental missiles.

1.2

The net force on the (belt+person) system is zero. This can be seen as follows. To pull the rope up, the person also pushes the ground and therefore the belt on which he is standing. This gives zero net force on the belt. For the person, the ground pushes him up on the feet but the belt pulls him down when he pulls it, giving a zero net force on him.

1.3

A vector between coordinates (x1, y1, z1) and (x2,y2,z2) is given by ( x 2  x1 )iˆ  ( y 2  y1 ) ˆj  ( z 2  z1 ) kˆ . Thus (i), (ii) and (iv) are equal.

1.4

The vectors are (i)

2iˆ  3 ˆj  5kˆ

4iˆ  3 ˆj  kˆ

(ii)

(iii)

2iˆ  9 ˆj  5kˆ

(iv)

 3iˆ  3 ˆj  2kˆ

1.5

4iˆ  6 ˆj  10 kˆ

(i) The resultant vectors are 6iˆ  6kˆ , (ii) The resultant vectors are

1.6

2iˆ  6 ˆj  4kˆ

 A  B

,

2iˆ  6 ˆj  4kˆ

  A B

  B A

and  iˆ  3kˆ

 A

and 7iˆ  3kˆ

 B

1.7 On each reflection, the sign of the vector component perpendicular to the reflecting mirror changes. z

y 1.8

 v 2

O

x

The fly is flying along the vector from (2.5, 2, 0) to (5, 4, 4). This vector is 2.5iˆ  2 ˆj  4kˆ .

The unit vector in this direction is

2.5iˆ  2 ˆj  4kˆ 2.5iˆ  2 ˆj  4kˆ  . 6.25  4  16 26.25

The velocity of the fly is therefore 0.5 

2.5iˆ  2 ˆj  4kˆ 26.25

 0.25iˆ  .20 ˆj  0.39kˆ .





1.9 After time t, the position vectors rA and rB of particles A and B, respectively, are   rA  l sin t iˆ  l cos t ˆj rB  l sin t iˆ  l cos t ˆj   Their velocities v A and v B are given by differentiating these vectors with respect to

time to get   v A  l cos t iˆ  l sin t ˆj v B  l cos t iˆ  l sin t ˆj    Velocity v AB of A with respect to B is obtained by subtracting v B from v A    v AB  v A  v B  2l cos t iˆ

1.10

For rotation about the z-axis by an angle  v x '  v x cos   v y sin 

v y '   v x sin   v y cos 

v z'  v z

It is given that   30  . Therefore v x' 



1 vx 3  vy 2



v y' 

3



1  vx  v y 3 2



v z'  v z

1.11



Component of a vector A along an axis is given by its projection on that axis. This is obtained by taking the dot product of the vector with the unit vector along that axis. Thus   Ax  A  iˆ  A cos  1

  Az  A  kˆ  A cos  3

  Ay  A  ˆj  A cos  2

Also 2 A  Ax2  A y2  Az2

Substituting the expression for Ax, Ay and Az completes the proof. 1.12





(i) Dot product of two vectors A and B is   A  B  Ax B x  Ay B y  Az B z

This gives the dot product of the first vector of problem 1.4 each of the other vectors to be 4, 2 and 25. 



(ii) Cross product between two vectors A and B is r r A  B  (Ay Bz  Az B y )iˆ  (Az Bx  Ax Bz )ˆj  (Ax B y  Ay Bx )kˆ 



Taking A to be the fourth vector and B to be the first, second and the third vector gives the cross products to be  9iˆ  11 ˆj  3kˆ  9iˆ  5 ˆ j  21kˆ ˆ  33iˆ  11 ˆ j  24 k

1.13

If the angle between two vectors is , the cosine of this angle is given by

cos  

  A B   A B

. Thus

Between (i) and (ii) cos   Between (i) and (iii) cos   Between (i) and (iv) cos   Between (ii) and (iii) cos   Between (ii) and (iv) cos  

4

 0.127    82.7 

38 26 2

38 110  25 38 22

 0.037    88.2   0.865    149.8 

40  0.748    41.6  26 110 5 26 22

4

 0.209    102.1

11

Between (iii) and (iv) cos  

1.14

 0.380    67.6 

38 22

Vector A can be written as A  A cos  1iˆ  cos  2 ˆj  cos  3 kˆ  





Similarly B  B cos 1iˆ  cos  2 ˆj  cos  3 kˆ  



Taking the dot product between the two vectors and using the formula

cos  

  A B   A B

,

where  is the angle between the two vectors, we get the answer.

1.15

Magnitude

  A B 

  AB 

Similarly

2   2 A  B  2 A  B cos  2   2 A  B  2 A  B cos 

Equating the two gives

  A  B  0 which

implies that the two vectors are perpendicular

to each other.

1.16

  B  C   B y C z  B z C y iˆ   B z C x  B x C z  ˆj   B x C y  B y C x  kˆ





   A  B  C  Ax  B y C z  B z C y   A y  B z C x  B x C z   Az  B x C y  B y C x 



1.17

From the expression for







      C  A  B and B  C  A    A  B  C it is clear that it is equal

This comes out to be equal to





Ax

Ay

Az

Bx

By

Bz

Cx

Cy

Cz

to the determinant

Interchange of two rows in a determinant changes the sign of the determinant. This implies

Bx

By

Bz

Bx

By

Bz

Ax

Ay

Az

Cx

Cy

C z   Ax

Ay

Az  B x

By

Bz

Ax

Ay

Az

Cy

Cz

Cy

Cz

Cx

thereby proving the equalities in problem 1.16. 5

Cx

1.18

  B  C   B y C z  B z C y iˆ   B z C x  B x C z  ˆj   B x C y  B y C x  kˆ

Therefore





   A  B  C   Ay  B x C y  B y C x   Az  B z C x  B x C z  iˆ

  Az  B y C z  B z C y   Ax  B x C y  B y C x  ˆj

  Ax  B z C x  B x C z   Ay  B y C z  B z C y  kˆ

The x-component of

A B C y

x

y



   A B  C



 B y C x   Az  B z C x  B x C z   B x ( Ay C y  Az C z )  C x ( Ay B y  Az BZ )

On the right hand side above, add and subtract Ax B x C x to get

A B C y

x

y

 B y C x   Az  B z C x  B x C z    B x ( Ax C x Ay C y  Az C z )  C x ( Ax B x Ay B y  Az BZ )      B x ( A  C )  C x ( A  B)

Do the same manipulation for the other components to get



 







         A B C  AC B  A B C

1.19

( i)



 

     

AB  aiˆ  aˆj  aiˆ  akˆ  a ˆj  kˆ AC  aˆj  akˆ  aiˆ  akˆ  a ˆj  iˆ BC  aˆj  akˆ  aiˆ  aˆj  a kˆ  iˆ

 

(ii)

   











Body diagonal from the origin to the opposite corner is a iˆ  ˆj  kˆ . This vector is perpendicular to the vectors AB, AC and BC in the plane, because its dot product with each one of them vanishes. This shows that the diagonal is perpendicular to the plane.

6

Another way to see this is to find a vector perpendicular to the plane by taking cross product of any of the two vectors from AB, AC or BC, and show that it is parallel to the body diagonal. For example taking the cross product of AB and AC gives



AB  AC  a 2 kˆ  iˆ  ˆj



which is parallel to the body diagonal. (iii) If the angle between OA and AB is  then cos  



   

OA  BA aiˆ  akˆ  akˆ  aˆj 1   OA BA 2 a 2 a 2



This gives   60 . Note that we have taken dot product with the vector BA rather than AB because we wish to keep  less than 90 . In the same manner we get the angle between OA and AC also to be 60 . 1.20

The problem is to be solved exactly in tha same manner as done in example 1.3 by replacing the position vector by the velocity vector and the velocity vector by the acceleration.

1.21

Position vector R(t )  R  cos t iˆ  sin t ˆj  

  dR (t )  R  sin t iˆ  cos t ˆj Velocity vector v (t )  dt



  dv (t )   2 R cos t iˆ  sin t ˆj Acceleration a (t )  dt



1.22





(i) In reference frame 2, the components V x 2 , V y 2 , V z 2 of vector

 V at

time t are

given in terms of its components V x1 , V y1 , V z1 in frame 1 by formula (1.10) as V x 2 (t )  V x1 cos t  V y1 sin t V y 2 (t )  V x1 sin t  V y1 cos t V z 2  V z1

Here V x1 , V y1 , V z1 are time-independent because vector Differentiating V x 2 , V y 2 , V z 2 with respect to time, we get

7

 V is

constant in frame 1.

dV x 2 (t )  V x1 sin t  V y1 cos t  V y 2 dt dV y 2 (t )  V x1 cos t  V y1 sin t  V x 2 dt dV z 2 0 dt

(ii) From (i) it is clear that   dV  kˆ  V dt

1.23

Done in later chapters

1.24

Magnitude of a vector quantity

 A(t ) is fixed. This means   A(t )  A(t )  constant

Differentiating both sides with respect to time we get   A(t ) A(t )  0 dt

Since

 A(t )

is not zero, the equation above implies that

 A(t )

 dA(t ) and are dt

perpendicular to each other. An everyday example is a particle moving in a circle. The magnitude of its position vector is a constant and therefore its velocity, which is the time-derivative of its position vector, is perpendicular to the position vector.

1.25

 



OA  R1 sin t iˆ  cos t ˆj OB  R2  sin t iˆ  cos t ˆj



The area of triangle OAB is given as RR 1 1 OA  OB  R1 R2  2 sin t cos t   1 2 sin 2t 2 2 2

This is maximum at 2t  1.26

 2

or t 

 4

Let the angle between the z axis and the vector be . Then the component OB is OA sin  .

Thus the magnitude of OB is

kˆ  OA

.

However, its direction is

perpendicular to the plane containing the z axis and the vector OA. To get the proper

8

direction we again take cross product of kˆ  OA with kˆ . The component in the z direction is given as kˆ  OA . Thus the vector OA is general kˆ can be replaced by nˆ .

9

kˆ  OAkˆ  kˆ  OA  kˆ .

In

Chapter 2 2.1

(i)

y

L

T(y)

y

T(y+y)+

(ii) Since the element of length y is in equilibrium, we have T ( y )  T ( y  y ) 

M yg L

Using Taylor series expansion for T(y+y), which gives T ( y  y )  T ( y ) 

dT 1 d 2T y  ( y ) 2   dy 2 dy 2

And taking limit y  0 leads to the differential equation for T(y). The equation is dT M  g dy l

Solution of this equation is T ( y)  

M gy  C L

where C is the integration constant. This is determined by using the fact that at the lose end (y=L) of the rope, the tension is zero. This gives C  Mg

and T ( y ) 

10

M ( L  y) g L

2.2



Torque   r  F It is given that of vector

 r  2iˆ  ˆj

3iˆ  2 ˆj

and the force has magnitude 50N and acts in the direction

. Thus the force is 50 times the unit vector in the direction of the

given vector. This gives   3iˆ  2 ˆj   F  50 13   50 ˆ 3i  2 ˆj   50 kˆ With this the torque is  2iˆ  ˆj   13

2.3

13

(a)

100N

100N

(b) The centre of the rod is at (3, 2) The right end is at position   F  100 

 r  8iˆ  2 ˆj

and the force at this end is

3 ˆ 1 ˆ i  j  2 2 

The left end is at position

 r  2iˆ  2 ˆj

and the force at this end is

11

  F  100 

3 ˆ 1 ˆ i  j  2 2 

Torque with respect to the origin =

8iˆ  2 ˆj   100 

 3 ˆ 1 ˆ  i  j    2iˆ  2 ˆj  100 2 2  





 3 ˆ 1 ˆ  i  j   10iˆ  100 2 2    500kˆ

3 ˆ 1 ˆ  i  j 2 2 

Torque with respect to the centre of the rod =  8iˆ  100 

 3 ˆ 1 ˆ i  j    2iˆ  100 2 2  





 3 ˆ 1 ˆ i  j   10iˆ  100 2 2   ˆ  500k

3 ˆ 1 ˆ i  j  2 2 

Torque with respect to the left end of the rod =  10iˆ  100  ˆ  500k

3 ˆ 1 ˆ i  j 2 2 

Torque with respect to the right end of the rod =   10iˆ  100  ˆ  500 k

3 ˆ 1 ˆ i  j 2 2 

(c) Torques about all the points are equal because the net force on the rod is zero.

12

2.4 (a)

TA

TB

3m A

B 1m 70N

(b)

F

y

0.5m

30N

120N

 0 gives

T A  TB  220

(i)

Net torque about A is zero, which gives 3TB  1  70  1.5  30  2.5  120  415

(ii)

Equation (ii) gives TB  138 N This substituted in equation (i) gives T A  82 N 2.5 Component of force in the plane perpendicular to the axis is F cos  at a distance of R from the axis. Therefore the torque about the axis is RF cos  2.6 Free-body diagram of the block

13

N1 N2

W

N1, N2 and W are three forces in a plane. Thus they must pass through one common point for equilibrium. So the equilibrium conditions are only the force conditions.

F

horizontal

 0 gives

N 1 sin   N 2 cos 

F

vertical

 0 gives

N 1 cos   N 1 sin   W

Solution of these two equations is N 1  W cos 

and

14

N 2  W sin 

2.7 Free-body diagram of the plank

N Ry

2m

1m

100N 0.2m Rx

Free-body diagram of the block N

F

Nground

If the rod makes angle  with the horizontal then sin   0.2 and cos   0.98

(a) To get the horizontal force F, we first calculate the normal force N on the rod. To do so, we calculate the total torque about the hinged end of the plank   1  N  3  100 cos   and equate it to zero. This gives N  3  100 cos   3  98  294 N

Now we balance the horizontal forces

 F

horizontal

15

 0  on the block to get

F  294 sin   0.2  294  59 N

(b) Force balance on the plank

F

horizontal

F

vertical

 0  R x  N sin   59 N  0  R y  N cos   100

This gives R y  100  288  188 N

Minus sign in front implies that the direction is opposite to that shown in the free-body diagram above. 2.8

Free-body diagram of the rod

N2

N1 F

W

Balancing the vertical forces gives N1 = W = 50N Balancing the horizontal forces gives N2 = F Balancing the torque about the centre of gravity gives F

8  50  0.5

leading to F 

2.9

25  8.8 N 8

Free body diagram of the painting

16

T

N1

N2

N1+N2 Fx W

W

Force balance equations give Fx  T and N 1  N 2  W

N1 and N2 are equal because the component of torque perpendicular to the wall must vanish. This gives N1=N2=25N Balancing the component of torque parallel to the wall taken about the lower end of the painting gives 20 3  T  10  50

giving T 

25 3

17

 14.4 N

2.10

We first calculate the forces at the ends of the rod. These forces are applied by the supports. After finding the forces on the rod, we then calculate the forces and the torques applied by the wall on the supports. Free body diagram of the rod

N1

N2

140cm 60cm 35N

Free body diagrams of the left and the right supports

F1 1

F2 2

5cm

5cm

N1

The forces on the rod satisfy

F

N2

y

 0 which gives

N 1  N 2  35

Taking torque about the left end and using



 0 gives

140  N 2  60  35  N 2  15 N

This gives N 1  20 N

Now balancing vertical forces and the torque on the supports gives

18

For the left support F1=20N and  1  0.05  20  1Nm For the right support F2=15N and  2  0.05  15  0.75 Nm 2.11 Ry

60cm

Rx 40cm

N 40N To find the force applied by the plastic block, we balance torque about the upper left corner. This leads to 40  N  30  40



N  30 N

Balancing the vertical forces gives Ry = 40N Balancing the horizontal forces gives Rx = 30N Negative sign means that the direction of Rx is opposite to that assumed in the free-body diagram above. Free-body diagram of the pole

19

30N 40cm

40N

30N

N

 Balancing the vertical forces on the pole gives N = 40N There is no net horizontal force and the two horizontal forces give a couple = 300.4 = 12Nm Balancing the torques on the pole about the ground gives  = 300.4 = 12Nm 2.12

Free-body diagram of the table

90cm

ˆ j

Rx

iˆ

Ry

Nx

Ny 20N

To find Ny, we balance the torque on the table about its left hand edge to get 90  Ny  45  20

20



Ny  10 N

By balancing the vertical forces, we get Ry  10 N . The negative sign tells us that the force is direction opposite to that shown above. Free-body diagram of one of the rods

Ry 2 30

Rx 2

Sx

Sy Free body diagram of the entire system 90cm Nx

30 Ny

20N

2Sy

2Sx

To get Nx, we balance the component of the torque coming out of the paper on the entire system about the lower hinge. This gives 30 3  Nx  45  20



Nx  10 3 N

The negative sign again tells us that the direction of the force is opposite to that shown. Balancing the horizontal component of the force on the table then gives 21

