• Author / Uploaded
• Eric Campbell

• Categories
• Dinámica (Mecánica)
• Cinemática
• Filosofía natural
• Mecánica
• Geometría

Citation preview

EGR 232 Engineering Dynamics Lecture 32: Constrained Plane Motion

Fall 2012

Today: Homework Questions Read Chap 16 Section 8 Homework: Chap. 16, Problems 78, 89, 97, 109 Lecture Outcomes: Following today's lecture you should be able to -- apply Newton's 2nd Law to rigid bodies -- be able to set up problems where the rotation is about a noncentroidal point -- be able to set up problems for rolling motion. Constrained Plane Motion: Constrained plane motion refers to the fact that many times, locations on a rigid body will be constrained to move in a prescribed fashion. Where a pin moves along a slot or a body rotates about a pin at one end, or wheel must roll along the curve of the road, certain points on the body will have locations required of them. The dynamics of these constrained bodies may be found by applying Newton's 2nd Laws as in previous problems, but these constraints may have some effect on how the accelerations (both linear and angular) are related to each other. Common types of Constrained Plane Motion: Noncentroidal Rotation: Rolling Motion of Wheel: ( centroid at center of wheel) F 1

aGt

G O

F2

Active Constraint equations:

aGt  rG / O

aGn  rG / O 2

α aGn

F

G

O C

α aCn

Active Constraint Equations:

aG  rG / C

C

aG

General procedural steps for solving these types of problems: 1) Find the kinematic relationships: a) Locate the relative positions of contact points and centroids. b) Identify effect of any constraints (such as pulley problems) c) Determine the angular velocities of each rigid body. d) Determine the relationships between the angular acceleration and the centroidal acceleration of the rigid bodies. 2) Apply Newton's 2nd Law to each rigid body: a) Draw the FBD and Acceleration Diagram of each rigid body b) Write out the equations to match the diagrams

Fy  m aGy Fx  m aGx r r r r M P  I G   r  maG 3) Apply friction equations if appropriate a) for impending motion:

Ff   s N

b) for friction below the point of impending motion:

Ff   s N

c) for sliding:

Ff  k N

4) Compare the number of unknowns to number of independent equations. If equal, solve for the unknown quantities.

Example 7: An 80 kg disk is released at rest from the position shown. Determine the horizontal and vertical support reactions at A.

G A

Given: m = 80 kg Radius: r = 1.5 m Mass moment of inertia of a disk about its centroid:

I

r = 1.5 mm

1 mR2 2

Velocity Analysis: Released from rest

v A  vG  0

 0

and

Acceleration analysis:

r r r r r  aA  0 aG  a A    rG / A   2 rG / A r aG  0   kˆ  riˆ  0  0 ˆi   r ˆj

Force/Moment analysis:

FBD

Fx  m aGx

AD

FAy

aGy

FAx  maGx FAx  0

FAx A

Fy  m aGy

G

r = 1.5 m

G W

FAy  W  maGy FAy  W  maGy  mg  mr  m( g  r )

r M G  I

 FAy r  I



2 1 I mr 2 mr FAy         r r 2

therefore:

mg  mr  

mr  2



 

2g 3r

then the reaction force may be found as

FAy  

mr mr  2 g 1        mg 2 2  3 r 3

so:

FAx  0

1 1 F  mg  ( 80kg )( 9.81m / s2 )  262 N Ay and 3 3

α

aGx

Example 8: A long slender bar of mass, m, and length, L, is inclined at an angle of β with respect to the flat horizontal plane. The bar is not to slide as it is released from rest, determine a) the support reactions N and Ff. b) the required coefficient of friction. c) the acceleration of the centroid G.

m L G β A

Given: no slipping Length, L Mass, m Angle, β Mass moment of inertial for long slender rod: I 

1 mL2 12

r L L rG / A  cos  ˆi  sin  ˆj 2 2

Geometry

Velocity analysis: Body released from rest: vA = vG =0 Acceleration analysis:

aA  0



r r r r r aG  a A    rG / A   2 rG / A

r L  L aG  0   kˆ   cos  ˆi  sin  2  2 L L   cos  ˆj   sin  ˆi 2 2

ˆj  0  

I

L

aGy

Force/Moment Analysis:

G

Fx  m aGx 

L  Ff  m    sin   2 

G

Ff

β

β

A

N

 L   cos   2 



 L   L  cos   Ff   sin   2   2 

 I

M G  I N 

α

W

Fy  m aGy N W  m 

m

N  mg 

1 mL  cos  2

aGx

therefore:

1 1   L      mg  mL  cos   cos    mL  sin   2 2   2   

L   sin 2 

Solving for α

1 1 1  mgL cos   mL2 sin2    mL2 cos2    I 2 4 4 1 1 1  mgL cos   mL2  mL2 sin2   cos2   2 12 4 1  1 1  mgL cos      mL2 2  12 4













1  mgL cos  3 cos  g  2  1 2 2 L mL 3

so:

N  mg 

1  3 cos  mL cos    2 2 

g  L

 3 cos2  3 cos2  mg   1   mg 4 4   L 3 cos   g   sin     2 2 L 

N  mg  

Ff  m  

Ff 

s 

Ff N

3 cos  sin  mg 4

 





3 cos  sin  3 cos  sin  mg 3 cos  sin  4 4   4 3 cos2  4  3 cos2  3 cos2  1  mg 4  4 4 

