Chapter 15 Kinematics of Rigid Bodies PDF

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Dr Mahbuboor Choudhury...

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Chapter 15 Kinematics of Rigid Bodies

Dynamics Dr. Mahbuboor Choudhury Concordia University

Contents Introduction Translation Rotation About a Fixed Axis: Velocity Rotation About a Fixed Axis: Acceleration Rotation About a Fixed Axis: Representative Slab Equations Defining the Rotation of a Rigid Body About a Fixed Axis Analyzing General Plane Motion Absolute and Relative Velocity in Plane Motion Instantaneous Center of Rotation

Absolute and Relative Acceleration in Plane Motion Analysis of Plane Motion in Terms of a Parameter Rate of Change With Respect to a Rotating Frame Plane Motion Relative to a Rotating Frame Motion About a Fixed Point General Motion Three Dimensional Motion Relative to a Rotating Frame Frame of Reference in General Motion 15 - 2

Applications The linkage between train wheels is an example of curvilinear translation – the link stays horizontal as it swings through its motion.

15 - 3

Applications How can we determine the velocity of the tip of a turbine blade?

15 - 4

Applications Planetary gear systems are used to get high reduction ratios with minimum weight and space. How can we design the correct gear ratios?

15 - 5

Introduction • Kinematics of rigid bodies: relations between time and the positions, velocities, and accelerations of the particles forming a rigid body. • Classification of rigid body motions: - translation: • rectilinear translation • curvilinear translation - rotation about a fixed axis - general plane motion - motion about a fixed point - general motion

15 - 6

Translation • Consider rigid body in translation: - direction of any straight line inside the body is constant, - all particles forming the body move in parallel lines. • For any two particles in the body,    rB  rA  rB A • Differentiating with respect to time,     rB  rA  rB A  rA   vB  v A All particles have the same velocity. • Differentiating with respect to time again,     rB  rA  rB A  rA   aB  a A All particles have the same acceleration. 15 - 7

Rotation About a Fixed Axis. Velocity • Consider rotation of rigid body about a fixed axis AA’   • Velocity vector v  dr dt of the particle P is tangent to the path with magnitude v  ds dt

Ds  BP Dq  r sin f Dq Dq ds  lim r sin f   r q sin f v dt Dt 0 Dt

• The same result is obtained from   dr    w r v dt    w  w k  qk  angular velocity

15 - 8

Concept Quiz What is the direction of the velocity of point A on the turbine blade?

w

a) → b) ←

A y L

c) ↑ d) ↓

x

15 - 9

Rotation About a Fixed Axis. Acceleration • Differentiating to determine the acceleration,   dv d   a  w  r  dt dt   dw   dr  r w  dt dt  dw      r w  v dt  dw   a  angular acceleration • dt     a k  w k  qk • Acceleration of P is combination of two vectors,       a  a  r  w w  r   a  r  tangential acceleration component    w  w  r  radial acceleration component 15 - 10

Rotation About a Fixed Axis. Representative Slab • Consider the motion of a representative slab in a plane perpendicular to the axis of rotation. • Velocity of any point P of the slab,      v w r wk r v  rw • Acceleration of any point P of the slab,       a  a  r  w w  r     a k  r w 2r • Resolving the acceleration into tangential and normal components,    at  ak  r a t  ra   an  w 2 r an  rw 2 15 - 11

Concept Quiz What is the direction of the normal acceleration of point A on the turbine blade? a) → b) ←

w A

y L

c) ↑ d) ↓

x

15 - 12

Equations Defining the Rotation of a Rigid Body About a Fixed Axis • Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration. • Recall w 

dq dt

or

dt 

dq

w

dw dw d 2q  w a dq dt dt 2

• Uniform Rotation, a = 0:

q  q 0  wt • Uniformly Accelerated Rotation, a = constant: w  w0  a t

q  q0  w0t  12 at 2 w 2  w 02  2a q q 0  15 - 13

Sample Problem 15.3 STRATEGY: • Due to the action of the cable, the tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration.