Rx  10 3N

Note: The net force on each rod on its upper end is 

Rx ˆ Ry ˆ i j  5 3iˆ  5 ˆj which is along 2 2

the rod as it must be for the equilibrium of a rod held at its ends. Balancing the horizontal and vertical components of forces on each rod gives Sx 

Rx  5 3N 2

and

Sy  5 N

Thus the net force on each rod is 10N compressive. 2.13 Free body diagrams of the two side portions and the portion AC over the pulley: N

TA

TA

TC

RM g L

TC

L2 M g L

L1 M g L

F

Tension TA and TC at both ends of the portion over the pulley is the same because the torque about the centre must vanish. This gives TA  TC  Free body diagrams of the portion AB and BC

22

L1 Mg L

Neffy

Neffy

Neffx

TB

Neffx

TB

RM g 2L

RM g 2L

TA

TC

Notice that Torque of the normal reaction about the centre of the cylinder vanishes because for each small portion of the rope over the cylinder, the normal reaction is radial. Thus T A (or TC) and TB cannot be equal because they together provide a torque to balance the torque due to the weight of the rope. Balancing the torque about the centre on AB gives R  TA 

R 2



RM g  R  TB 2L

 R  M  TB   L1  g  2 2 L 

Thus if the net force by the cylinder on the rope is Neff at an angle  from the horizontal then by force balance R  M  N eff sin    L1  g  2  L 

R  M  N eff cos    L1  g  2 2 L 

Note that Neff acts at a point different from the centre of BC because on different infinitesimal portions it is different. 2.14 The support does not apply any torque about the x-axis. All other components and torques are balanced by the support. 2.15 When forces are applied at two points of the rod, force balance demands that the force be equal and opposite. However two such forces acting at two different points will give rise to a couple moment. The couple moment is zero only if the forces point along the rod (see figure below)

23

Couple moment non-zero

Couple moment zero

2.16 Let cables OA and OC make angle 1 and OB and OD angle 2 with the vertical. Then balancing the vertical forces gives 2T (sin 1  sin  2 )  45000

The sine of the angles is easily calculated to be sin 1 

1 2

sin  2 

1 11 4



2 5

This gives T=14050N 2.17 The torque direction is given by the direction of cross product

 ˆF n

, which is

perpendicular to nˆ . This implies there is no component of the torque in the direction of ˆ. n

2.18 The net force on the plate is

50iˆ  50 ˆ j  70iˆ  120iˆ  50 ˆ j

Therefore the force that must be applied to the plate to keep it in equilibrium is  F  120iˆ  50 ˆ j

. Since there are only three forces acting on the body, they must all pass

through the same point so that their net torque is zero. This is shown in figure below.

24

A

B

O

D



The force F

C

 120iˆ  50 ˆ j

 50    22.6 from the line DC. Thus  120 

1 is at an angle   tan 

it does not pass through O and intersects with side AD and diagonal BD of the square. Therefore: (i)

It is not possible to keep the square in equilibrium by applying the third force at O.

(ii)

It is possible to keep the square in equilibrium by applying the third force at a point on BD. Equation of BD with O as origin is y  x Equation of line along which the third force acts is y   Solving the two equations gives y  x 

a 5  3a    x  2 12  2 

3 a 34

This gives the distance of point O = 0.125a And from B = (iii)

 1 3    a  0.58a  2 34 

2 

It is clear that for equilibrium, the force can be applied only on AD and BC.

25

Chapter 3 3.1 For the three trusses shown, m = 21, j = 12. Thus they all satisfy 2j-3 = m. Thus they are all simple trusses. 3.2 Showing that pin E is in equilibrium There are five forces acting on E of which two (FE and ED) are horizontal, two (CE and the external load) are vertical and one (BE) is at an angle. We wish to check if the horizontal and vertical forces add up to zero. It is solved in 3.1 that 5000 5000 2 N  Tensile  , FBE  N  Tensile  3 3 10,000 10,000  N  Tensile  , FDE  N  Tensile  3 3

FFE  FCE

Since all the forces are tensile, they all pull the pin. In addition there is the external load of 5000N vertically down. The net horizontal force is

F

x

  FFE  FBE cos 45  FDE 

5000 1 5000 2 10000   3 3 3 2

0

Similarly the net vertical force is

F

y

 5000  FBE sin 45  FCE  5000 

1 5000 2 10000  3 3 2

0

3.3

(i) The truss has 4 members and 4 joints. Number of force balance equations therefore is 8 (2number of joints).

On the other hand, number of forces available is only 7

(3+number of members), which implies that the truss will not be stable and will collapse. In terms of stability condition 2 j  3  m which implies that the truss will collapse. (ii) If we add one more member to the truss, i.e make m = 5, then 2 j  3  m is satisfied and the truss becomes stable and a simple truss. Let us add a member across AC. To find forces in each member we start by first finding the forces applied by the external supports. The free-body diagram of the truss is as follows:

26

B

C

NAy

NAx

5000N A

D ND

The direction of the forces applied by the external supports has been anticipated as shown. To find ND, we balance torque about A to get

2  N D  (2  1.5 cos 60)  5000

  N D  6875N

 2.75  5000 

Now balancing the vertical and horizontal forces on the truss gives NAx = 0

and

NAy = 1875N

The negative sign again tells us that the direction of the force is opposite to that shown. We begin to apply the method of joints from point D since at this point there are two unknown forces FAD and FCD. Assuming these forces to be tensile gives the free body diagram of joint D as follows FCD

60 FAD 6875N

27

Balancing the vertical forces on point D gives FCD sin 60  6875  0 

FCD  7939 N

The negative sign shows that the force FCD is compressive and not tensile as assumed. Balancing the horizontal forces on point D gives FCD cos 60  FAD  0 

FAD  3969 N

The negative sign again shows that the force FAD is compressive and not tensile as assumed. Next we go to point A and balance the forces there. The free body diagram of point A is FAB FAC 60



3969N 1875N

In drawing the figure above, we have shown the direction of FAD according to it being a compressive force. The length of rod AC is =

 2  1.5 cos 60 2

 1.5 sin 60

2

 3.04 m

So sin  

1.5 sin 60  0.427 3.04

and

cos  

2  1.5 cos 60  0.904 3.04

Now balancing the horizontal forces at A gives FAB sin 60  FAC sin   1875

and balancing the vertical forces at A gives FAB cos 60  FAC cos   3969

Solving these two equations gives FAB = 0 and FAC = 4390N Since the sign of FAC is positive it is in the same direction as assumed and therefore tensile.

28

Now we can easily see that force FBC will be zero because point B is under equilibrium under only two forces FAB and FBC and FAB has already been determined to be zero. Thus FBC = 0. Thus all the forces are now determined. They are FCD  7939 N (compressive)

FAB = 0

FAD  3969 N (compressive)

FAC = 4390N (tensile) and FBC = 0.

Finally to check our answer we make the forces at point C and see if they all balance. The free body diagram of point C is

7939N 60  5000N 4390N

Balancing the horizontal forces at C gives 7939 cos 60  4390 cos   3969  3969  0

Balancing the vertical forces at C gives 7939 sin 60  5000  4390 sin   6875  5000  1875  0

This indicates that our answers are correct. Note: We see that FAB and FBC both vanish. This implies that members AB and BC may not bee needed for the truss. This is true because with just three members AD, AC and CD (m=3) there are only three joints (j=3) and the truss satisfies the condition 2 j  3  m for it to be a stable structure. 3.4 Free body diagram of the truss

29

Ry 30cm

B

Rx

C 

20cm

30N N A

Since point C is in equilibrium under one known and two unknown forces, both unknown forces can be determined easily. The forces on C look as follows FAC 

FBC

30N Balancing the vertical forces at C gives FAC sin   30

30

With sin  

2 13

this implies FAC = 54N (compressive)

Balancing the horizontal forces at C gives FBC  FAC cos   54 

3  45 N (tensile) 13

The only force left is at AB. We calculate this by balancing forces acting on pin A, which look as follows.

FAB N

 FAC This gives FAB  FAC sin   30 N

Additionally we can also solve for the normal reaction N and the forces Rx and Ry. These are N = 54N, Rx = 45N, and Ry = 30N 3.5 Rod AB provides a vertical force to hold pin A. However if it is removed and the vertical force is provided by a fixed pin joint, the structure will remain stable because we need 3j=6 forces for equilibrium of 3 joints; two of these are provided by the fixed supports and two by the two members. The forces in the members remain the same. So do the forces by the two support except that the fixed point at A also provides a vertical fore of 30N that was earlier provided by member AB. 3.6 Free-body diagram of the truss

31

Ry B

C

Rx

N A

D

100N Since pin D has only two unknown forces acting on it, we can start our calculations from this point onwards. The forces on D are

FCD

FAD 100N

It is immediately clear that FAD = 0 and FCD = 100 N (tensile) Next we go to pin C and balance the forces there. The forces acting on C are

32

FAC

FBC 100N

Balancing the vertical forces at C gives FAC  100 2 N (compressive)

Balancing the horizontal forces at C then gives FBC  100 N (tensile)

Next we go to pin A. The forces there are as follows

FAB N

100 2 N

Balancing the vertical forces at A gives FAB  100 N (tensile)

Balancing the horizontal forces at A gives N=100N Finally balancing forces at pin B will give the external forces Rx = 100N and Ry = 100N 3.7 Free body diagram of the truss is as follows

33

Ry Rx

C

D

B 500N

Ny Nx A (i)

E

There are 4 reaction forces at the supporting pins at A and B. In addition the forces generated by the members of the truss equal 6. This makes the total number of forces available = 10. The number of joints in the truss is 5 that require exactly 10 number of forces for equilibrium. Thus the truss is a stable one.

(ii)

It is also statically determinate since the number of forces available is equal to the number of equations to be satisfied for equilibrium.

(iii)

First we find Nx by balancing the torque about point B. This gives 0.75  Nx  1.5  500



Nx  1000 N

We now begin by balancing the forces at point D

34

FED

FCD 500N

Balancing the vertical forces at D gives FED  500 2 N (compressive)

Balancing the horizontal forces at D then gives FCD  500 N (tensile)

Nest we go to pin E because it has two unknown forces acting on it. The forces are as follows

FCE FAE

N

Balancing forces at E gives FAE  500 N (compressive) FCE  500 N (tensile)

35

Next we go to pin at A. The forces there are

Ny

1000N

500N FAC

Balancing the forces gives FAC  500 2 N (compressive)

and Ny  500 N

(tensile)

Finally we go to point C where only one force FBC is unknown. The forces on C are

FAC

FBC

500N

500N

Balancing the horizontal forces at C gives FBC=1000N As a final check, the value for FBC gives the horizontal force Rx by pin B on the truss to be 1000N and vertical force Ry to be zero. This is consistent with the overall equilibrium of the truss when it is treated as a system by itself.

36

3.8 Since each member of the truss weighs 50N, at each pin we take the load by each pin at that point to be 25N. The free body diagram of the truss is as follows; here each small arrow pointing down indicates the weight of the truss member, acting at its centre.

NBy

NBx

C

B

D

A

E

1000N

NE

We firs find NE. To do this we balance the torque about B. This gives l  NE 

l 3l  150  l  50   100  2l  1000  2 2

N E  2275N

This immediately gives, by balancing forces on the entire truss NBx = 0 and NBy = 925N The negative sign showing that the force I opposite to the direction assumed in the figure above. We begin at pin D as there are two unknown forces there. The force diagram on pin D is as follows (there are two members meeting at pin D that give a load of 225=50N there)

37

FCD

FDE 1050N

Balancing the forces gives FCD  1050 2 N  1485N (tensile)

and FDE  1050 N (compressive)

Next we go to pin E. The forces acting there are (including 325=75N from 3 members)

2275N

1050N

FAE FCE +75N

Balancing the forces gives FCE  2200 N (compressive)

and FAE  1050 N (compressive)

Next we go to pin A. The forces acting there are (including 325=75N from 3 members)

38

FAB

1050N

FAC

75N

Balancing the forces gives FAC  1050 2 N  1485 N (tensile)

and FAB   75  1050 N  975 N  FAB  975 N (compressive)

Net we move to point B where the forces are (including 225N=50N from two members) FAB

FBC

50N 925N

This immediately gives FAB = 975N (compressive) and FBC = 0 As a final check, one balances all the forces at C and sees that they all balance properly. This implies that the answers obtained by us are correct. 3.9 Free body diagram of the truss with the weight of each member included. The free body diagram is then as follows.

39

C

B

NA

A

F

Nx

ND

D

E 5000N

By balancing the torque about A we get

12  N D  2 1000  4  500  6 1500  8   500  5000  10 1000

  ND 

 67000 

16750 3

Balancing the vertical and horizontal forces on the truss, this gives NA 

11750 and Nx  0 3

For calculating force in members, we take the weight of each member shared equally at each joint. The forces on A are (including the weight of two members)

11750 N 3 FAF

500N

FAB Balancing the forces at point A

F

0

y

F

x

0

gives gives

FAB

11750  500 or FAB  4832 N (compressiv e) 3 2 F 10250 FAF  AB  N (tensile) 3 2 

Next we go to point F. The forces at point F are (including the weight of three members) 40

FBF

10250 N 3

FFE

750N Balancing the forces here gives FBF  750 N ( tensile) FFE 

10250 N ( tensile) 3

Next we go to point B since now there are only two unknown forces there. At point B the forces look as follows (including the weight of four members) 4832N

FBC 750N

FBE

1000N Balancing the forces

F

y

F

x

0

gives

FBE

0

gives

FBC 

2

 1000  750  4832 2



2357 2

4832 2



0 

FBE  2357 N (tensile) FBC  5083 N

Negative sign above means that the direction of the force is opposite to the one assumed. So FBC = 5083N (compressive)

41

We next consider point C and balance the forces there. The forces at point C are (including 750N form the weight of three members)

FCD 5083N 750N FCE Balancing the forces

F

x

F

y

 0 gives

FCD 2

 5083  FCD  7188 N (compressiv e)

 0 gives FCE  750 

FCD 2

 5083  FCE  4333 N (tensile)

Next we go to pin D where the normal reaction is

16750 N and balance the forces there. The 3

force diagram there is (including 500N form the weight of two members)

16750 N 3 FED 500N 7188N

It is easily seen that the vertical forces balance at this point. This points to the correctness of our calculations so far. Balancing horizontal forces then gives FED = 5083 N (tensile)

42

As a final check we should check whether all the calculated forces balance at pin E. The forces at pin E are (including the weight of four members)

2357N

4333N

N

5083N 1000N

5000N

All these forces balance as can be seen by calculating the net x and y components of the forces. Thus our calculations are correct. 3.10

Free body diagram of the truss 2000N

NBy C NBx

B

D

A

1000N

E NE

Taking torque about B we get 3  N E  4  2000 

NE 

8000 3

Balancing the horizontal and vertical forces now gives NBx = 1000 N

and

NBy = 

2000 N 3

Negative sign above means that the direction of the force is opposite to the one assumed.

43

We begin at pin D. The forces there are FED 

FCD

1000N

2000N

In the diagram above cos  

1  0.667 and sin   1.5

1.5 2  1  0.745 1.5

Balancing the vertical forces gives FED sin   2000 

FED  2683N (compressive)

Balancing the horizontal forces gives FCD  FED cos   1000  2789N

Next we go to pin E. The forces there are N

FAE 

 FCE

2683N Balancing the vertical forces gives

 2683  FCE  sin 



8000 3



FCE  895 N (compressive)

Balancing the horizontal forces gives FAE  2683 cos   FCE cos   1192 N (compressive)

44

Next we go to pin A. The forces there are FAC  1192N

 FAB

Balancing the vertical forces gives FAC = FAB. Balancing the horizontal forces gives

 FAB  FAC  cos  2 FAB cos  1192

 FAB  895N(compressive)

This also means that FAC = 895N (tensile) Now we go to pin C. The forces there are 895N 

2789N

FBC  895N

The vertical forces are already balanced here. Balancing the horizontal forces gives FBC  2  895  cos   2789 

FBC  1596 N(tensile)

To check our answers, we finally balance the forces at pin B and see if they all balance there. At pin B the forces are as follows

45

895N 

1596N

1000N N

It is easily seen that all the forces above balance. So our answers are all consistent. 3.11

The weight of the road = volume of the roaddensityg = 128.3200010 = 576000N This weight is divided equally between the two trusses on the sides. 288000N is supported by each truss. Weight of the members of the truss = 135000 = 65000N Total weight supported by each truss therefore is = 353000N Free body diagram of the truss is B

C

D

NAy

NAx

NE

A

H

G

F

353000N From the balance of forces, is clear that N Ax  0

N Ay  N E 

353000  176500 N 2

46

E

Thus only

Let us now consider forces at each pin one by one. Each pin has the following forces acting on it: The weight of the road divided over 5 pins, which is

288000  57600 N ; 5

the weight of the members at that pin; and the forces applied by the members. Let us now balance forces at point E. The forces on pin E are (including the weight of the members) 176500N

FEF  57600N

5000N

FDE

In the figure above sin  

3  0 .6 5

cos  

4  0 .8 5

Balancing the vertical forces in the figure above gives FDE cos   57600  5000  176500  FDE  142375 N (compressive)

Balancing horizontal forces then leads to FFE  FDE sin   85425N (tensile)

Next we consider pin D.