L 3 cos  r aG    2 2

g 3 cos   g L  cos  ˆj     sin  ˆi L 2 2 L  

3 cos  sin  ˆ 3 cos2  ˆ r aG  g i g j 4 4

 I

Example 9: A force P of 18 N is applied to the cord which wraps around the central hub of the wheel shown. Hub radius is 80 mm and wheel radius is 160 mm. The mass of the wheel and hub is 5 kg and it had a radius of gyration of 120 mm. Static and kinetic coefficients of friction are 0.20 and 0.15 respectively. Determine a) if the wheel rolls or slides along the horizontal plane. b) the angular acceleration and centroidal acceleration of the wheel. Given: P =1 8 N rhub=80 mm r wheel = 160 mm k = 120 mm m = 5 kg μs=0.20 μk=0.15

G P

B A

Radius of gyration: I  k 2 m = 72000 kg-mm2

m, I

Case 1: Assume no slippage.

aGy

W

At point A:

vA  0 r a A  0ˆi  an ˆj

α G

G P

B

At Point G:

Ff

r aG  aGˆi  0 ˆj

N

Acceleration analysis:

r r r r r aG  a A    rG / A   2 rG / A aGˆi  0 ˆj   0ˆi  an ˆj    kˆ  rwheel ˆj   2 rwheel ˆj aGˆi  0 ˆj  an ˆj   rwheel ˆi   2rwheel ˆj

i:

aG   rwheel

j:

an   2 rwheel

aGx

Force Analysis:

Fx  m aGx P  F f  maG

Fy  m aGy N W  0



N  W  mg

M G  I Prhub  F f rwheel  I  Solve to find α, N, and Ff

P  Ff  m(  rwheel )



Ff  P rwheel m Ff  P

Prhub  F f rwheel  I 

 

Prhub  F f rwheel I

Prhub  F f rwheel

rwheel m I IFf  IP  Prhub rwheel m  F f rwheel rwheel m

I

f

rwheel rwheel m  Ff   rhub rwheel m  I  P Ff 

rhub rwheel m  I P I  rwheel rwheel m

so:

rhub rwheel m  I P I  rwheel rwheel m 160* 80* 5  72000 Ff  ( 18 )  1224 . N 72000  160* 160 * 5 Ff 

 s _ needed 

Ff N



N  W  mg N  5* 9.81  491 . N

1224 .  025 .  0.20   s _ available so the wheel slips. 491 .

Case 2: Slip at A: At point A:

At Point G:

r a A  a Axˆi  an ˆj

r aG  aGˆi  0 ˆj

Acceleration analysis:

r r r r r aG  a A    rG / A   2 rG / A aGˆi  0 ˆj  a Axˆi  an ˆj   kˆ  rwheel ˆj   2rwheel ˆj





aGˆi  0 ˆj  a Axˆi  an ˆj   rwheelˆi   2rwheel ˆj i:

aG  a Ax   rwheel

j:

an   2 rwheel

Force Analysis:

Fy  m aGy

Fx  m aGx P  F f  maG

N W  0 N  W  mg

M G  I

Friction Eq Ff  k N

Prhub  F f rwheel  I  Combine to find aAx, α, N, and Ff

N  W  mg  5* 9.81  491 . N Ff  k N  015 . * 491 .  7365 . N Prhub  F f rwheel  I  



Prhub  Ff rwheel

I 18* 60  7.365* 120 N mm 1kg m / s2 1000mm  72000 mm2 kg 1N 1m 2  2725 . rad / s

P  Ff  ma Ax  mrwheel  

a Ax 

P  Ff m

 rwheel

.  18  7365  N  0120 . m* 2725 . rad / s2  2.454m / s2

5kg therefore the disk does slip and the acceleration of point G is given by

r aG  aGˆi  0 ˆj   a Ax   rwheel  ˆi  0 ˆj

ˆi  0 ˆj m / s   2.454  2725 . * .120  ˆi  0 ˆj  2127 .

EGR 232 Eng. Dynamics HW Set 32 Fall 2012 Problem 16:78 A uniform slender rod of length L = 900 mm and mass m = 4 kg is suspended from a hinge at C. A horizontal force P of magnitude 75 N is applied at end B. Knowing that r = 225 mm, determine a) the angular acceleration of the rod, b) the components of the reaction at C.

EGR 232 Eng. Dynamics HW Set 32 Fall 2012 Problem 16: 89 Two uniform rods, ABC of mass 3 kg and DCE of mass 4 kg, are connected by a pin at C and by two cords BD and BE. The T shaped assembly rotates in a vertical plane under the combined effect of gravity and a of a couple M which is applied to rod ABC. Knowing that at the instant shown the tension is 8 N in cord BD, determine a) the angular acceleration of the assembly, b) the couple M.

EGR 232 Eng. Dynamics HW Set 32 Fall 2012 Problem 16:97 A homogeneous sphere S, a uniform cylinder C, and a thin pipe P are in contact when they are released from rest on the incline shown. Knowing that all three objects roll without slipping, determine after 4 s of motion the clear distance between a) the pipe and the cylinder b) the cylinder and the sphere.

EGR 232 Eng. Dynamics HW Set 32 Fall 2012 Problem 16:109 Two uniform disks A and B each of weight 4 lb, are connected by a 3 lb rod as shown. A counterclockwise couple M of moment 1.5 lb-ft is applied to disk A. Knowing that the disks roll without sliding, determine a) the acceleration of the center of each disk b) the horizontal component of the force exerted on disk B by pin D.