Cable C has a constant acceleration of 9 in/s2 and an initial velocity of 12 in/s, both directed to the right. Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in position of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0.

• Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s. • Evaluate the initial tangential and normal acceleration components of D.

15 - 14

Sample Problem 15.3 MODELING and ANALYSIS: • The tangential velocity and acceleration of D are equal to the velocity and acceleration of C. aD t  aC  9 in. s  v    v   12 in. s  D 0

C 0

v D 0  rw 0 vD 0 w0 

r



aD t  ra aD t a

12  4 rad s 3

r



9  3 rad s 2 3

• Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s.





w  w 0 a t  4 rad s  3 rad s 2 2 s   10 rad s





q  w0 t  12 at 2  4 rad s 2 s   12 3 rad s 2 2 s 2  14 rad  1 rev   N  14 rad  number of revs  2 rad  v B  r w  5 in. 10 rad s  Dy B  r q  5 in.14 rad 

N  2.23 rev

 v B  50 in. s  DyB  70 in. 15 - 15

Sample Problem 15.3 • Evaluate the initial tangential and normal acceleration components of D. aD t  aC  9 in. s 

aD n

 rDw 20   3 in. 4 rad s2  48 in s2

aD t

 9 in. s 2 

aD n  48 in.

s2 

Magnitude and direction of the total acceleration,

aD t2  aD n2

aD 

 9 2  48 2 tan f  

aD  48.8 in. s2

a D n aD t 48 9

f  79.4 15 - 16

Sample Problem 15.3 REFLECT and THINK: • A double pulley acts similarly to a system of gears; for every 3 inches that point C moves to the right, point B moves 5 inches upward. This is also similar to the rear tire of your bicycle. As you apply tension to the chain, the rear sprocket rotates a small amount, causing the rear wheel to rotate through a much larger angle.

15 - 17

Group Problem Solving STRATEGY: • Using the linear velocity and accelerations, calculate the angular velocity and acceleration. • Using the angular velocity, determine the normal acceleration.

A series of small machine components being moved by a conveyor belt pass over • Determine the total acceleration a 6-in.-radius idler pulley. At the instant using the tangential and normal shown, the velocity of point A is 15 in./s to acceleration components of B. the left and its acceleration is 9 in./s2 to the right. Determine (a) the angular velocity and angular acceleration of the idler pulley, (b) the total acceleration of the machine component at B.

15 - 18

Group Problem Solving MODELING and ANALYSIS:

v= 15 in/s

B

at= 9 in/s2

Find the angular velocity of the idler pulley using the linear velocity at B. v  rw 15 in./s  (6 in.)w

w  2.50 rad/s

Find the angular velocity of the idler pulley using the linear velocity at B. a  ra 9 in./s  (6 in.)a 2

a  1.500 rad/s2

Find the normal acceleration of point B. an  rw 2  (6 in.)(2.5 rad/s)2

a n  37.5 in./s2

What is the direction of the normal acceleration of point B? Downwards, towards the center 15 - 19

Group Problem Solving B

at= 9 in/s2

an= 37.5 in./s2

Find the total acceleration of the machine component at point B.

at  9.0 in./s 2

an  37.5 in./s 2

Calculate the magnitude

a  9.02  37.52  38.6 in./s2 at= 9 in/s2

Calculate the angle from the horizontal

 37.5   76.5o   9.0 

q  arctan 

Combine for a final answer

aB an= 37.5 in/s2

a B  38.6 in./s 2

76.5 15 - 20

Example – General Plane Motion As the woman approaches to release the bowling ball, her arm has linear velocity and acceleration from both translation (the woman moving forward) as well as rotation (the arm rotating about the shoulder).