The forces on pin D are (including the weight of the

members)

47

142375N  FCD FDF 7500N

Balancing horizontal forces then leads to FCD  142375 sin   85425 N (compressive)

Balancing the vertical forces gives FDF  142375  0.8  7500  106400 N (tensile)

Next we look at pin F. The forces there are (including the weight of the members) 106400N

FGF

85425N  57600N

10000N

FCF

Balancing the vertical forces in the figure above gives FCF cos   57600  10000  106400 

FCF  48500 N (compressive)

Balancing horizontal forces then leads to FGF  FCF sin   85425 N  114525N (tensile)

By symmetry of the problem, forces on the members to the left of member CG will be exactly the same as on the corresponding members to its right. The only force that we now have to calculate is on member CG. For this we consider point G. Two horizontal

48

forces at G are by HG and GF and are equal to 114525N each. The forces at point G are then given as FCG

114525N

114525N  57600N

7500N

This gives FCG = 65100N Finally we check our answer at point C. The forces there (including the weight of 5 members meeting there) are 48500N

48500N

85425N

85425N

65100N

12500N

FCF

As is easily seen, the forces at C balance and therefore our calculations have been consistent throughout. 3.12

Free body diagram of the truss is as follows

49

2000N NA

1000N NE

C B

RA

D

 A

H

G

F

E

Here the distances and the angles are AH = HG = GF = FE = 2.5m tan  

BH = DF = 1.25m (similarity of triangles)

1.25  0.5    26.6  sin   0.447 and cos   0.894 2.5

Balancing torque about point A gives 10  N E  5  1000  2.5  2000 

N E  1000 N

Thus NA = 2000N and RA = 0 Now that the external reactions have been determined, we can go about calculating the forces in the members. We start with pin E because there are only two unknown forces there. The forces at E are 1000N

FEF

E

 FDE

Balancing the vertical forces in the figure above gives FDE sin   1000 

FDE  2236 N (compressive)

Balancing the horizontal forces gives FEF  FDE cos   2236 * 0.894  2000 N (tensile)

50

Nest we go to pin D. The forces there are FDF

2236N

D FCD Balancing the forces gives FCD  2236 N (compressive)

FDF = 0 Next we go to point F where the forces are as shown below. FCF 0N

FGF

F

2000N

Balancing the forces gives FGF = 2000N (tensile)

FCF = 0

Pin G is considered next. Since the forces there are only vertical and horizontal, even without making the forces there, we immediately can write FCG = 0 and FGH = 2000N (tensile) Next we consider point A where two members AB and AH meet and therefore there are two unknown forces. The forces there are

51

2000N



FAH

A

FAB Balancing the vertical forces gives FAB sin   2000 

FAB 

2000  4472 N (compressive) 0.4472

Balancing horizontal forces then gives FAH  4472 cos   4000 N (tensile)

Next we go to point B. Here there are four forces acting and each pair (FBH and 2000N; and FBC and 4472N) has two forces in opposite directions. Thus without solving the detailed force balance equations, we can directly write FBH = 2000N (compressive) and FBC = 4472N (compressive) Next we go to point H and balance the forces there. The forces there are as follows FCH

45 4000N

2000N

H

2000N By balancing the forces at H, it becomes clear that FCH  2000 2 N (tensile) Finally the answers are checked at C. The forces at C are as shown below

52

2236N

4472N

D

1000N

2000 2 N

As is easily seen, the horizontal a n vertical forces all balance at C. Thus our answers are all correct. To calculate the forces by method of sections, we make a cut through the truss so that it passes the concerned members. In the present case we take the following section of the truss and show various forces on that section. 2000N 2000N FCB

B  A

H

FCH

FGH

In the figure above, FCH is determined easily by taking torque about A since the torque due to FCB and FGH both vanish about A. This gives AH 

FCH 2

 AH  2000 

FCH  2000 2 N (tensile)

To find FCB, we balance the vertical component of the forces to get FCH 2

 FCB sin   0  FCB  

2000   4472 N sin 

Negative sign here means that the force is opposite to the direction assumed and therefore is compressive in nature. Finally, balancing the horizontal forces leads to

53

FGH 

3.13

FCH 2

 FCB cos   0 

FGH  2000  4472  0.894  2000 N (tensile)

The free-body diagram of the truss on one side is as follows (Notice that the weight of the truck is equally divided between the two trusses) C

B

D

NA RA

A

NE E

H

G

F

50kN We first calculate NE by balancing torque about A. This gives 16  N E  12  50 kN 

N E  37.5 kN

This gives NA = 12.5 kN and RA = 0 To find forces in members CD and DG, we make a cut through CD, DG and GF. This looks like the following D

FCD

37.5 kN

E FGF

F FGD 50kN

To find FGF, we balance torque about point D about which the torques due to FCD and FGD vanish. This gives

54

4  37.5  5  FGF



FGF  30kN (tensile)

To obtain FCD, we take torque about point where FDG and FGF intersect, which is point G. This leads to 5  FCD  8  37.5  4  50



FCD  20kN (compressive)

Now we balance the horizontal and vertical forces on the truss. Balancing horizontal forces gives FGD 

4 41

 20  30  0 

FGD  16kN

Negative sign here means that the force is opposite to the direction assumed and therefore is compressive in nature. To find the forces in the members BC and BG, we make a cut through the members BC, BG and HG as follows and then calculate the forces. B

FBC

12.5 kN A

FGH

H

FBG To obtain FBC, we take torque about point where FGH and FBG intersect, which is point G. This leads to 5  FBC  8  12.5 

FBC  20kN (compressive)

Next we find FGH by taking torque about B. We get 5  FGH  4  12.5 

FGH  10kN (tensile)

Finally we get FBG by balancing vertical and horizontal forces. Horizontal force balance gives FBG 

4 41

 10 

FBG  16kN (tensile)

Finally to find FCG, we make the following cut through the truss

55

FCG

B

D

12.5kN A

37.5kN

E H

G

F

50kN Since the vertical forces all balance, this implies FCG = 0. Further, the horizontal forces are also balanced.

56

Chapter 4 4.1

N s

Frictional force

F

4.2

Applied force max

Since the block is in equilibrium under three forces, the three forces must pass through the same point. Thus the normal reaction will be at the point where the arrow showing the weight meets the inclined plane. This is shown below. f

N

mg θ

4.3

The free body diagram of the block is as follows

57

N 

F 

f mg Balancing the horizontal forces gives f  F sin 

Balancing the vertical forces gives N  mg  F cos 

Since the maximum frictional force f max  N , for equilibrium we should have F sin     mg  F cos  

4.4



F

 mg sin    cos 

We consider two different situations when the weight on the table is about to move to the left or to the right. When it is about to move to the left, its free body diagram will look as follows N

10g

mg f

50g By equilibrium conditions, we have N = 50g f + mg = 10g 58

Since f  N

We have f  10 g  mg  0.1  50 g



5m

Thus the minimum value of m is 5kg when the frictional force is at its maximum pointing to the right. As m is increased above 5kg, frictional force becomes less and less, eventually changing direction and attaining its maximum value pointing left. In that situation, the free body diagram of the block on the table is as follows. N

10g

mg f

50g In this situation, the equation for horizontal force balance is f + 10g = mg This coupled with f  N leads to f  mg  10 g  0.1  50 g



m  15

Thus 5  m  15

4.5 Taking the x axis along the plane and the y axis perpendicular to the plane, the free-body diagram of the block looks as follows.

59

Y

N

F

X

30º

30º

F

100g The equations describing equilibrium in the X and the Y directions are

F

0

x

F

y



0



F cos 30  F   100 g sin 30  0 N  F sin 30  100 g cos 30  0

The first equation implies that F    F cos 30  100 g sin 30

Taking g = 9.8ms2, the value of F for different values of F is

4.6

F  600 N



F  500 N



F   29.6 N F   57.0 N

F  100 N



F   403.4 N

Free body diagram of the box when it is about to move (i.e. the frictional force is at its maximum) is shown below N

F h

b a

N mg

60

When the box is about to move, the friction is at its maximum and is equal to N. The force F also equals N at this point. This creates a couple that is counterbalanced by the couple formed by the weight of the box mg and N (=mg). This is the reason that N shifts towards the direction of the push. However, the maximum couple moment that can be created by mg and N is

a mg . 2

Thus for the box not to topple, the couple created by F and the friction should remain less than a mg . Thus implies 2 h  mg 

a mg 2



h

a 2

4.7 suppose each break show makes an angle  at the centre as shown below



The force F is assumed distributed uniformly over the shoe. Then the torque due to the frictional force will be b

   a

Fr

 r dr 

 2 b  a 2  2

2 F  b 3  a 3  2 F  a 2  ab  b 2   3 b 2  a 2  3  a  b

With two shoes therefore, the torque would be 4 F  a 2  ab  b 2   3  a  b

4.8 It is given that mass M is balanced by mass m. The contact angle is π. Since each time the string is wound once more around the rod, the mass M that can be balanced by m becomes twice as large, we have

61

M  m exp( )   2M  m exp(3 )  2  exp(2 ) 4M  m exp(5 ) This gives  = 0.11

4.9 Neglecting the length of the rope passing over the pulley, we have mass

side of the pulley that is balanced by mass

L2 M on the other side. Thus we have L1  L2

L1 L2 Mg  exp(  ) Mg L1  L2 L1  L2

4.10

L1 M on one L1  L2



L1  exp(  ) L2

As the weight is put, it has a tendency to move down. Hence the frictional force will be in the counterclockwise direction. Thus if the tension in the rope on the spring balance side is T1 and that on the weight side is T2 then   T2  T1 exp   2 

Now it is given that T1 = 5g and T2 = mg. Thus we get m  5 exp(0.2   / 2)  6.85kg

An interesting possibility exists if a person had pulled the weight down and then slowly brought it to equilibrium. In that case the tension will work in the other direction and m  5 exp( 0.2   / 2)  3.65kg

However we have not considered this possibility. 4.11 There is a range of M2 that exists because frictional force can act with its maximum value in one direction to the maximum in the other direction.

Largest value of M2 is when the

mass M1 is about to slide up the plane. The free body diagram of M1 in that case is as follows

62

N

T

f 

M1g

When the mass M1 is about to slide up, we have T  M 1 g sin   1 M 1 g cos   M 1 g (sin   1 cos  )      2  

The contact angle between the rope and the pulley is 

Since the rope has a tendency to move clockwise, the frictional force due to the pulley will be acting counterclockwise. Thus we have      M 2 g  T exp  2       2    Thus

      M 2 g  M 1 g  sin   1 cos   exp  2       2   

     M 2  M 1  sin   1 cos   exp  2       2   

The other extreme is when the mass M1 is about to slide down the plane. In that case the free body diagram of M1 is

N

T

f 

M1g

Thus we have T  M 1 g sin   1 M 1 g cos   M 1 g (sin   1 cos  )

63

Now the rope has a tendency to move counterclockwise, the frictional force due to the pulley will be acting clockwise. Thus we have      M 2 g exp   2       T  2    Thus

      M 2 g  M 1 g  sin   1 cos   exp    2       2   

     M 2  M 1  sin   1 cos   exp   2       2   

4.12 Free body diagram of the tire when it is loaded and is about to roll is as follows

F

f W

N

Balancing the vertical forces gives N  abP  W

 a

W bP

Balancing the horizontal forces gives F = f Balancing torque about the centre of the wheel gives FR 

a W 2

 F

64

W2 2bPR

Chapter 5 5.1

Consider a composite surface of total area A made up of N different surfaces. Then the coordinates  X C , YC  of the centroid satisfy

 xdA   ydA

AX C  AYC

Ai If the area of each surface is Ai (i  1 N ) then A   i

Now in the definition of the centroid, the integrals can be performed separately over each surface so that we can write AX C    xdA   Ai X Ci i

AYC 

i

i

  ydA   A Y i

i

Ci

i

i

This immediately gives

XC 

5.2

 Ai X C i A

  Ai X C i   Ai

 

 

and YC  

 Ai X C i A

  Ai YC i   Ai

 

  

By symmetry it is clear that XC = 2. We are nevertheless going to prove it below. We first calculate the area of the surface. It is 4

A

 ydx  0

 4   x  2 dx 4

2

0

4

 16    x  2  dx 2

0

Substituting z  x  2 we get 2

A  16   z 2 dz  16  2

Y

16 32  3 3

To calculate Xc, we take vertical strips of width dx on the surface at distance x from the origin and then calculate XC 

O x

X 65

 xdA A

Thus

 x 4   x  2  dx 4

XC 

2

0

32

2

3

Similarly to calculate YC, we take horizontal strips of width dy at height y and calculate YC 

Y

 ydA A

At height y, the strip extends from x1 to x2. These points are given by the equation y  4   x  2

y 1 O thusxhave We

X

x2

2



x1  2 

4 y

x2  2 

4 y

Therefore dA  ( x 2  x1 )dy  2 4  y dy

4

32 YC  2  y 4  y dy 3 0

Substituting y  4 sin 2  so that dy  8 sin  cos  d , we get  2

1

32 YC  2  4 sin 2   2 cos   8 sin cos  d  128 sin 2  cos 2  d  cos   3 0 0 1

 128  cos 2   cos 4   d  cos   0



256 15

This gives YC  

8 5

8 5

Thus the centroid is at  2,  

5.3

One curve (call it curve 1) y  4  ( x  2) 2 in this problem is the same as that in the problem above. The other curve (curve 2) is



y  16  4( x  2) 2  4 4   x  2 

66

2



The y-axis of curve 2 is thus 4 times curve 1. The area of curve 2 is therefore

128 . Similarly 3

the x coordinate the centroid of curve will remain at 2 but the y coordinate will be 4 

8 32  . 5 5

Thus we have A1 

32 3

 X C1 , YC1   

8 2,   5

;

A2 

128 3

 X C 2 , YC 2    

2,

32   5 

The area A for which we wish to obtain the centroid  X C , YC  is obtained by removing surface formed by curve 1 from the surface formed by curve 2. We thus have 128 32   32 3 3 AX C  A2 X C 2  A1 X C1 A

AYC  A2YC 2  A1YC1

5.4

 32 X C   A2  A1   2  32YC 



128 32 32 8    3 5 3 5



XC  2 YC  8

Trapezoidal loading is shown in the figure below

w2

f(x) w1

X1

X2

X

The total force on the beam will be equal to the area under the curve. Thus the total force is equal to

 w1  w2   X 2

2

 X1 

This load will be acting at the centroid of the area. Thus it acts at 67

X2

XC 



 x

X1

w1 

 w2  w1  x  X 1   dx X 2  X1 

 w1  w2  X 2  X 1 

X

2

  w





 w1  X 12  X 22  X 1 X 2  w2  w1  X 1  X 2  X 1 w1  3 2   w1  w2  X 2  X 1  2 2 2 2 2 3w1 X 2  3w1 X 1  2w2 X 1  2w2 X 2  2w2 X 1 X 2  2w1 X 12  2 w1 X 22  2 w1 X 1 X 2  3w2 X 12  3w2 X 1 X 2  3w1 X 12  3w1 X 1 X 2  3 w1  w2  X 2  X 1  2 2

X 2

2 1

2



w1 X 22  2 w1 X 12  w2 X 12  2 w2 X 22  w2 X 1 X 2  w1 X 1 X 2 3 w1  w2  X 2  X 1 

X 22  w1  2 w2   X 12  2 w1  w2   w2 X 1 X 2  w1 X 1 X 2  3 w1  w2  X 2  X 1 

Adding and subtracting w2 X 1 X 2 and w1 X 1 X 2 in the numerator we get

5.5



X 22  w1  2 w2   X 12  2 w1  w2   2 w2 X 1 X 2  w1 X 1 X 2  2 w1 X 1 X 2  w2 X 1 X 2 3 w1  w2  X 2  X 1 



 2w1  w2  X 1  X 2  X 1    w1  2w2  X 2  X 2  X 1  3 w1  w2  X 2  X 1 



1  2w1  w2  X 1   w1  2 w2  X 2  w1  w2  3

From figure 5.14, for a plate of width w w1  gh1 w

w2  gh2 w

Similarly, from figure 5.15 X 1  Y1 

h1 cos 

X 2  Y2 

68

h2 cos 

This gives from formula 5.9 X C  YC 



1 (2h1  h2 )h1  (h1  2h2 ) h2 3 (h1  h2 ) cos 



2 h12  h1 h2  h22 3  h1  h2  cos 



which is equivalent to the depth given by formula (5.17) 5.6

Loading on the tank door is triangular as shown below

N1 NA 0.25m 0.5m 153.125N NB

19.6N The average pressure is the pressure of water at the centroid of the submerged part. Thus the average pressure will be ghcentroid ( plate )  1000  9.8  0.125  1225 Nm 2

Thus the total force due to the water pressure is F  0.5  0.25  1225  153.125 N

This force acts at the centroid of the loading that is triangular in this case. Thus it is at a distance 1  0.25  2  0.50  0.417 m 3

below point A. We now apply equilibrium conditions to the door. This leads to 69

N 1  19.6 N 0.5  N B  0.417  153.125  N A  N B  153.125 

5.7

N B  127.7 N

25.4 N

The rectangular surface looks as follows Y

b X

a

To find Ixx, we take a strip (see figure above) of width dy parallel to the x-axis and calculate b 2

2  y  ady  

I xx 

b 2

ab 3 12

Similarly to find Iyy, we take a strip (see figure above) of width dx parallel to the x-axis and calculate a 2

I yy 

2  x  bdx  

a 2

a 3b 12

To find Ixy, we take a small square (see figure above) of size dxdy parallel and calculate a 2

I xy 

b 2

  xy dxdy   0

 a 2 b 2

by symmetry of the inegrand.

