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• Rhys Lim YH

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1. The two small 0.2-kg sliders are connected by a light rigid bar and are constrained to

move without friction in the circular slot. The force P=12 N is constant in magnitude and direction and is applied to the moving slider A. The system starts from rest in the position shown. Determine the speed of slider A as it passes the initial position of slider B if (a) the circular track lies in a horizontal plane and if (b) the circular track lies in a vertical plane. The value of R is 0.8 m. a) In horizontal plane

U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1 2 12 cos 300.8  12 sin 300.8  2 1 m v  2

 2 0.2 

v2  8.1 m / s b) In vertical plane Reference line for part (b)

Nothing changes.

2. Determine the constant force P required to cause the 0.5 kg slider to have a speed v2 = 0.8 m/s at position 2. The slider starts from rest at position 1 and the unstretched length of the spring of modulus k = 250 N/m is 200 mm. Neglect friction.

m=0.5 kg v2 = 0.8 m/s rest at position 1 and k = 250 N/m lo=200 mm

Length of cable

l1

l2 2

1

1

l1  0.4 2  0.252  0.472 m

2

l2  0.2 2  0.252  0.32 m

h2=0.2sin15=0.052 m

Reference line

Work by the cable

U1 2  Pl  P0.472  0.32  0.152 P U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1 1 2 1 1 2 2 mv2  mgh2  k x2   k x1  2 2 2 1 1 1 2 2 2 0.152 P  0.50.8  0.59.810.052  2500.45  0.2   2500.25  0.2  2 2 2

0.152 P 

P=52.07 N

3. The 2 kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of kinetic friction is 0.4. calculate (a) the velocity v of the collar as it strikes the spring and (b) the maximum deflection x of the spring.

FBD of Collar y

mg x Motion

F

N

y

0

N  9.81 N

Ff=mkN



N  mg cos 60  0



F f  m k N  0.39.81  3.924 N

(a) the velocity v of the collar as it strikes the spring

1

F f  3.924 N

U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1  3.9240.5 

1 2v22  29.810.5 sin 60 2 v2  2.556 m / s

2

Reference

(b) the maximum deflection x of the spring.

x

3

U 2 3  T3  T2  Vg 3  Vg 2  Ve3  Ve 2  3.924 x   

1 22.5562  29.81 x sin 60  1 1600x 2 2 2 800 x 2  13.066 x  6.533  0

x1,2

13.066  13.066 2  4800 6.533  2800



0.0989  xmax  0.0989 m  0.08

4. The 1.2 kg slider is released from rest in position A and slides without friction along the vertical-plane guide shown. Determine (a) the speed vB of the slider as it passes position B and (b) the maximum deflection d of the spring.

m=1.2 kg determine vB and the maximum deflection d of the spring.

a)

1

U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1 3

d 2

T2  Vg1  0



v2  9.396 m / s Datum

b) U13  T3  T1  Vg 3  Vg1  Ve 3  Ve1 1 mg 1.5  mg 4.5  kd 2  0 2 1.29.811.5  1.29.814.5  1 24000d 2  0 2 d  0.0542 m  54.2 mm

Vg 3  Vg1  Ve3  0



1 2 mv2  mg 4.5  0 2

5. The light rod is pivoted at O and carries the 2- and 4-kg particles. If the rod is released from rest at q =60o and swings in the vertical plane, calculate (a) the velocity v of the 2 kg particle just before it hits the spring in the dashed position and (b) the maximum compression x of the spring. Assume that x is small so that the position of the rod when the spring is compressed is essentially horizontal.

released from rest at q =60o (a) the velocity v of the 2 kg particle just before it hits the spring in the dashed position and

(b) the maximum compression x of the spring.