©My Make OU/123RF 15 - 21

Analyzing General Plane Motion

• General plane motion is neither a translation nor a rotation. • General plane motion can be considered as the sum of a translation and rotation. • Displacement of particles A1 and B1 to A2 and B2 can be divided into two parts: - translation to A2 and B1 - rotation of B1 about A2 to B2 15 - 22

Absolute and Relative Velocity in Plane Motion

• Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A.   

vB  v A  vB

A

   v B A  w k  rB A

v B A  rw

    v B  v A  w k  rB A

15 - 23

Absolute and Relative Velocity in Plane Motion

• Assuming that the velocity vA of end A is known, wish to determine the velocity vB of end B and the angular velocity w in terms of vA, l, and q. • The direction of vB and vB/A are known. Complete the velocity diagram. vB  tan q vA v B  v A tan q

vA v  A  cosq vB A lw

w

vA l cosq 15 - 24

Absolute and Relative Velocity in Plane Motion

• Selecting point B as the reference point and solving for the velocity vA of end A and the angular velocity w leads to an equivalent velocity triangle. • vA/B has the same magnitude but opposite sense of vB/A. The sense of the relative velocity is dependent on the choice of reference point. • Angular velocity w of the rod in its rotation about B is the same as its rotation about A. Angular velocity is not dependent on the choice of reference point. 15 - 25

Sample Problem 15.6 STRATEGY: • The displacement of the gear center in one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities. The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s. Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

• The velocity for any point P on the gear may be written as    vP  v A  v P

A

    v A  wk  rP

A

Evaluate the velocities of points B and D.

15 - 26

Sample Problem 15.6 MODELING and ANALYSIS • The displacement of the gear center in one revolution is equal to the outer circumference. For xA > 0 (moves to right), w < 0 (rotates clockwise). xA q  2 r 2 y

x A  r1q

Differentiate to relate the translational and angular velocities. x

v A  r1w v 1.2 m s w  A  r1 0.150 m







w  wk  8 rad sk

15 - 27

Sample Problem 15.6    • For any point P on the gear, vP  v A  v P

Velocity of the upper rack is equal to velocity of point B:      vR  vB  v A  w k  rB A     1.2 m s i  8 rad s k  0.10 m  j    1.2 m s i  0.8 m s i   v R   2 m si

   w    rP v k A A

A

Velocity of the point D:     vD  v A  wk  rD A     1.2 m si  8 rad s k   0.150 mi

   vD  1.2 m si  1.2 m s j vD  1.697 m s 15 - 28

Sample Problem 15.6 REFLECT and THINK: • Note that point A was free to translate, and Point C, since it is in contact with the fixed lower rack, has a velocity of zero. Every point along diameter CAB has a velocity vector directed to the right and the magnitude of the velocity increases linearly as the distance from point C increases.

15 - 29

Sample Problem 15.7 STRATEGY: • Will determine the absolute velocity of point D with    vD  vB  v D B  • The velocity vB is obtained from the given crank rotation data.  velocity v D The crank AB has a constant clockwise • The directions of the absolute  and the relative velocity vD B are angular velocity of 2000 rpm. determined from the problem geometry. For the crank position indicated, • The unknowns in the vector expression determine (a) the angular velocity of are the velocity magnitudes v D and v D B the connecting rod BD, and (b) the which may be determined from the velocity of the piston P. corresponding vector triangle.

• The angular velocity of the connecting rod is calculated from v D B . 15 - 30

Sample Problem 15.7 MODELING and ANALYSIS: • Will determine the absolute velocity of point D with    v D  vB  vD B

 • The velocity vB is obtained from the crank rotation data.  min  2 rad    209.4 rad s   min  60 s  rev   v B   AB w AB  3 in. 209.4 rad s

rev w AB   2000

The velocity direction is as shown.

 • The direction of the absolute velocity vD is horizontal.  The direction of the relative velocity v D B is perpendicular to BD. Compute the angle between the horizontal and the connecting rod from the law of sines. sin 40 sin b  8 in. 3 in.

b  13.95

15 - 31

Sample Problem 15.7

• Determine the velocity magnitudes vD and vD from the vector triangle.