70

5.8

Y’ X’



b O a

From the figure sin  

b

cos  

a2  b2

a a2  b2

2ab a  b2 a2  b2 cos 2  1  sin 2 2  2 2 a b

sin 2  2 sin  cos  

2

From the formula for transformation of area moments (taking X and Y axis as in the problem above) we get I x'x' 

I xx  I yy





2



ab a 2  b 2 24



a 3b 3 6 a2  b2





I xx  I yy

cos 2  I xy sin 2 2 ab b 2  a 2 a 2  b 2  24 a2  b2





Similary

71

 

 

I y' y ' 

I xx  I yy



2





ab a 2  b 2 24



ab a 4  b 4 12 a 2  b 2

 

  

I xx  I yy

cos 2  I xy sin 2 2 ab b 2  a 2 a 2  b 2  24 a2  b2



 

 

and I x'y'  

I xx  I yy 2



sin 2  I xy cos 2



ab b 2  a 2 2ab 2 24 a  b2







a b a b 12 a 2  b 2 2

2



2

2







5.9

Y

b

X

a

To calculate IXX we take a horizontal strip of width dy at y (see figure) for dA and calculate

72

b

I XX   y 2 dA 2  y 2 b

a 2 b  y 2 dy b

Taking y  b sin  , we get  2

I XX  2



b 2 sin 2 

 2

 2

a 2 sin 2 2 ab 3 b cos 2 d  2ab 3  d  b 4 4  2

Similarly for IYY, we take a vertical strip at x for dA and calculate a

I YY   x 2 dA  2  x 2 a

b a 2  x 2 dx a

Taking x  a sin  , we get I YY 

a 3 b 4

And by symmetry I XY  0

We now calculate the moments and product of inertia about a set of axes rotated by an angle with respect to the original one and with the same origin. Thus I x'x' 

I xx  I yy



I xx  I yy

cos120   I xy sin 120 

2 2 2 2 ab(b  a ) 1 ab(b 2  a 2 )   8 2 8 ab  3a 2  b 2  16

Similarly I y'y ' 

I xx  I yy



I xx  I yy

cos 120   I xy sin 120 

2 2 2 2 ab b  a  1 ab b 2  a 2    8 2 8 2 2 ab a  3b   16

Product of inertia is calculated using the formula

73

 3

I x' y ' 

I xx  I yy

sin 120   I xy cos120 

2 3 ab b 2  a 2   2 8 ab  3a 2  b 2  16

5.10 Y

y O

x2

x1

X

To calculate IXX we take a horizontal of width dy strip at y (see figure) for dA and calculate I XX 

y

R

2

dA  2  y 2 R 2  y 2 dy 0

To evaluate the integral, we substitute y  R sin  so the integral is transformed to  2

I XX  2 R 4  sin 2  cos 2  d 0



4  2

R 2

 sin

2

2 d

0

Now substituting z  2 , we get I XX  



R4 sin 2 z dz 4 0

 R4 8

74

Y

x X O

Similarly for IYY, we take a vertical strip of width dx at x for dA (see figure) and calculate I YY   x 2 dA 2R



x

2

R 2   x  R  dx 2

0

Taking  x  R   R cos  we get 0

I YY  R 4  1  cos   sin    sin  d 2





 R 4  1  2 cos   cos 2   sin 2  d 0



2 Now  sin  d  0

 , 2





0

0

2  cos  sin  d  0 and

I YY 

 sin

2

 cos 2  d 

 . This gives 8

5 4 R 8

5.11 Product of area about the origin O is given as I xy 

 x y A i

i

i

i



 xy dA

If the centroid is at O’ which has the coordinates  x0 , y 0  and the coordinates of a point with respect to O’ are  x , y  then x  x0  x

y  y0  y

75

Y

Y’

X’

(x0,y0)

X

O Therefore I xy 

 xy dA    x

0

 x  y 0  y  dA  x 0 y 0 A  x 0  y  dA  y 0  x dA   x y  dA

However, by definition of the centroid

 xdA  0

 y dA  0

Thus I xy  x 0 y 0 A 

 x y  dA

5.12 Consider the moments and product of inertia of a square about a set of axes parallel to its sides and passing through its centre.

Y

X

a For this set of axes

76

 a4   I xx  I yy    12 

I xy  0

Now by the formula I x'y' 

I xx  I yy 2

sin 2  I xy cos 2

Ix’y’ will always remain zero because of the equality of Ixx and Iyy irrespective of the angle of rotation of the new set of axes x’y’. Thus any set of axes passing through the centre is the principal set of axes. 5.13 The formulae for the moments of inertia in rotated frames are

I x'x' 

I y'y' 

I xx  I yy 2

I xx  I yy 2





I xx  I yy 2

I xx  I yy 2

cos 2  I xy sin 2

cos 2  I xy sin 2

Taking the second derivative of these expressions with respect to  , we get

I xx  I yy d 2 I x 'x '  4 cos 2  4 I xy sin 2 2 2 d d 2 I y'y' d 2

4

I xx  I yy 2

cos 2  4 I xy sin 2

Thus the two derivatives have opposite signs. This implies if one of them is a maximum, the other one will be a minimum.

77

Chapter 6 6.1 (i) (a) and (d) are the virtual displacements because these are the only ones consistent with the constraint that the block can move only in the vertical direction. (ii) Suppose the strech is y0. In that situation, the forces on the block are as shown

ky0

mg

Now a virtual displacement gives a displacement of y in the vertical direction. Taking it in the vertically up direction gives the virtual work to be

W   ky 0  mg y Equating this to zero gives y0 

mg k

6.2 Figure below shows the students and the plank on a wedge and a possible virtual displacement of the system.

x

(3-x)

 40g

30g 50g

It is clear that as long as the point on the wedge does not move, the only possible displacement is the rotation of the plank about this point and the system has only one degree of freedom. For the

78

virtual displacement shown, the displacement and the virtual work done by various forces is as follows: 40kg :

virtual displacement  x ;

Center of gravity of plank : 50kg :

virtual work  40 xg

virtual displacement   x  1.5 ;

virtual displacement   3  x  ;

virtual work  30 x  1.5 g

virtual work  50 3  x  g

Since the net virtual work must vanish for equilibrium, we have

 50 3  x   40 x  30 x  1.5 g

 0   120 x  195  0 

x

195  1.625 120

6.3 (i) Since the number of parameters required to describe the system is 1, the number of degrees of freedom is 1. We choose it to be the angle , the rod makes from the vertical. A virtual displacement will be to change  by .

1.5 m



2kg

T 20kg

(ii) To apply the principle of virtual work, we need to calculate the virtual work done by various force when  is changed to  +.

For this we first write the coordinates (x, y) of the tip of the

rod and yCG of the centre of gravity of the rod with respect to the pivot point. These are

79

x  1.5 sin 

y  1.5 cos 

y CG  0.75 cos 

As  is changed to  +, these coordinates change and the changes are given by x  1.5 cos  

y  1.5 sin  

y CG  0.75 sin  

Thus the total virtual work done by the external forces – 2g at the CG, 20g at the tip, both in the positive y direction, and T at the tip in the positive x direction – is W  1.5 cos   T  1.5 sin   20 g  0.75 sin   2 g 

Equating this to zero gives T  21g tan 

6.4 To find the forces applied by the bricks, we treat these forces as external. For only vertical motion, there are two degrees of freedom. One is the vertical displacement of the centre of gravity and the other the rotation of the plank about the CG. Equivalently, we can take the vertical displacements of the ends A and B as the virtual displacements. We choose the second option because this is related directly to the forces applied by the bricks. The plank in equilibrium and virtually displaced is shown below

1.5m

y2

1m

y1

NA

100N 2m

NB 500N

Now the vertical virtual displacement of different points is Point A :

y1

Point B :

y 2

Centre of gravity of the plank :

y1  y 2 2

80

  y 2   y1  y1  1.5   0.25 y1  0.75 y 2 2  

Athlete : Thus the total virtual work done is

 y  y 2  W  N A y1  N B y 2  100 1   500 0.25y1  0.75y 2 2     N A  50  125y1   N B  50  375y 2

Equating the virtual work to zero and therefore the coefficients of each independent displacement (y1 and y2) to zero gives NA=175N and NB = 425N 6.5 (i) Constraints on the system: length of the strings holding the masses is fixed (ii)

There is only one degree of freedom. This can be understood as follows. There are three variables needed to describe the system: The distance of two masses and one pulley from the ground. However there are two constraints: To of the strings have fixed lengths. Thus only one variable is left to change freely.

(iii)

In terms of the lengths shown in the figure

h1

M1 M2

y1

h2

y2

The constraint that the longer string has fixed length is expressed as

 h1  y1   2 h1  h2   3h1  2h2  y1  const.

81

The constraint that the shorter string has fixed length is expressed as h2  y 2  const.

(iv)

The constraint forces are the tension in the two strings. It is by these tensions that the constraints are maintained.

(v)

To apply the method of virtual work, we make a virtual displacement y1 of mass 1. The corresponding displacement of mass 2 is then y 2 . However, since there is only 1 degree of freedom, we will finally express y 2 in terms of y1 to apply the method of virtual work. The virtual work done in these processes is W   M 1 gy1  M 2 gy 2

Notice that the two virtual displacements are not independent. Therefore we should not conclude that M1=M2=0 for equilibrium. To apply the method, we first express the virtual work in terms of only y1 . From the first constraint equation, since h1 if constant,  2h2  y1  0  h2  

y1 2

From the second constraint we have h2  y 2  0  y 2  h2  

y1 2

Substituting these in the expression for the virtual work gives W   M 1 gy1  M 2 g

y1  M     M 1  2  gy1 2 2  

Equating this to zero then gives, for equilibrium M 2  2M 1

6.6 (i) Since the two blocks are free to move in one dimension without any constraint, the number of degrees of freedom for the system is 2. (ii)

The system in equilibrium is shown below. At equilibrium spring on the left is stretched by x1 and the total stretch of the two springs together is by x2. Thus spring on the right is stretched by (x2x1).

Thus force on mass m1 is k1 x1 to the left and

k 2 ( x 2  x1 ) to the right. These two forces must be equal for equilibrium but we wish

82

to obtain this by applying the principle of virtual work. Similarly the two force acting on m2 are F and k 2 ( x 2  x1 ) to the left. Again these two forces must be equal for equilibrium and we will obtain this by applying the principle of virtual work. k1

m1

k2

m2 F

x1

x2

As the mass m2 is moved by a virtual displacement  x 2 , let us assume that mass m1 moves by  x1 . Then the virtual work done will be W   k1 x1x1  k 2 ( x 2  x1 )x1  Fx 2  k 2 ( x 2  x1 )x 2

Equating this to zero and therefore the coefficients of  x1 and  x 2 to zero gives k1 x1  k 2 ( x 2  x1 ) and F  k 2 ( x 2  x1 )

This gives x1 

F  1 1    . and x 2  F  k1  k1 k 2 

6.7 (i) There is only one degree of freedom. Although the piston moves in the vertical direction and the wedge in the horizontal direction, their movements are connected because the piston moves on the surface of the wedge. (ii) Normal reactions on all the surfaces are the constraint forces. In the present context, the constraint force specific to the piston’s movement on the surface of the wedge is the normal reaction of the wedge surface on the piston. (iii)

if the distance of the middle line of the piston is at a distance a from the origin, its height is y and the distance of the left edge of the wedge is x (see figure) then the constraint is expressed as

83

m

x



F

y

a

y   a  x  tan 

  y   x tan 

Now the method of virtual work is applied by considering x as the free variable, varying it by x, and equating the total virtual work to zero. The virtual work is W   F x  mg y    F  mg tan   x

Equating the coefficient of x to zero gives F  mg tan  6.8 When the equilibrium angle is , the distances of various points (see figure) , taking A as the origin are as follows x B  l sin

 2

x D  l sin

 2

y C  2l cos

The force on the two points B and D due to the spring is   k  l 2  2l sin  2 

to the left on B and to the right on D

84

 2

A

x B

D

y



C W It is clear that we need only one parameter  to specify the system. Thus there is only one degree of freedom. Now let us make a virtual displacement by changing  by  . In that situation l  x B    cos 2 2

l  x D   cos 2 2

y C  l sin

 2

Thus the total virtual work done as  is changed by  is    l   l    W   k  l 2  2l sin  cos  k  l 2  2l sin  cos  mgl sin   2 2 2 2 2 2 2   

Vanishing of the virtual work then implies    l   l     k  l 2  2l sin 2  2 cos 2  k  l 2  2l sin 2  2 cos 2  mgl sin 2   0      

85

Or equivalently 2  2 sin

If

  mg     tan 2  kl  2

mg  mg   0 then the solution is   . For very small value of therefore we have    x kl 2 kl 2

, where x is very small.

Substituting in the equation above, we get x   mg  x    2  2 sin  45      tan  45   2   kl  2  

Since

x is small, we get by Taylor series expansion 2 x x 1  x  sin  45    sin 45   cos 45    1  2 2 2 2  x x  tan  45    tan 45   sec 2 45   1  x  2 2 

Thus the equation for equilibrium is 2 2

1  x   mg   1     (1  x) 2   kl  2



 1

x 

2



mg  mg   kl  kl

Or to a good approximation x

mg kl

2

Thus 

 mg  2 2 kl

6.9 (i) Even if we consider only one dimensional motion, we would require two variables, one the angle that the bar makes with the vertical and the other describing the position of the mass. However, the two are connected by a rope. Hence the two variables cannot vary independently since the length of the rope is foxed (constraint). Therefore there is only one degree of freedom in the system. (ii)

The constraint is the length of the rope remaining constant. It is enforced by the tension that develops in the rope. This tension makes the movement of the bar and the mass restricted. Thus it is the tension in the rope that is the force of constraint.

86

(iii)

A virtual displacement would be to displace the bar by an angle  from its equilibrium position. Since the length of the rope is fixed, the midpoint of the bar moves by the same distance as the mass connected to the rope. Therefore the tension does positive work at one end of the rope and exactly equal but negative work at the other end. The sum of the work done by the tension then vanishes.

(iv)

Let us say we make a virtual displacement of the bar by turning it counterclockwise by an angle from the vertical. Then the end of the rod moves by l in the direction of the force. At the same time, the virtual displacement of the mass is equal to the movement of the midpoint of the bar. Thus the mass moves by

l  2

opposite to the gravitational force (see figure) F 

m

The net virtual wok therefore is l W  Fl  mg  2

Equating this to zero then gives F

mg 2

6.10 Initially the bars are at 90. The weight has a tendency to fall down so the torque applied is such that it tends to pull the weight up. Thus is a virtual displacement of angle  is made in the direction of the torque, the weight will be lifted up by the corresponding distance y.