1

3 (maximum compression)

2

A 

Reference

B

U1 2  0



T1  Vg1  Ve1  T2  Vg 2  Ve 2

a) T1  Vg1  Ve1  T2  Vg 2  Ve 2

49.810.3 sin 60  29.81 0.45 sin 60  1 4v A2  1 2vB2 v A  rA  0.3

vB  rB  0.45

2 

2   2.58 rad / s

b) T1  Vg1  Ve1  T3  Vg 3  Ve3

49.810.3 sin 60  29.81 0.45 sin 60  1 xmax  0.01207 m  12.07 mm

2 k xmax

2 35000

 vB  1.16 m / s

6. Two springs, each of stiffness k=1.2 kN/m, are of equal length and undeformed when q =0. If the mechanism is released from rest in the position q =20o, determine its angular velocity 𝜃ሶ when q =0. The mass m of each sphere is 3 kg. Treat the spheres as particles and neglect the masses of the light rods and springs.

k=1.2 kN/m, are of equal length and undeformed when q =0. mechanism is released from rest when q =20o, determine 𝜽ሶ when q =0. m =3 kg.

1

l1

b

l2

l1  20.25 sin 35  0.287 m

d1  0.25 2  0.287  0.0668 m

Reference line

a

l2  20.25 sin 55  0.41 m

d1  0.41  0.25 2  0.056 m U1 2  0

c



T1  Vg1  Ve1  T2  Vg 2  Ve 2

We ignore the equal and opposite potential energy changes for masses (a) and (b).

2

Ref.

1 1 E1  mg 0.25 cos 20  kd12  kd12 2 2 1 1 2 2 E1  39.810.25 cos 20  12000.0668  12000.056 2 2

1  2  E2  3 3 q  0.25   mg 0.25   2 2  v   E E  q  4.22 rad / s



1

2



7. The two right-angle rods with attached spheres are released from rest in the position q = 0. If the system is observed momentarily come to rest when q = 45°, determine the spring constant k. The spring is unstretched when q =0. Treat the spheres as particles and neglect friction.

released from rest when q = 0. System is momentarily stationary when q = 45°, determine the spring constant k. Spring is unstretched when q =0.

1

 180  o   71.56  60 

a  a tan 180 mm

a

U1 2  0



T1  Vg1  Ve1  T2  Vg 2  Ve 2

Reference

60 mm Deformation of spring in Case 2:

2

x2  20.1897 cos 26.56  0.06  0.219 m

71.5645=26.56o a

45o

From conservation of energy:

1 2 22 9.810.18  22 9.810.1897 sin 26.56  k 0.219 2 k=155.7 N/m

8. The light quarter-circular rod is pivoted at O and carries the 3 kg particle. When the system is released from rest at the position (1), it moves to position (2) under the action of the constant force F=250 N applied to the cable. The spring of stiffness k=1500 N/m has an unstretched length of 200 mm. Calculate the speed of the particle and the angular velocity of the circular rod as the particle passes the position (2).

9. The 0.6-kg slider is released from rest at A and slides down the smooth parabolic guide (which lies in a vertical plane) under the influence of its own weight and of the spring of constant 120 N/m. Determine the speed of the slider as it passes point B and the corresponding normal force exerted on it by the guide. The unstretched length of the spring is 200 mm.

Work and Energy Principle 1

2

Datum

U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1 T1  Vg1  Ve1  T2  Vg 2  Ve 2





2 1 2 2 0.69.810.5  120 0.25  0.5  0.2  1 0.6v22  1 1200.25  0.22 2 2 2

v2  5.92 m / s

Newton’s Second Law (Normal&Tangential Coordinates) FBD of slider (at point B) +n

Fspring

mg +t N

Equation of parabolic guide : y  kx2

y  2x

2



0.5  k 0.52



  dy  2  1       dx    d2y

dy d2y  4x  0 , 4 2 dx x  0 dx

k 2

3/ 2

 1  0 3 / 2  4

 0.25 m

dx 2

 Fn  man N  84 N



N  Fspring  mg  m

v2





5.92 2 N  1200.25  0.2   0.69.81  0.6 0.25

10. Calculate the horizontal velocity v with which the 20 kg carriage must strike the spring in order to compress it a maximum of 100 mm. The spring is known as a “hardening” spring, since its stiffness increases with deflection as shown in the accompanying graph. 1

2

v1

U1 2  T2  T1  Vg 2  Vg1  Ve 2  Ve1

 T1  Ve 2  0



1  20v12  2

 10 x  20 x 1000dx  0

0.1

2

0

0.1

20 3    10v12  10005 x 2  x  0 3 0 



v1  2.38 m / s