B

vD B vD 628.3 in. s   sin 53.95 sin 50 sin76.05

v D  523.4 in. s  43.6 ft s

   vD  vB  vD

B

vD

B

 495.9 in. s

vD

B

 lw BD vD

495.9 in. s l 8 in.  62.0 rad s

w BD 

B

vP  vD  43.6 ft s







w BD  62.0 rad s k 15 - 32

Sample Problem 15.7 REFLECT and THINK: Note that as the crank continues to move clockwise below the center line, the piston changes direction and starts to move to the left. Can you see what happens to the motion of the connecting rod at that point?

15 - 33

Concept Quiz In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE. Which of the following is true? a) The direction of vB is ↑ b) The direction of vD is → c) Both a) and b) are correct

15 - 34

Group Problem Solving STRATEGY: • The displacement of the gear center in one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities. • The velocity for any point P on the gear may be written as    vP  v A  v P

In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.

A

    v A  wk  rP

A

Evaluate the velocities of points B and D.

15 - 35

Group Problem Solving y x

wAB= 4 rad/s

MODELING and ANALYSIS: Determine the angular velocity of bars BD and DE. How should you proceed? Determine vB with respect to A, then work your way along the linkage to point E. Write vB in terms of point A, calculate vB. vB  v A w AB  r B/A

w AB  (4 rad/s)k

rB/ A   (7 in.)i

vB  w AB  rB/ A  ( 4k )  ( 7i )

vB  (28 in./s)j

Does it make sense that vB is in the +j direction? 15 - 36

vD,

Group Problem Solving y

Determine vD with respect to B. x

w BD  wBDk rD/B  (8 in.)j vD  vB  wBD  rD /B  28 j  (wBD k )  ( 8 j) vD  28 j  8wBD i

wAB= 4 rad/s

Determine vD with respect to E, then equate it to equation above. w DE  wDE k rD /E  (11 in.)i  (3 in.)j vD  w DE  rD / E  (w DE k)  ( 11i  3 j) vD  11w DE j  3w DE i

Equating components of the two expressions for vD j:

28  11wDE

i : 8w BD  3w DE

wDE  2.5455 rad/s 3 8

wBD   w BD

w DE  2.55 rad/s w BD  0.955 rad/s 15 - 37

Instantaneous Center of Rotation • Plane motion of all particles in a slab can always be replaced by the translation of an arbitrary point A and a rotation about A with an angular velocity that is independent of the choice of A. • The same translational and rotational velocities at A are obtained by allowing the slab to rotate with the same angular velocity about the point C on a perpendicular to the velocity at A. • The velocity of all other particles in the slab are the same as originally defined since the angular velocity and translational velocity at A are equivalent. • As far as the velocities are concerned, the slab seems to rotate about the instantaneous center of rotation C.

15 - 38

Instantaneous Center of Rotation • If the velocity at two points A and B are known, the instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B . • If the velocity vectors are parallel, the instantaneous center of rotation is at infinity and the angular velocity is zero. • If the velocity vectors at A and B are perpendicular to the line AB, the instantaneous center of rotation lies at the intersection of the line AB with the line joining the extremities of the velocity vectors at A and B. • If the velocity magnitudes are equal, the instantaneous center of rotation is at infinity and the angular velocity is zero. 15 - 39

Instantaneous Center of Rotation • The instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B . v v v w A  A v B  BC w  l sin q  A l cos q AC l cos q  vA tan q • The velocities of all particles on the rod are as if they were rotated about C. • The particle at the center of rotation has zero velocity. • The particle coinciding with the center of rotation changes with time and the acceleration of the particle at the instantaneous center of rotation is not zero. • The acceleration of the particles in the slab cannot be determined as if the slab were simply rotating about C. • The trace of the locus of the center of rotation on the body is the body centrode and in space is the space centrode. 15 - 40

Instantaneous Center of Rotation At the instant shown, what is the approximate direction of the velocity of point G, the center of bar AB? a) G

b) c) d)

15 - 41

Sample Problem 15.9 STRATEGY: • The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation. • Determine the angular velocity about C based on the given velocity at A. The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.

• Evaluate the velocities at B and D...