87

The corresponding virtual work done by the torque  is  and the corresponding work by the gravity is Wy. By the method of virtual work then we have   Wy  0

This gives  W

y 

Next we calculate the relationship betweeny and . As the rod is rotated by angle , the length of the horizontal diagonal decreases by

p  . If the corresponding angle at the 2

vertical corner changes from 90 to 90+α, then we have (see figure)

(90+α)



W

Change in the length of the horizontal diagonal 

= 2 l sin 

This should equal

90    l    l  l   2 l sin 45   l  cos 45      2 2 2 2 2 

p  so we have 2 l p    2  2

As the angle changes the length of the vertical diagonal changes by  90    l    l  l   2 l cos 45   l 2 l cos  sin 45      2 2 2 2 2  

Thus we have

88

y 

l 2

 

p  2

y p   2



This gives  

6.11 (a)

pW 2

For the motion in a plane, the orientation of the rod can be described by the

displacement of its centre of mass and the angle it has rotated by about its CM. Or equivalently the displacement of its two ends is sufficient to describe its orientation. Thus the degrees of freedom is 2. (b) We take the vertically down displacement of the two ends as the virtual displacement as shown in the figure.

yCM y1 y2

Let the centre of mass be at a distance

2L from the left hand end of the rod. In that case, 3

the centre of mass moves down by y CM  y1 

2  y 2  y1   1 y1  2 y 2 . The virtual work 3 3 3

done by the springs in such a virtual displacement is  ky1y1 and  ky 2y 2 respectively while

89

that by the weight of the rod is Wy CM 

W 2W y1  y 2 . Thus equating the net virtual work to 3 3

zero gives W 2W y1  y 2  ky1y1  ky 2y 2  0 3 3

Now equating the coefficient of each independent displacement to zero gives y1 

W 3k

y2 

90

2W 3k

Chapter 7 7.1

For Cartesian coordinates (x,y), the planar polar coordinates are

r

x2  y2

and

 y  . If y180 and if y0, >270.  x

  tan 1 

Unit vectors are given by rˆ 

7.2

xiˆ  yˆj  yiˆ  xˆj and ˆ  r r   At each point the velocity v is given as  v .rˆ  rˆ   v .ˆ  ˆ . Thus for (iv)



 v  2iˆ  3 ˆj

for a particle at (-2,-3), the velocity is

  2iˆ  3 ˆj     2i  3 j  rˆ    2iˆ  3 ˆj   3i  2 j  ˆ   5rˆ  12 ˆ

ˆ

ˆ

13

7.3

ˆ

ˆ

13

The two points with polar coordinates

13

 r1 , 1  and  r2 ,  2 

and the corresponding

vectors are shown in the figure below

 r2

ˆ2

  r1  r2

ˆ1 2

 r1

(21)

1

(i) As is clear from the figure, the angle between the vectors r1 and r2 is   2  1  . This also 



the angle between unit vectors rˆ1 and rˆ2 and unit vectors ˆ1 and ˆ2 . Therefore rˆ1  rˆ2  ˆ1  ˆ2  cos  2  1 

91



Similarly, angle between r1 and ˆ2 is

     2  1  and that between r2 and ˆ1 is 2

    2  1  . Therefore 2  rˆ1  ˆ2  cos   rˆ2  ˆ1  cos 

(ii) (iii)

7.4

    2  1    sin   2  1  2     ( 2  1 )   sin   2  1  2 

  Similarly, from the figure r1  r2  sin   2  1  zˆ   r1  r2 

r12  r22  2r1 r2 cos  2  1 

The trajectory of the projectile is shown schematically in the figure below. The horizontal distance from the origin to the point of highest elevation (height 1.25m) is 2.5 3m .

The vertical component of the velocity at this point vanishes while the

horizontal component is 5 3ms -1 .

Similarly, the vertical component of the

acceleration is 10ms-2 vertically down. Also shown in the figure are unit vectors in the radial and the tangential directions at the highest point and on the ground.

ˆ

rˆ ˆ

1.25m



rˆ

2.5 3m From the figure it is clear that for the point of highest elevation tan  

1.25 2.5 3



1 2 3

;

sin  

1 13

;

cos   2

3 13

Therefore the radial and tangential components of the velocity at the highest point are

92

v r  5 3 cos   5 3  2

3 30  ms 1 13 13

v  5 3 sin   5

3 ms 1 13

Similarly the radial and tangential components of the acceleration are a r  10 sin   

10 13

a  10 cos   20

ms  2

3 ms 2 13

On the ground, the radial unit vector points towards positive x direction and the tangential unit vector is in the negative y direction.

Thus the radial component of the

velocity is its x component and the tangential component is its y component. Therefore v r  5 3ms 1

v  5ms 1

Similarly, since the gravitational acceleration on the ground is in the negative y direction, its radial and tangential components are a  10ms 2

ar  0

7.5

The position of the particle at time t is shown in the figure below.

2ms-1

1

 2t The polar coordinates of the particle at time t are given as  1   2t 

  tan 1 

r  4t 2  1

Therefore r 

4t

  

4t  1 2

This gives the velocity in polar coordinates as

93

2 4t 2  1

 v  rrˆ  rˆ 

4t 4t  1 2

rˆ 

2

ˆ

4t 2  1

Differentiating the velocity vector gives  dv  dt

4 4t  1 2

Now substituting rˆ  ˆ  

rˆ 

 4t

16t 2 2



1

32

4t

rˆ 

4t  1 2

rˆ 

 4t

8t 2



1

32

ˆ 

2 4t  1 2

 ˆ

2 2  ˆ and ˆ   rˆ  2 rˆ , we get the acceleration above to 4t  1 4t  1 2

be zero. 7.6

(i) Kepler’s second law states that the rate of the area swept (

) A

by the radius vector is

constant. This can be expressed as (see figure for the symbols used)   1 r 2  constant A 2

 r 

Sun

Now differentiating the equation above with respect to time gives 1 2  r   rr  0 2



r 2  2rr  0

The expression on the right is the tangential acceleration. Thus Kepler’s second law gives tangential acceleration to be zero. (ii)

Since the tangential acceleration is zero, the force is only in the radial direction. Acceleration in the radial direction is

94



a r  r  r 2 We differentiate the orbit equation r  r 



r0 with respect to time to get 1  e cos 

r0 e sin 

1  e cos   2



r 2e sin  2 A e sin   r0 r0

 is constant, differentiating the equation above once more with respect to time, we Since A

get

 r  2 A  0  1  r 2 A e cos  4 A 2  r0    r     1  2 r0 r0 r0 r  r  This gives 4 A 2  r0   1  r 2  2 r  r  2 4 A  r0 4 A 2    1  r   r0 r 2  r r4   4 A 2  1  2   r  0  r

ar 

This shows that the force is proportional to r2. 7.7

It is given that mx   F sin  t my  F cos  t

which gives F  my sec  t

Now write y  x tan t and differentiate it twice with respect to t to get y  x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t

This gives





F  my  m x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t sec t

Substitute this in mx   F sin  t to get





x   x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t sin t sec t

Multiplying and rearranging terms gives

95

x  2 x tan t  2 2 x tan 2 t

The equation does not appear to be easily ingrable. 7.8 It is given that r  2t and   1rad s 1 Since r (0)  1 , integrating the equation for r gives

(i) r (t )

t

1

0

2  dr   2t ' dt '  r (t )  1  t or equivalently r (t )  t 2  1



(ii) v  rrˆ  rˆ  2trˆ  (t 2  1)ˆ If the velocity vector is at 45 to the radius vector, we have

2trˆ  (t

  v .rˆ  v cos 45 

2



1  1)ˆ .rˆ  2



4t 2  (t 2  1) 2



This gives 2t 

4t 2  t 4  1  2t 2 t 4  6t 2  1  2 2

Squaring both sides gives 8t 2  t 4  6t 2  1 OR t 4  2t 2  1  0

This is equivalent to

t (iii)

2



2

 1  0  t  1s

Radial acceleration zero implies





a r  r  r 2  0 r  2t  r  2 which leads to





2  t 2  1  0 giving t  1s

This also gives the distance from the origin to be r (t  1)  1  1  2m 7.9 Free body diagram of the bead when its radius vector is making an angle  (increasing as the particle slides down) from the vertical is shown below

96

N R



 mg

Taking the components of the forces in the radial and tangential directions and equating these to mass times the radial and tangential components of the acceleration, respectively, gives



In the radial direction

N  mg cos   m r  r 2

In the tangential direction

mg sin   m r  2r







Since the particle moves on a path of constant radius r  R , we have r  r  0 When substituted in the equations above, this gives N  mg cos    mR 2 g sin   R

1 d 2 Now using   , we get from the second equation above 2 d

R 

R d 2  g sin  2 d

which gives upon integration 

R 2   g  sin  d  g 1  cos    R 2  2 g 1  cos   2 0

When substitutes in the equation N  mg cos    mR 2 , this gives N  m 3 g cos   2 g 

7.10

The free body diagram of the bead at equilibrium is shown in the figure below.

97

N

 mg



The horizontal components of the normal reaction N provides the centripetal force while the vertical component balances the weight of the bead. Thus N sin   mx 2 N cos   mg

Dividing the first equation by the second one gives tan  

x 2 g

The slope of the curve is also equal to tan  . Thus x 2 dy   4cx 3 g dx

This gives 2   4cg 2 cg

x

7.11

Since  

4 and y  cx 

4 16cg 2

1 2 t , we have 2

r  2  t 2 ,

r  2t





and

r  2 ;   t

and

  1

2 2 2 Thus a r  r  r 2  2  t 4 and a   r  2r   t  4t  5t

7.12

We know that rˆ  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ

ˆ  cos  cos  iˆ  cos  sin  ˆj  sin  kˆ

ˆ   sin  iˆ  cos  ˆj

Therefore 98

 ˆ   sin  cos  iˆ   cos  sin  iˆ   sin  sin  ˆj   cos  cos  ˆj   cos  kˆ







  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ   cos   sin  iˆ  cos  ˆj   rˆ   cos  ˆ

Similarly  ˆ    cos  iˆ  sin  ˆj 



  sin  rˆ  cos  ˆ

7.13



Since m3 y 3  T2  m3 g

We get from the solution y3 

4m1 m2   m1  m2  m3 4m1 m2   m1  m2  m3

T2  m3



g

4m1 m2   m1  m2  m3 g  m3 g 4m1 m2   m1  m2  m3

8m1 m2 m3 g 4m1 m2   m1  m2  m3

From T2  2T1  0 , we get

T1 

4m1 m2 m3 g 4m1 m2   m1  m2  m3

Now from m1 y1  T1  m1 g we get

y1 

3m2 m3  4m1 m2  m1 m3 g 4m1 m2   m1  m2  m3

And from m2 y 2  T1  m2 g we get

y 2 

3m1 m3  4m1 m 2  m2 m3 g 4m1 m 2   m1  m2  m3

7.14

99



m

l

 T2

T1

(i) The tension in the outermost string provides the centripetal force for the outermost bead. Let the tension in this string be T1. Then T1  Nml 2

Similarly, centripetal force for the second bead from outer end is provided by the difference in the tension T2 in the second string and tension T1 in the first string. Thus T2  T1  ( N  1) ml 2

 T2   2 N  1 ml 2

T3  T2  ( N  2)ml 2

 T3   3N  3 ml 2

T4  T3  ( N  3) ml 2

 T4   4 N  6 ml 2

Extending further

and so on. In general we can write for the ith string from the outside, Ti  Ti 1   N  (i  1) ml 2

 Ti  2   N  (i  2) ml 2   N  (i  1) ml 2  

This is then easily seen to be Ti   iN   (i  1)  (i  2)  (i  3)    2  1  0  ml 2 i (i  1)     iN  ml 2  2   2 iml   2 N  i  1 2

(ii)

Now we generalize the result of part (i) to a rope of length L and mass per unit length λ. To make the transformation from the problem in part (i) to this problem, we consider the mass of each bead to be distributed over the connecting string whose length we take to be vanishingly small i.e. l  0 . Thus we have L  Nl , x  ( N  i )l ,  L  x   il and m  l

Substituting this in the expression for Ti above, we get

100

( L  x) 2 (2 Nl  il  l ) 2  2  ( L  x )(2 L  L  x  l ) 2  2  ( L  x )( L  x  l ) 2

T ( x)  Ti 

On taking limit l  0 , we then get T ( x) 

 2 2 (L  x 2 ) 2

To get this answer by considering the rope directly, we take a small portion of the rope of length x at a distance x from the centre O. The centripetal force to it is provided by the difference in the tension at its two ends, as shown in the diagram below T(x+x)

T(x) Then T ( x)  T ( x  x)  x 2 x 

T ( x)  T ( x  x)   2 x x

This gives the differential equation 

dT   2 x dx

The solution of this equation is T ( x)  C 

 2 x 2 2

where C is the integration constant. With the condition that the tension vanishes at x=L, we then get C

 2 L2 2

and

T ( x) 

 2 2 L  x2  2

Consider a small section of the rope making a small angle  at the centre of the

7.15

loop. The force at its two ends due to the tension in the rope is shown in the diagram below

101

/2  T

T Tsin(/2)

For small angle the forces at the ends give a net force towards the centre which is equal to      2T  T 2  2

2T sin 

This provides the required centripetal force for the segment. Therefore T  R  R 2

7.16

 T  R 2  2

The relevant coordinates and the free body diagrams of the two masses are shown below

r T m

 y

T

mg

M Mg

102

We treat the mass m using the polar coordinates as shown in the figure. The total number of unknowns in the problem are : y coordinate of mass M, polar coordinates of mass m and tension T in the string. The corresponding equations are M y  Mg  T m r  r 2  mg cos   T m r  2r  mg sin 

 





And the constraint equation r  y  constant

 r  y  0 and r  y  0

7.17 (i) After time t, the end of the rod that was at the origin has moves by a distance . Thus the relationship between the x and y coordinates will be y  tan  1 2 x  At 2



y cot   x 

1 2 At 2

(ii) Free body diagram of the bead is shown below

θ θ

N

Thus the equations of motion are m x  N sin  m y   N cos 

From the constraint equation A  y cot   x

Thus N sin   m A  y cot  



N  mA cos ec  my cot  cos ec

which gives N cos   mA cot   my cot 2 

103

1 At 2 2

Thus my   mA cot   my cot 2 

 my cos ec 2   mA cot 

Or y   A cos  sin   

A sin 2 2

Integrating the equation above with the condition y (0)  0 and y (0)  d gives

(iii)

y (t )  d 

Thus the bead will take time t 

A sin 2 t 2 4

4d to reach the lower end. A sin 2

Suppose the lower corner with angle  is at the origin at t=0. Then if the position of

7.18

the mass is (x, y) at time t, the relationship between these coordinates is y  tan  1 2 x  At 2



y cot   x 

1 2 At 2

if the acceleration of the wedge is A. This implies A  y cot   x

(i)

Now if the mass falls vertically down, y   g which gives A  g cot  i.e. the wedge accelerates to the right with acceleration g cot  .

The interpretation is

quite simple: when the wedge moves by horizontal distance should move vertically down by

minimum acceleration is (ii)

1 At 2 , the mass 2

1 2 gt and the relationship between the two at 2

1 gt 2 2  tan   A  g cot  . 1 At 2 2

For the particle not to move with respect to the wedge, we have y  0

which implies x  A .

104

The free body diagram of the mass is as follows Y N

θ

X

Thus we have by the conditions above N cos   mg  N sin   mA

which gives A   g tan 

7.19

From example 7.9 mx2  N 1 sin  M x1   N 1 sin  y 2  ( x1  x2 ) tan 

N1 

mg cos  m   2  1  sin   M  

This gives (assuming x 1 (0)  x1 (0)  0 ) m g cos  sin  x1   M m   sin 2    1 M  

Similarly x2 

g cos  sin  m   sin 2    1 M  

This gives

105



m 2  g sin  M m  1 sin 2   M 

 1

y 2   

  

Thus if mass m starts from height h, it will take time 2h (m  M cos ec 2 ) (m  M ) g

2h  y 2

t

And at this time the speed of the wedge will be x1t  m cot 

2 gh (m  M )( m  M cos ec 2 )

7.20 The coordinates used in solving the problem are shown in the figure below P1 P2 T h1

T m

T T

P3

y1

h2

m y2

y3

Let the tension in the string be T. The equations of motion for the two masses are my1  T  mg my 2  T  mg

And since the pulley P3 is massless, we have 2T  k  y 3  h   0

Thus we have four unknowns y1, y2, y3 and T but only three equations. One more equation is provided by the constraint equation. The constraint is that the length of the string is a constant. If the heights of the two fixed pulley are h1 (for P1) and h2 (for P2), the constraint is expressed as

106

 h1  y1    h1  y3    h2  y3    h2  y 2   const. or equivalently as y1  2 y 3  y 2  const. 

y1  y 2  2 y 3

Now adding the equations of motion for the two masses gives

m y1  y 2   2T  2mg On substituting y1  y 2  2 y3 and 2T  k  y 3  h  from the equations above, we get  2my 3  ky 3  kh  2mg

This gives y 3 

k kh   y3   g   2m 2m  

The general solution of the equation above is the sum of its homogeneous solution yh(t) and the particular solution yp(t). We have y h (t )  A cos t  B sin t y p (t )  h 

2mg k

Here A and B are two constants to be determined by the initial conditions and   the general solution is 2mg   y 3 (t )  A cos t  B sin t   h   k  

Now the initial condition is y 3 (t  0)  h

y 3 (t  0)  0

These give A

2mg k

B0

Thus y 3 (t )  h 

2mg  1  cos k 

k 2m

This immediately gives through the equation 2T  k  y 3  h   0  T  mg  1  cos 

107

k 2m



t  



t  

k . Thus 2m

This gives k t 2m

y1   g cos

This is easily integrated. Since it is gives that y 1 (t  0)  0 and y1 (t  0)  h , we get y1 (t )  h 

2mg   1  cos k 



k 2m

t 

k 2m

t 



In exactly tha same manner we get y 2 (t )  h 

2mg k



 1  cos  

 

7.21 (i) Wedge m3 is free to move only in the horizontal direction; m1 and m2 move both horizontally as well as vertically. Thus we would have had 5 degree of freedom. However, there are three constraints: The length of the string is fixed, mass m 1 moves only in the vertical shaft and mass m 2 moves on the plane of the wedge.

These

constraints reduce the degrees of freedom to 2. Thus there are only two degrees of freedom. (ii) The origin and the coordinate axes chosen to describe the motion are given in the figure below. Also shown are the free-body diagrams of the three masses

m1 m2

y1

m3

x1 x2

x3

108

y2



T

T

N2

T N1

N1

T

N2

m3

m1g m2g

m2g N3

We see that there are in total 8 unknowns: x1, y1, x2, y2, x3, N1, N2, N3 and T. The equations of motion are: mx1  N 1 (i) my1  T  m1 g (ii) m3 x3   N 1  N 2 sin   T cos  m2 x2  N 2 sin   T sin  (iv) m 2 y 2  N 2 cos   T sin   m2 g

N 3  T 1  sin    N 2 cos   m3 g

(iii) ( v) ( vi)

These are the equations of motion. In addition there are three constraint equations.

y 2   x3  x 2  tan 

y 2   x3  x2  tan  ( vii)  h  y1    x2  x1  sec  const.   y1   x2  x1  sec  0  x3  x1   const.  x1  x3 (ix ) 

( viii)

In the above, h is the height of the wedge. There are a total of nine variables and nine equations. Of these equation (vi) is not relevant for the motion since the wedge moves only in the horizontal direction. Equations (i), (iii), (iv) and (ix) give m3 x1   m1 x1  N 2 sin   T cos   m1 x1  m2 x2

which gives

 m1  m3  x1  m2 x2  0 This is the equation expressing momentum conservation in the horizontal direction. Similarly equation (iv) and (v) along with (ii) give

109

m2 x2 cos   m2 y 2 sin   T  m2 g sin   m1 y1  m1 g  m2 g sin 

Now substituting for y1 from (viii) and y 2 from (vii) and using (ix), we get

m2 x2 cos   m2  x1  x2  tan  sin   m1  x2  x1  sec   m1 g  m2 g sin  Which is equivalent to m2 x2 cos 2   m2  x1  x2  sin 2   m1  x2  x1    m1  m2 sin   g cos  

 m1  m2  x2   m1  m2 sin 2   x1   m1  m2 sin   g cos 

Now using  m1  m3  x1  m2 x2  0 , we eliminate x2 from the equation above to get



 m1  m2  m1  m3  





x1  m1  m2 sin 2  x1   m1  m2 sin   g cos 

m2

This gives

x1 

m2  m1  m2 sin   g cos   x3 m12  2m1m2  m1m3  m2 m 3  m22 sin 2 

And using  m1  m3  x1  m2 x2  0 , we then get

x2  

 m1  m3  m1  m2 sin   g cos  m  2m1 m2  m1m3  m2 m 3  m22 sin 2  2 1

Now using equation (viii) we get

y1  

 m1  m2  m3  m1  m2 sin   g m  2m1 m2  m1 m3  m 2 m 3  m22 sin 2  2 1

Using equation (vii), we get

y 2 

 m1  m2  m3  m1  m2 sin   g sin  m12  2m1 m2  m1 m3  m2 m 3  m22 sin 2 

As m3   , we get

x1  x3  0 x2  

 m1  m2 sin   g cos  m1  m2 

y1  

 m1  m2 sin   g  m1  m2 

y 2 

 m1  m2 sin   g sin  m1  m2 

7.22 Consider two parts of the rope x and (L-x) in length. The free body diagrams (showing only the horizontal forces) of the rope and these two parts are shown below

110

x

F

Friction

T(x)

F Friction

T(x) Friction

Since the rope is moving with constant speed, there is no net force on the rope or any part of it. Thus from the free body diagram of the rope F  Mg

From the free body diagram of the left portion of the rope F  T ( x)  

 L  x  Mg M xg  0  T ( x)   L L

Or from the free body diagram of the right portion of the rope T ( x)  

 L  x  Mg L

It is also instructive to solve the problem by considering a small portion of length dx at distance x from the left and balancing the forces there to get a differential equation for T(x). The free body diagram of such a portion is as given below dx T(x)+dT(x)

T(x) Friction T ( x )   T ( x )  dT ( x )  

dx Mg L



dT ( x ) M  g dx L

With the condition that T(L) = 0, the equation above can be integrated to get

111

T ( x)

x

M 0 dT   L g L dx



T ( x) 

 L  x  Mg L

7.23 The force applied should be such that the frictional force on mass m is sufficient to balance its weight. Free body diagrams of the two blocks are shown below

friction F

N

M

mg

N friction Mg

If the entire system is moving with acceleration a then N  Ma F  N  ma

This gives a

F M m

N

and

M F M m

If mass m is not falling then friction  N  

 M  m  mg M F  mg  F  M m M

Thus Fmin 

 M  m  mg M

For m=16kg, M=88kg, =0.38 and g=9.8ms-2, we get Fmin 

104  16  9.8  487.7 N 0.38  88

112

Normal reaction

7.24 The forces on the bead are: Its weight, normal reaction NV in the vertical direction and normal reaction NH in the horizontal direction. Thus the free body diagram of the bead is as follows (assuming the rotating arm is going into the page)

NV friction

rˆ

NH mg

These force provide the radial and the tangential acceleration given by









 a  r  r 2 rˆ  r  2r ˆ

Since the rod is rotating with a constant angular speed , we have





 a  r  r 2 rˆ  2rˆ

(i) If the bead is stationary at r=R, we have  a   R 2 rˆ

This gives N V  mg

NH  0

friction  mR 2

Since friction    N , we get for stationary bead mR 2  mg   02 

(ii)

g R

If    0 and negligible weight of the bead, we have N V  mg

N H  2mr

N  m g 2  4r 2 2  2r

friction   N  2 mr

The minus sign for the friction shows that since the bead slides outwards, the frictional force in inwards. The equation describing the motion of the bead is then r  r 2  2 r 

OR

r  2 r   2 r  0

Assuming a solution of the form r (t )  e t and substituting it in the equation above we get 2  2    2  0

Solution of this equation gives

113



1   1   2  



and





 2   1   2  

Thus the general solution is

 

r (t )  A exp t



 

1   2    B exp  t

1  2  



Here A and B are to be determined by the initial conditions that r (t  0)  R and r (t  0)  0 . Thus A B  R



 

1  2   A 



1  2   B  0

The solution is A

R  2 

1   2    1  2

B

and  

R  2

1   2    1  2



 

Thus the distance of the bead from the center is given as

r (t ) 

R 2 1 

2



  

1   2   exp t

 

1  2   

  

1   2   exp  t

1  2  

 

7.25 The solution for the distance travelled by the particle is x (t ) 

F k





m  k m  t   t  k 1  e  

For k=0, we can’t substitute this directly in the formula since we are dividing by k in the expression above. Thus we take the limit k  0 . In this limit, the first nonzero term we get is  m k 1 k2 2  1  1  t  t  higher order terms  2 k  m 2m  1 k 2  t 2m

t

This gives x (t ) 

7.26 The equation of motion for the particle is 114

1F 2 t 2m

y   g 

k y OR m

y 

k y   g m

The solution of the homogeneous equation y 

k y  0 m

is k   y  A exp  t   m 

Here A is a constant to be determined by the initial conditions. The particular solution is y  

mg k

Therefore the full solution is k  mg  y  A exp  t   k  m 

Now the initial condition is y (t  0)  v0 . This gives A  v0 

mg k

Thus mg  k  mg   y (t )   v 0   exp  t   k  k   m 

This is easily integrated to get y(t) also. Thus y (t )  

mg m mg  t   v0   k k k 



 k    1  exp  m t      

When the ball reaches the highest point, its speed is zero. If this time is tup, then mg    k  mg 0   v0   exp  t up   k  k   m 

mg k 1  k   exp  t up    mg    kv   m   v0    1  0  k  mg   

OR t up 

kv  m  ln 1  0  k  mg 

This gives the height h to be kv0  mg m  m mg   h  ln 1   v0  k k mg k k   

115

 1    1 kv0   1  mg  









 

mv0 kv 0  m2 g    ln 1  k mg  k2 

If the total time of flight is T then T=tup+tdn, where tdn is the time taken to come down from the highest point. At time T, y(T)=0. Therefore 0

mg  t up  t dn   m  v0  mg  k k k 



 k   k    1  exp  m t up  exp  m t dn        

Thus tdn will be given by solving 

mg mg t up   t dn k k

mg k m mg   k     v0  exp  t dn    1 mg  k  k   m     v0    k   

     

OR t up  t dn 

1 mg  k      1  exp  t dn     v0  g k  m      t dn 

v0 m  k      1  exp  t dn   g k  m   

To understand whether tup is larger or tdn is larger, let us see the time change in the limit of very small k. In that case (up to order k) t up 

 kv  m  kv 1 k 2 v02 v 0 1 kv02 m  ln 1  0    0   ......    k  mg  k  mg 2 m 2 g 2 g 2 mg 2 

And tdn is given by solving t up  t dn 

v0 m   k     1  exp  t dn   g k   m  

v0 m  k 1 k2 2    1  1  t dn  t dn  g k  m 2 m2  v 1 k 2  0  t dn g 2m  t dn 

This gives 2mv0 2m 2  kv  t   2 ln 1  0  kg mg  k   2mv0 2m 2  kv0 1 k 2 v02 1 k 3 v03   2     ....  2 2 3 3 kg 3m g k  mg 2 m g  2 dn



v02 2 kv03  g 2 3 mg 3

Therefore

116

t dn

v  0 g



2 kv 0   1   3 mg  

12



v 0 1 kv02  g 3 mg 2

A comparison shows that tup is smaller than tdn. This makes sense because the while coming down, the average speed is smaller since the particle has lost energy due to viscosity and continues to do so. This also gives the total time of flight to be (up to order k) 2v0 5 kv02   g 6 mg 2

T  t up  t dn

However, this approximation will be valid only if

kv02  1 mg 2

1 5.27 It is given that v 0  100ms and   45 . Therefore v 0 sin   v0 cos   50 2ms 1 . It is

also given that g  10ms 2 . v02 sin 2   250m (i) Height = 2g

Range =

2v 0 sin  v cos   1000m g

(ii) When k  0 , from the expression derived in the problem above, we get with the initial vertical speed v 0 sin  mv0 sin  m 2 g  kv sin   2 ln 1  0 Height = k mg k    

Substituting the values , we get For k=0.1, Height = 202m For k=0.2, Height = 172.3m To find the range, we first fine the total time of flight and then use formula derived in example 7.13 to find the horizontal distance travelled. From the solution of the previous problem, we know that the time of flight T is given by solving mg m mg  T   v0 sin    k k k 



 k    1  exp  m T      

For k=0.1, this gives 200T  54141  exp(0.05T )

OR

117

T  27.071  exp(0.05T )

Since the time of flight without drag is

2v 0 sin   14.14 s , and as the result of the problem g

shows the time of flight becomes smaller for k  0 , we tabulate T and the right hand side of the equation above to find T for T  14

It is clear from the Table above that the expression on the right was smaller than T till T=13s and becomes larger at 12.5s. Thus the time of flight will be between 13 and 12.5s. A little more tabulation gives T=12.8s. Substituting this in x(T )  Since t up 





mv o cos   k / m  T 1 e , we get Range = 669m k

kv sin  m  ln 1  0 k  mg 

  6.05s , we get tdn= 6.75s. This confirms numerically that time 

taken to come down is greater. For k=0.2, the equation to determine the time of flight is 100T  1071  exp(0.1T )

Making a table like above

x (T ) 



T 14 13.5 13 12.5

OR

T  17.071  exp(0.1T )

27.071  exp(0.05T )

13.62 13.29 12.93 12.58



mv o cos   k / m  T 1 e  490m k

118

gives T=11.8s and Range

(iii)

k=0

k=0.1

k=0.2

(iv) When drag is introduced, it is the range that is affected much more than the height. (7.28) In the problem above, we have already found the range for   45 . Let us now take the case of k=0.1 and find the range for   40 and   50 . For   40 , the equation

119

mg m mg    k   T   v0 sin     1  exp  T   k k k    m  

beomes 200T  52861  exp  0.05T   or T  26.431  exp  0.05T  

and gives T=11.7s. This gives a range of x (T ) 





2  100 cos 40 1  e 0.0511.7 = 678m. 0 .1

Thus for   40 , the range increases. For   50 , the equation mg m mg  T   v0 sin    k k k 



 k    1  exp  m T      

beomes 200T  55321  exp  0.05T   or T  27.661  exp  0.05T  

and gives T=13.8s. This gives a range of x (T ) 





2  100 cos 50 1  e 0.0513.8 = 641m. 0.1

Thus for   50 the range decreases. The two calculations above show that for maximum range, the projectile should be launched at an angle less than 45 . The reason for this is as follows. Since the horizontal speed reduces as the projectile moves, it should cover a larger distance in the initial part of the flight. For this it is better to have a relatively larger horizontal component of the velocity compared to the case when there is no drag. Thus the angle should be smaller than 45 . (7.29) In this case the drag force is proportional to the square of the speed. So the equation of motion will be given as follows for the motion up and motion down (taking vertically up direction to be the positive y direction)

120

Motion up

my   mg  ky 2

Motion down

my   mg  ky 2

2 Since we are only interested in height, we change y to 2 dy  y  to get the speed as a

1 d

function of the vertical distance of the ball from the ground. The first equation in that case is

 

1 d k y 2  y 2   g 2 dy m

The solution of this equation is the sum of the solution

 

1 d k y 2  y 2  0 and a particular solution y 2 2 dy m

p

( y)

2 y

h

( y ) of

the homogeneous equation

. These solutions are

mg  2k  y 2 ( y )  A exp  y  and y 2 ( y )   h p m  k 

Here A is a constant to be determined from the initial conditions. The full solution therefore is mg  2k  y 2 ( y )  A exp  y  m  k  2 2 The initial condition is that y ( y  0)  vi . This gives

A  vi2 

mg k

This the speed of the ball as it moves up is  2k y 2 ( y )  vi2 exp  m 

 mg y  k 



 2k  1  exp  m  

  y   

At the maximum height h, the speed becomes zero. Therefore  2k  mg   2k   0  vi2 exp  h  1  exp  h  m  k  m    

This gives

h

kvi2  m  mg k  m     ln  ln 1  2k  mg k  vi2  2k  mg 

Now we consider the motion for downward motion. This can be rewritten as

 

1 d k 2 y 2  y   g 2 dy m

121

2 y

Again the solution of this equation is the sum of the solution 1 d k 2 2 2 equation 2 dy  y   m y  0 and a particular solution y

p

( y)

h

( y ) of

the homogeneous

. These solutions are

mg  2k  y 2 ( y )  A exp y  and y 2 ( y )  . Thus the complete solution is h p k  m  mg  2k  y 2 ( y )  A exp y  k  m 

Now the initial conditions are y ( y  h)  0 . This gives, with h 

kvi2  m    ln 1  2k  mg 

 kvi2  mg mg k    0  A 1   A mg  k kv 2  1 i mg This gives y 2 ( y )  

mg k  2k  mg exp y  2 k kvi  m  1 mg

If the final speed is vf, then v 2f  

vi2 mg k mg   k kv 2 kv 2 1 i 1 i mg mg

2 2 Note that of k=0 then v f  vi The answer can also be written as

1 1 k  2  2 mg v f vi

122

Chapter 8 8.1 Since there is no external force on the system in the horizontal direction, the total momentum in the horizontal direction is conserved. Initial momentum in the horizontal direction = momentum of the carriage + momentum of rain = Mv  0 = Mv Final momentum of the system after time t

=  M  mt  v f

Here vf is the final velocity. Equating the two moment gives vf 

Mv  M  mt 

8.2 Since the water leaking out of the carriage still has a horizontal velocity equal to the velocity of the carriage, total momentum of water after it came out for time t is = mtv If the initial amount of water in the carriage was m0, then the initial momentum of the system (carriage + water in it) =  M  m0  v If the aped of carriage (with left over water in it) after time t is vf, then by momentum conservation

 M  m0  mt  v f

 mtv   M  m0  v  v f  v

8.3 Exactly like in problem 8.2, there will be no change in the speeds of the two bicyclists. This is easily done by considering the momentum of the two friends before and after the books are given by one of them to the other person. Consider the person giving the books. Her momentum before transferring the books is Mv . After she gives the books, let her speed by vf. Then by momentum conservation Mv  mv   M  m  v f

 vf  v

Similarly, for the person receiving the books Mv  mv   M  m  v f

 vf  v

8.4 Conserve momentum after the first bullet has been fired. Initial momentum is 0. Let the velocity (since the motion is one dimensional, we write only the symbol for it, the direction is taken care of by the sign) of the gun after the bullet is fired be v1. Since the relative velocity of the bullet when it leaves the gun is u, and the bullet leaves the gun when the gun

123

is already moving with v, bullet’s speed ug with respect to the ground is calculated as follows: u  u g  v1

 u g  u  v1

Therefore momentum conservation gives

 M 0   N  1 mv1  m u  v1   0

 v1  

mu M 0  Nm

Now the momentum of the gun and  N  1 bullets in it is   M 0   N  1 m 

mu  M 0  Nm 

Now let the speed of the gun after the second bullet is fired be v2. Then momentum conservation gives

 M 0   N  2 mv2  m u  v2    M 0   N  1 m  v2  

mu  M 0  Nm 

mu mu  M 0  Nm M 0   N  1 m

Similarly one can now show that if the speed after the third bullet is fired is v3 then v3  

mu mu mu   M 0  Nm M 0   N  1 m M 0   N  2  m

Generalizing this we get after N bullets have been fired

v final 

k  N 1

 M k 0

0

mu  ( N  k )m 

8.5 (i) Momentum of the system = sum of the momentum of each particle = 0.2iˆ  0.4 ˆj kg ms1 (ii)

velocity of the centre of mass = total momentum/total mass



1 0.2iˆ  0.4 ˆj 0.3 2 4 ˆ  iˆ  j ms 1 3 3 

8.6 (i) acceleration of the CM

124



 Fnet  total mass iˆ  ˆj  0.3 10 ˆ ˆ  i j 3



(ii)



No, the acceleration is not in the same direction as the momentum of the CM.

8.7 If the base of the cylinder is in the xy plane, as shown in the figure, the x and y coordinates of the CM are (0, L/2). We thus have to calculate the z coordinate of the CM.

z y L x R

To obtain the z coordinate of the CM, consider a rectangular sheet of thickness dz at height z, as shown in the figure below.

z

R

If the density of the material that the cylinder is made of is , the z coordinate of the CM, by definition, is

125

R

z CM 

  2 L R 2  z 2 zdz 0

L R 2 2



4 R 2

R



R 2  z 2 zdz

0

To evaluate the integral, we substitute z  R sin  and dz  R cos d . This gives z CM 

4 R 2

 2

 R cos   R sin   R cosd 0

1



4R cos 2 d  cos    0



4R 3

8.8 (i) CM of a cone shown in the figure below

h

r z

R 2

The CM is on the axis of the cone by symmetry. To calculate its height, we take a thin disc of thickness dz at height z. By similarity of triangles, it radius r is given by r R  hz h

 r   h  z

R h

If the density of the material that the cone is made of is , then the position of the CM is given by h

z CM 

  z   r 2 dz 0

1 2 R h 3

h





3 R2 2 h  2  2 h  z 2  2hz zdz  4 R h0h

It is reasonable that the location of the CM is more towards the base since larger mass of the cone is concentrated there. 126

(ii)

Hemispherical bowl of radius R is shown in the figure below.

r R

z



The CM will be on the line passing through the centre of the base. To calculate its height zCM, we take a ring of height dz at height z. According to the figure z  R sin  and dz  R cos  d ,

z , R

sin  

cos  

r  R

R2  z2 R

If the mass per unit area for the shell is , then the mass dm of the ring is dm  2 r  R d  2 R 2  z 2 

dz  2 R  dz cos 

Thus zR

z CM 

 zdm

z 0 zR

2R 

zR

 zdz

z 0 2

2R 

 dm



R 2

z 0

N



8.9 Given N particles of masses mi (i=1-N) with total mass M   mi at positions ri (i=1-N), i 1

position of their CM

 RCM

is given as N

 RCM 

  mi ri i 1 N

m i 1

i

127

N





m r i 1

M

i i

Now let us make m subsystems of these masses with number of particles N1, N2, N3……Nm in Ni

them. Then we have the mass of each subsystem as M i   mi . The position of the CM can i 1

then be written as N1

N

 RCM 

  mi ri i 1 N

m i 1



 N2  m r  i i   mi ri  ....... i 1

i 1

M

i

By definition of the CM we have for the position

 RCMi of

Ni   each subsystem M i RCMi   mi ri . i 1

Thus the expression above can be written as N

 RCM 

  mi ri i 1 N

 mi i 1



m



M 1 RCM 1  M 2 RCM 2  ..... m

Mi



M R i

i 1

i 1

CM i

m

M i 1

i

This shows that the CM of the system can be calculated by treating each susbsystem as a point particle of mass Mi located at the CM of each subsystem. 8.10

To find the CM, we will treat the cone and the hemisphere as two subsystems. It is also clear by symmetry that the CM will be on the extended axis of the cone. Taking the axis as the z direction with z = 0 at the base of the cone, we have z CM 

mass of the sphere  z CM ( sphere)  mass of the cone  z CM (cone) total mass

Assuming the entire system is made of a material of uniform density, we get

z CM 



3R1 2R13 H R22 H    4 2 2 8 3 4 3   3R1  R2 H 2R13 R22 H 8R13  4 R22 H  3 3

8.11Since there is no external force on the system in the horizontal direction, the position of the CM will remain unchanged as the small block moves from one side to the other. Taking

128

horizontal direction to be the x-direction, let the position of the CM when the block in on the left be X1 and let it be X2 when the block is on the right. Then the poison of the CM of the block is (X1R) and (X2+R), respectively. Since the CM does not move, we have X CM  m X 1  R   MX 1  m X 2  R   MX 2

This immediately gives

 X 2  X1   

2mR m  M 

8.12 When the ball is compressed, it looks like shown in the picture below

R

R x

The radius r of the circular area of contact for xm, we have v1  3V If the balls are dropped from a height h then V



2 gh

If after the collision the smaller ball goes to height H then

3V 

2 gH

These two equations give H  9h 9.14 By momentum conservation we have

 m1  m2  v  m1v0 Totla initial energy = Final energy =

 v

m1v o m1  m2

1 m1v 02 2

m1 1 m12 v 02    initial energy  2 m1  m2 m1  m2

As is clear, the final energy is less than the initial energy. Therefore some energy is lost in the process. Now consider a pile of beads, each of mass m, connected with strings of length l between them. As the first bead falls over the edge, its initial speed is zero bu by the time the string connecting it to the second bead is taut, it gains a velocity of

v

2 gl

. However, as soon

as it pulls the second bead, by momentum conservation at that instant (neglecting the impulse due to gravity) the speed of two beads is v

m 2 gl  2m

gl 2

Now the two beads fall together and as the string connecting the second bead to the third bead becomes taut, the energy of hse two beads is 1 gl 5  2m   2mgl  mgl 2 2 2

This gives the speed of the two beads to be

147

5gl 2

Thus when the third bead is pulled over, momentum conservation at that instant gives the speed of the three beads to be =

2m 5 gl 2 3m



2 3

5 gl 2

One can go on like this and build up the solution up to p th bead falling by recognizing a pattern. We can also do the problem in the following way. Suppose when the (p-1)th bead has just fallen of the edge, the speed of the beads becomes v p 1 .

Then when these beads fall and the string between the (p-1) th and pth string becomes

taut, the energy of the system will be 1  ( p  1) m  v 2p 1  ( p  1) mgl 2

Thus the speed of these beads just before the pth bead falls off the edge is given by 1 1 ( p  1) mv 2   ( p  1) m  v 2p 1  ( p  1) mgl 2 2



v 2  v 2p 1  2 gl

By momentum conservation, therefore, we get the speed when the pth bead falls off as follows pmv p  ( p  1) mv



pv p  ( p  1) v 2p 1  2 gl



p 2 v 2p  ( p  1) 2 v 2p 1  ( p  1) 2 2 gl

We know that v1  0 i.e. when the first bead just falls off the edge of the table, its speed is zero. Now we can write ( p  1) 2 v 2p  ( p  2) 2 v 2p 1  ( p  2) 2 2 gl

And so on so that continuing in this manner we get ( v1  0 )





p 1

p 2 v 2p  v12  ( p  1) 2  ( p  2) 2  ( p  3) 2  ..........  1 2 gl  2 gl  i 2  i 1

( p  1) p( 2 p  1)  2 gl 6

This gives vp 

9.15

( p  1)(2 p  1) gl 3p

From problem 8.18, the total force required to move the chain is gh  v 2 . Thus the power delivered by the person pulling the chain is = ghv  v 3 On the other hand, the energy of the chain is = kinetic energy + potential energy If the length of the chain on the table is x then

148

dx  v and dt

Kinetic energy =

1 ( x  h)v 2 2

Thus Total energy E = This gives

potential energy =

1 gh 2  xgh 2

1 1 ( x  h)v 2  gh 2  xgh 2 2

dE 1 3  v  vgh dt 2

Thus the rate of change of the energy of the chain is not equal to the power delivered. As shown in the previous problem, the energy is lost in the inelastic collision. One may ask a question: what if instead of the chain, it s a rope or a strin that is being pulled? Where does the energy go in that case? The answer is that the energy difference is then used in stretching the rope to generate the required force. 9.16 Let the length of the chain be l. If its length y is hanging from the table, the force pulling it down is =

Mg y L

Taking the vertically down direction to be positive, the coordinate of the end of the hanging portion of the chain, when its length is y , is also y. Thus the equation of motion of the chain is M y 

Mg y l

or

y 

g y0 L

The equation can also be derived by considering the two protions, the one hanging from the table and the other on the table, separately. 1 dy 2 Writing y  we get from the equation above 2 dy

1 dy 2 Mg M  y0 2 dy l

v



1 Mg M  d y 2  2 0 l

y

 y' dy'  0

y1

This gives 1 1 Mg 2 Mv 2   y  y12   0 or 2 2 l

1 1 Mg 2 1 Mg 2 Mv 2  y  y1 2 2 l 2 l

The second term in the equation above is the change in the potential energy of the chain as the length of its hanging portion changes from y1 to y. Thus the total energy of the chain

149

remains unchanged as it slips off or the energy is conserved. This can also be seen as follows When length y1 is hanging, kinetic energy = 0

potential energy = 

1 Mg 2 y1 2 l

Total energy = 

1 Mg 2 y1 2 l

When length y is hanging, kinetic energy =

1 Mv 2 2

potential energy = 

Total energy =

1 Mg 2 y 2 l

1 1 Mg 2 Mv 2  y 2 2 l

It is clear from the above that the total energy remains unchanged. 9.17 As the chain unfolds, the mass of the hanging (and moving) portion keeps on changing. Thus the problem is like the variable mass problem. We take the mass per unit length of the chain to be λ. Taking the vertically down direction to be positive, the coordinate of the end of the hanging portion of the chain, when its length is y , is also y. The external force on the chain is yg ,

dm  y , and the relative velocity of the dt

mass that is added to it is  y . Thus the equation of motion my 

dm u rel  Fext of dt

the chain is yy  y 2  yg

OR

yy   y 2  yg

1 dy 2 Using y  we write this equation as 2 dy

1 dy 2 y 2  g 2 dy y

A

2 We now solve this equation for y 2 as the sum of the solution y  y 2 of the homogeneous

part and the particular solution

2 yg to get 3 y 2 

A 2  yg y2 3

150

Here A is a constant. Using the condition that y 2 ( y  y1 )  0 , we get A  

2 3 y1 g and 3

therefore

2 y 3  y13  y  g 3y 2 2

The same result can be obtained by letting p   and l  0 such that pl  y in the solution for a coiled set of beads in problem 9.14. Now let us compare the energy of the chain at the beginning o fthe motion and agter it has slipped so that y of its length is hanging. At the beginning, when length y1 is hanging, kinetic energy = 0

potential energy =  

1 gy12 2

Total energy =

1 gy12 2

When length y is hanging, Kinetic energy = Potential energy = 

 y 3  y13  g  1 y 2 g  1 y13 g 1 yy 2   2 3y 3 3 y

1 gy 2 2

Total energy = 

 2 y1  1 y13 1 1   Initial total energy  gy 2  Initial total energy g  gy 2   3 y 6 6  3y 

Thus we see that the energy is lost during the motion. This happens due to the inelastic collision between the chain links as each new link is pulled by the moving chain (see problem 9.14).

151

Chapter 10 10.1 (a) The fields are shown on the axses. They are the same all over the place. y 

(i) F ( x, y )  y iˆ x

y 

(ii) F ( x, y )  x ˆj x

(b) If we think of the force arrows as velocity of a fluid, we see that a matchstick thrown in that fluid will tend to rotate clockwise in (i) and counterclockwise in (ii). Thus the curl of both the fields is not zero; it is negatve for (i) and positive for (ii). (c)

curlf of force field (i) iˆ     F  x y

ˆj  y 0

kˆ    kˆ z 0

Curl of force field (ii) iˆ     F  x 0

152

ˆj  y x

kˆ   kˆ z 0

10.2

Y (2, 1)

O

X



(i) Force field F ( x, y )  y iˆ . Path 1: Work done is zero as the particle moves along the x-axis. Similarly as it moves from (2,0) to (2,1), work done is again zero since the particle is moving perpendicular to the force field. Thus alongh path 1, the net work is zero. Path 2: Work done is zero as the particle moves along the y axis. However, as it moves from (0,1) to (2,1), it is under a constant force of 1 unit in the x direction. So the work done by the field is 2 units. Path 3: Along path 3 y 

 x x . Thus F ( x, y )  iˆ . The displacement along path 3 is given as 2 2

 1 dl  dxiˆ  dy ˆj  dxiˆ  dx ˆj . Thus 2

  12 dW  F .dl   xdx  1 unit 20

(ii)



Force field F ( x, y )  x ˆj

Path 1: Work done is zero as the particle moves along the x-axis since the particle is moving perpendicular to the force field. Similarly as it moves from (2,0) to (2,1), Field is constant and

153

equal to 2 units. Thus the work done by the field will be 2 units. Thus alongh path 2, the net work is 2 units. Path 2: Work done is zero as the particle moves along the y axis because the field is zero. As it moves from (0,1) to (2,1), it is is moving perpendicular to the force field so the work done is zero again. So the net work done by the field is zero. Path 3: Along path 3 y 

 x . Thus F ( x, y )  x ˆj . The displacement along path 3 is given as 2

 1 dl  dxiˆ  dyˆj  dxiˆ  dx ˆj . Thus 2

W 

  12 F  .dl  2 0 xdx  1 unit

In both the cases the work done depends on the path. It is therefore consistent with the curl of both the fields not being zero. r 10.3 Force field is F  2 y iˆ  3 x ˆj . The paths are shown below

Y

(1, 1) Path2 path1

O

X

We will calculate the work done along path 1 in two parts: first when the particle mones along the x-axis and secondly when it is moving from (1,0) to (1,1). Along the x-axis, the force is

 F  3 x ˆj

and the displacement is dxiˆ . Thus the work done is zero when the particle

moves along the x-axis.  F  y iˆ  3 ˆj

When the particle moves from (1,0) to (1,1), the force is

and displacement is

dy ˆj .

Thus the total work done is 1

W   3dy  3 units 0

154

This then is the total work done along path 1. Along path 2 y = x so that dy = dx. The work done is given by the integral 1 1 1   1 5 W   F .dl   2 ydx   3 xdy   2 xdx   3 ydy  units 2 0 0 0 0

Since the work done along two paths is different, the force field is NOT conservative. 10.4 The figure for the path si shown below. The way we have set up the transformations Y C

(2,1)

A



(0,1)

B X

x  (1  cos  ) and y  (1  sin  ) , the corresponding angle is as shown in the figure and

varies from π to 2π. Thus the integral W ACB 

F

x

dx 

ACB )

F

y

dy

ACB

from A to B over the

semicircular path is W ACB  A

2  x ydx  A

ACB ) 2



ACB

x3 dy 3

  A  1  cos   1  sin   sin  d  2



A 3

2

 1  cos  

2

cos  d

The integral can be carried out easily and gives W ACB 

8A 3



10.5 F   k1 y iˆ  k 2 x ˆj . The force field will be conservative if its curl vanishes. The curls of it is    F 

iˆ  x  k1 y

ˆj  y  k2 x

If the curl has to vanish, we should have k1  k 2 . 155

kˆ    k 2  k1 z 0

For the force field

 F   y iˆ  x ˆj

it is clear by inspection that the potential should be

U ( x, y )  xy . However, to derive it formally we put 

U ˆ U ˆ i j   y iˆ  x ˆj x y

This gives U  y x

Which is integrated to U ( x, y )  xy  f ( y ) U

Differentiating this with respect to y and writing y  x gives df  0 OR dy

f  const.

By the condition U (0,0)  0 , the const. is zero. Thus U ( x, y )  xy 10.6 (i) Curl of the field iˆ     F  x 3A

ˆj  y Az

kˆ    Aiˆ z 0

is not zero. Therefore the field is not conservative. (ii) The curl of the field    F 

iˆ  x Axyz

ˆj  y Axyz

kˆ    A x y  z  iˆ  y  z  x  ˆj  z  x  y  kˆ z Axyz





is not zero. Therefore the field is not conservative.

10.7 (i)



The function f ( x, y )  2 exp   

( x  2) 2  ( y  1) 2   will be a constant where its 4 

argument ( x  2) 2  ( y  1) 2 is a constant. Thus the contours are given by ( x  2) 2  ( y  1) 2  const.

These are concentric circles with centre at  2,  1 . (ii)

The gradient of the function is given by 156

  x  2  2   y  1 2    x  2  2   y  1 2   ˆ     i  2 exp    j   exp     y  4 4  x    

 x  2 iˆ   y  1 ˆj 

10.8 Kinetic energy of the system of N particles with masses mi  i  1 N  and positions  ri  i  1 N  is

 1 N mi ri 2  2 i 1 

Let the position of the CM be R so that

   V  R.

If the positions and velocities of these

  particles with respect to the CM are given respectively by ric  i  1 N  and ric  i  1 N 

then    r  R  ri

and

      r  R  ric  V  ric

Substituting this in the equation for kinetic energy above, we get  2 1 N  2 N   1 N 1 N 2 m r  m V  m r  m r  i i 2  i ic  i i ic  V 2 i 1 2 i 1 i 1 i 1 M

Denoting the total mass

 mi as M and using the property of the CM that i 1

N



m r i 1

i ic

 0,

we get  2 1  1 N 1 N 2 mi ri  MV   mi ric2  2 i 1 2 2 i 1 This is the desired result. 10.9 When some energy is released during collision, the equations during the collision are     p1  p 2  p1  p 2 (momentum conservation)     p12 p 22 p1'2 p 2'2   E   (energy conservation) 2m1 2m 2 2m1 2m 2 Here the unprimed quatities are before collision and the primed quantities are after collision. Further E  0 . The easiest way to prove that in the bove situation, the paricles cannot move stuck together is to write all quatities in the with respect to the CM. In that case (referring to quantities in the CM frame with subscript c)       p1c  p 2c  p1c  p 2 c  0  p1c   p 2 c and

157

  p1c   p 2 c

And using the splitting of the kinetic energy as in the problem above and the fact that     p1c   p 2c and p1c   p 2c     p12c p 22c p1' 2c p 2' 2c   E   2m1 2m2 2m1 2m2



 p12c 2

  1 p1' 2c  1 1  1     E      2  m1 m2   m1 m2 

Now if the particles are moving together, their momenta after the collision with respect to the CM will be zero. This is also seen mathematically as follows. By momentum conservation   p1c   p 2 c And if they are moving together, then   p1c  p 2 c

  The above two equation give p1c  p 2c  0

Putting this in the energy conservation equation gives   p12c  1 p12c  1 1  1     E  0  E       2  m1 m2  2  m1 m2  However that is not possible because two positive quantities cannot add up to zero. Thus the particles cannot move stuck together. 10.10

The picture of the carom coin and the striker is shown below Y

Direction of impulse on coin

2.05cm

1.55cm

2ms-1

2cm X

The coin will move in the direction of the impulse that it receives from the striker. This direction is perpendicular to their common surfaces and theefore along the line joining their centres. Thus the coin moves at an angle  such that

158

sin  

2 5  (2.05  1.55) 9



cos  

56 9

and   33.75

We will work in cgs units. Initial momentum of the system is that equal to the initial momentum of the striker which is

 p  3000iˆ

. 

If after the coliision, the velocity of the striker is V  V x iˆ  V y ˆj and the speed of the coin is  v 56 ˆ 5v ˆ v then its velocity would be v  v cos  iˆ  v sin  ˆj  i j and the momentum 9

9

conservation will lead to v 56  15V x  3000 9 5v 5  15V y  0 9 5

Since the collision is elastic, the total energy is conserved and we get 1 1 1  5  v 2   15  V x2  V y2    15  200 2 2 2 2

These equations simplify to v 56  3V x  600 9 5v  27V y  0 v 2  3V x2  3V y2  120000

These are 3 equations for 3 unknowns so all the answers can be obtained from these equations. Substituting V x  200 

56 v 27

and V y  

5v 27

In the energy conservation equation gives  4v 2993    0 9   3

v

This gives v  249.4cms 1  2.5ms 1 after collision. The trivial answer v  0 of course refers to the situation before collision. This then leads to V x  1.30ms 1 and Vy  0.46ms 1 . This gives the angle of deflection  0.46    19.5 below the x-axis.  1.30 

1 of the striker to be tan 

159

(ii) KE o fthe CM remains unchanged during the collision. It is therefore 3000 2  10  7  0.0225 J . 2 5  15

(iii)

Velocity of the CM =

15  2  1.5 ms 1 15  5

Therefore the striker is moving with velocity 0.5 ms1 and the coin with -1.5 ms1 along the xaxis in the CM frame.

The magnitude of momentum of each body is 0.0075 kgms1 in

directions opposite to each other (see figure). In the CM frame the magnitude of velicities does not change during an elastic collision. Thus the momentum and the velocity vectors of the striker and the coin just rotate. Further, the impulse J is in the direction (see figure) such that it makes and angle 

2    33.75  3.6 

  sin 1 

from the x-axis. As is clear from the figure, the magnitude of J must be such that the final momentum has the same magnitude as the initial momentum. This is seaily done by drawing a circle of radius equal to the magnitude of the momentum with its centre at the tail of the initial momentum vector. We then draw the impulse vector from the head of the initial momentum vector at the angle  and take its length such that its head touches the circle. The vector from the centre to this point gives the final momentum. immediately shows that    p f  pi  J

It is also clear from the figure that  J  2  0.0075  cos 33.75  0.0125kgms 1

It is also clear from the figure that the angular momenta vector srotate by  CM  180  2  33.75  112 .5

160

The construction

J 0.0075 kgms1

33.75 0.0075 kgms1

 CM

As a check we now calculate the velocities of the striker and the coin in the lab frame from the information above and compare our answer. As is clear from the figure, the x component of the striker’s velocity in the CM frame is V xCM  0.5 cos 112 .5  0.187ms 1

This then gives Vxlab as follows (keep in mind that the CM is moving in x direction only) V xCM  V xlab  VCM

 V xlab  VCM  V xCM  1.5  0.187  1.313ms 1

And we get from the figure (the CM is moving in x direction only) V ylab  V yCM  0.5 sin 112 .5  0.462ms 1

Thus V striker ,lab 

2 2 V xlab  V ylab  1.39ms 1

Similarly for the coin (see figure) V xCM  1.5 cos180  112 .5  0.574ms 1

This gives V xlab  VCM  V xCM  1.5  0.574  2.074ms 1

and we get from the figure V ylab  V yCM  1.5 sin 180  112 .5  1.386ms 1

This gives Vcoin ,lab 

2 2 V xlab  V ylab  2.49ms 1

These match with the previously obtained answers. 161

10.11 As in the problem above, the stationary balls will move in the direction of impulse after the collision.

This is going to be along the line joining the centre of the striking ball

and the centres of the stationary balls. At the time of colliding, the three balls will look as shown below. Direction of impulse on ball 30

Direction of impulse on ball Since the net impulse (due to the two stationary ball) will be along the line of its initial motion by symmetry (see figure), it will continue to move along its original direction. This is also seen by noticing that the statinary balls have a net momentum in the direction of v0. Thus there will be no component of the momentum to be balanced in the direction perpendicular to that.

Thus the striking ball will continue to move along its original

direction. Let the speed of the striking call be v1 after the collision and the speeds (equal by symmetry) of the other balls be v. Then by momentum conservation mv0  mv1  2  m  v

3 2

 v 0  v1  v 3

Since the collision is elastic, we have by energy conservation 1 2 1 2 1 mv0  mv1  2  m  v 2 2 2 2

 v 02  v12  2v 2

There are two unknowns v1 and v and two equations so we can get their values. Squaring the momentum conservation equation and subtracting it from the energy conservation equation gives





v v  2v1 3  0

One of its solutions is the trivial solution v  0 referring to the situation before collision. Theother solution is

162

v  2v1 3

Substituting this in the the momentum conservation equation gives v1  

v0 5

and v 

163

2v 0 3 5

Chapter 11 11.1 Disc rotating about a point on its periphery. Consider first a rotation about a point on the periphery by an angle. This is shown on the left in the figure below. The dashed circle shows the initial position of the disc while the solid circle shows the position after rotation.

 

On the right we show the same rotation carried out by translating the CM of the disc first to its new position, as shown by the straight arrow and then carrying out the rotation about the CM. We have made the disc in this position by dotted circle. As is evident from the figure, the magnitude and the sense of rotation is the same as in the figure on the left. 11.2 Now we generalize the result of the problem above. On the right in the figure below, we show the same rotation as in the figure on the left carried out by translating the opposite end of the diameter of the disc first to its new position, as shown by the straight arrow and then carrying out the rotation about the this point. We have made the disc in this intermediate position by dotted circle. As is evident from the figure, the magnitude and the sense of rotation is the same as in the figure on the left. Similarly, it can be shown about any other point.

164

 



11.3 The angular momentum L about the origin is given by

   L  mr  v .

Thus the answers

in different cases are 

(i) L  2iˆ  2iˆ  0

(ii)

(iii) L  2 ˆj   2iˆ  ˆj   4kˆ 

 L  2 ˆj  2iˆ  4kˆ

(iv) L   iˆ  2 ˆj    2iˆ  ˆj   kˆ  4kˆ  3kˆ 

11.4 The position of the wheel and the stone stuck to its periphery is shown in the figure below.

Y

Vt x(t) y(t)

t X

165







Thus we have for the position r (t ) , velocity v (t ) (obtained by differentiating r (t ) with 



respect to time once) and acceleration a (t ) (obtained by differentiating r (t ) with respect to time twice) of the stone  r (t )  Vt  R sin t  iˆ  R cos t ˆj

 v (t )  V  R cos t  iˆ  R sin t ˆj

 a (t )   2 R sin t iˆ   2 R cos t ˆj

As is evident, the acceleration is towards the centre of the wheel. It is the centripetal acceleration.

The force for this is provided by the force that keeps the stone stuck to the

wheel. This force is   F  ma (t )  m 2 R sin t iˆ  m 2 R cos t ˆj

The angular momentum

 L (t ) of

the stone with respect to the origin is



 

   L (t )  mr (t )  v (t )  m  Vt  R sin t  iˆ  R cos t ˆj   V  R cos t  iˆ  R sin t ˆj  mVRt sin t  R 2  VR cos t  kˆ



This gives  dL  mV 2 Rt cos tkˆ dt

The angular momentum changes due to the torque provided by the force that keeps the stone stuck to the wheel and is equal to the centripetal force. Thus the torque 

 m

 Vt  R sin t  iˆ  R cos t ˆj   

2

R sin t iˆ   2 R cos t ˆj



 mV Rt cos t kˆ 2

Thus it is equal to the rate of change of the angular momentum. 11.5 We assume that at time t = 0, the paricle is on the x axis. Thus after time t, its position vector is give as (see figure)  r (t )   R  R cos t  iˆ  R sin t ˆj

166

Y

R

t

O

X



Thus its velocity and acceleration are (obtained by differentiating r (t ) with respect to time once for the velocity and twice for the acceleration)  a (t )   2 R cos t iˆ   2 R sin t ˆj

 v (t )  R sin t iˆ  R cos t ˆj

As expected, the acceleration is the centriprtal acceleration. It is provided by the external force towards the centre. The angular momentum of particle with respect to the origin is



 

   L  mr (t )  v (t )  m  R  R cos t  iˆ  R sin t ˆj   R sin t iˆ  R cos t ˆj  mR 2 1  cos t  kˆ



Its time derivative is  dL   m 2 R 2 sin tkˆ dt

The torque due to the external force is



 

     mr (t )  a (t )  m  R  R cos t  iˆ  R sin t ˆj    2 R cos t iˆ   2 R sin t ˆj



  m 2 R 2 sin t kˆ

This is the same as the rate of change of angular momentum. 11.6 The angular momentum for a collection of particles about an origin O is    LO   mi ri  vi i



Now let is choose a different origin O’ such that the position vector of O’ from O is R (see figure) so that    ri  R  ri '





and vi  vi'

167

Z’ Z

 ri '

 ri

 R

O

O’

Y

Y’

X’

X Substituting the expressions above in the formula for



 LO



, we get

          LO   mi ri  v i  mi R  ri '  v i  R   mi vi   mi ri '  vi' i

i

i

i

 ' where we have used the fact that vi  vi .

Now the angular momentum about O’ is  LO ' 

'

m r

i i

  vi'

Thus if the total momentum of the system is zero, then 

'

m v  m v i

i

i

i

i

i

 0 and L  L O O'

11.7 At the maximum distance Rmax and the minimum distance Rmin from the sun, the velocity vector is perpendicular to the radius vector. If the speed of the earth at these distances is Vmax and Vmin, respectively, then the angular momentum of the earth at these two points is MearthRmaxVmax and MearthRminVmin , respectively. By the conservation of angular momrntum we have MearthRmaxVmax = MearthRminVmin This gives Vmax R 1.47  min   0.967 Vmin Rmax 1.52

168





11.8 Initial momentum p i , the final mpmentum p f and the change in the momentum    p  p f  pi are shown in the figure below.

 p

 pf

   

1 

2

Y

    X

 pi 

As is evident from the figure, the magnitude of p is 2 p sin

 where 2

  p  pi  p f

and

it is in the direction bisecting the angle      between the incident and the scattered direction.

Mathematically it can be seen as follows. Choose the direction of incoming

particle to be the x direction. Then  pi  piˆ and

 p f  p cos iˆ  p sin ˆj

 p  p cos   1 iˆ  p sin ˆj     2 p sin 2 iˆ  2 p sin cos ˆj 2 2 2      2 p sin   sin iˆ  cos ˆj  2 2 2  

Thus the magnitude of p is 2 p sin

    and it is in the direction   sin iˆ  cos 2 2 2 

ˆj  . 

(ii) Since the paticle is moving parallel to the x axis at a distance d, its angular momentum is mvd going into the plane of the paper. This can be seen as follows.

The position and

velocity vectors of the particle are  r  xiˆ  dˆj



and v  viˆ

This gives for the angular momentum    L  mr  v   mvd kˆ

Further, since the force is central, the angular momentum about the origin is a constant.

169

(iv)

  dp   F , we have p  Since dt





 Fdt . Therefore  Fdt is in the same direction as

 p .

To calculate



 Fdt , we change the integration over time to integration over the angle by

using the angular momentum conservation. Using polar coordinates, we write the angular 

2 momentum of a particle moving in the xy plane as L  mr

problem, it is a constant equal to

 mvd kˆ

mr 2

d ˆ k . Since in the present dt

, we have

d   mvd dt

 dt  

r2 d vd



The force F between the two charges is in the radial direction and is equal to  Qq F  k 2 cos  iˆ  sin  ˆj r





Thus  2  Qq ˆ  sin  ˆj   r d F dt  k cos  i   r2 vd 



k









Qq cos  iˆ  sin  ˆj d vd 





Qq  sin iˆ  1  cos   ˆj vd Qq    k  2 cos   sin iˆ  cos vd 2 2 2 k

Thus the magnitude of



 Fdt is 2k

ˆj  

Qq  cos . Comparison of this expression with that for vd 2

the momentum change shows that the two are in the same direction and 2 p sin

 Qq   2k cos 2 vd 2

Or d k

Qq  Qq  cot  k cot 2 pv 2 2 mv

This is the Rutherford formula.

170

11.9 (i) At angle  1 the potential energy of the girl is MgL1  cos 1  

1 MgL12 with 2

respect to the equilibrium point, i.e. when the swing is in the vertical position. The kinetic energy of the system at the lowest point , if the angular speed is 1 , is 1 ML212 . By energy conservation 2 1 1 ML212  MgL12 2 2

(ii)

 1  1

g L

Whe the child stands up at the lowest point, all the external forces on the her are passing through the pivot so her angular momentum remains unchanged. However 2 as she stands up her moment of inertia about the pivot is M  L  d   I CM where ICM

is her moment ofinertia about her CM. Neglecting it gives the moment of inertia to be M  L  d  2 . Then by conservation of angular momentum M  L  d   2  ML21 2

keeping only linear terms in (iii)

 2 

L2 2d   1   1   1 2 L  L  d 

d since d