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Copyright © 2012 Pearson Education Inc. Rigid body rotation Physics 7C lecture 13 Tuesday November 12, 8:00 AM – 9:20 AM Engineering Hall 1200.

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Presentation on theme: "Copyright © 2012 Pearson Education Inc. Rigid body rotation Physics 7C lecture 13 Tuesday November 12, 8:00 AM – 9:20 AM Engineering Hall 1200."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. Rigid body rotation Physics 7C lecture 13 Tuesday November 12, 8:00 AM – 9:20 AM Engineering Hall 1200

2 Copyright © 2012 Pearson Education Inc. Applying a torque Torque: τ = r × F

3 Copyright © 2012 Pearson Education Inc. Torque and angular acceleration for a rigid body The rotational analog of Newton’s second law for a rigid body is  z =I  z.

4 Copyright © 2012 Pearson Education Inc. Another unwinding cable What is the angular acceleration, assuming uniform disk?

5 Copyright © 2012 Pearson Education Inc. Another unwinding cable

6 Copyright © 2012 Pearson Education Inc. A. m 2 g = T 2 = T 1 B. m 2 g > T 2 = T 1 C. m 2 g > T 2 > T 1 D. m 2 g = T 2 > T 1 E. none of the above Q10.5 A glider of mass m 1 on a frictionless horizontal track is connected to an object of mass m 2 by a massless string. The glider accelerates to the right, the object accelerates downward, and the string rotates the pulley. What is the relationship among T 1 (the tension in the horizontal part of the string), T 2 (the tension in the vertical part of the string), and the weight m 2 g of the object?

7 Copyright © 2012 Pearson Education Inc. A. m 2 g = T 2 = T 1 B. m 2 g > T 2 = T 1 C. m 2 g > T 2 > T 1 D. m 2 g = T 2 > T 1 E. none of the above A10.5 A glider of mass m 1 on a frictionless horizontal track is connected to an object of mass m 2 by a massless string. The glider accelerates to the right, the object accelerates downward, and the string rotates the pulley. What is the relationship among T 1 (the tension in the horizontal part of the string), T 2 (the tension in the vertical part of the string), and the weight m 2 g of the object?

8 Copyright © 2012 Pearson Education Inc. Rigid body rotation about a moving axis The motion of a rigid body is a combination of translational motion of the center of mass and rotation about the center of mass. The kinetic energy of a rotating and translating rigid body is K = 1/2 Mv cm 2 + 1/2 I cm  2.

9 Copyright © 2012 Pearson Education Inc. Rigid body rotation about a moving axis What’s the center-of-mass’s motion? What’s the motion around center-of-mass?

10 Copyright © 2012 Pearson Education Inc. Rolling without slipping The condition for rolling without slipping is v cm = R .

11 Copyright © 2012 Pearson Education Inc. A yo-yo Find the angular speed of yo-yo.

12 Copyright © 2012 Pearson Education Inc. A yo-yo Find the angular speed of yo-yo.

13 Copyright © 2012 Pearson Education Inc. A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and the hoop is released from rest, the string unwinds and the hoop descends. How does the tension in the string (T) compare to the weight of the hoop (w)? A. T = w B. T > w C. T < w D. not enough information given to decide Q10.6

14 Copyright © 2012 Pearson Education Inc. A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and the hoop is released from rest, the string unwinds and the hoop descends. How does the tension in the string (T) compare to the weight of the hoop (w)? A. T = w B. T > w C. T < w D. not enough information given to decide A10.6

15 Copyright © 2012 Pearson Education Inc. The race of the rolling bodies These objects have the same mass. Which would acquire the largest speed at the bottom?

16 Copyright © 2012 Pearson Education Inc. The race of the rolling bodies These objects have the same mass. Which would acquire the largest speed at the bottom?

17 Copyright © 2012 Pearson Education Inc. Acceleration of a yo-yo We have translation and rotation, so we use Newton’s second law for the acceleration of the center of mass and the rotational analog of Newton’s second law for the angular acceleration about the center of mass. Find the acceleration.

18 Copyright © 2012 Pearson Education Inc. Acceleration of a yo-yo

19 Copyright © 2012 Pearson Education Inc. A solid bowling ball rolls down a ramp. Which of the following forces exerts a torque on the bowling ball about its center? A. the weight of the ball B. the normal force exerted by the ramp C. the friction force exerted by the ramp D. more than one of the above E. The answer depends on whether the ball rolls without slipping. Q10.7

20 Copyright © 2012 Pearson Education Inc. A solid bowling ball rolls down a ramp. Which of the following forces exerts a torque on the bowling ball about its center? A10.7 A. the weight of the ball B. the normal force exerted by the ramp C. the friction force exerted by the ramp D. more than one of the above E. The answer depends on whether the ball rolls without slipping.

21 Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere Find the acceleration.

22 Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere Find the acceleration.

23 Copyright © 2012 Pearson Education Inc. Rolling without slipping The condition for rolling without slipping is v cm = R .

24 Copyright © 2012 Pearson Education Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled to the right as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. Q10.8 F

25 Copyright © 2012 Pearson Education Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled to the right as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. A10.8 F

26 © 2012 Pearson Education, Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled straight up as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. Q10.10 F

27 © 2012 Pearson Education, Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled straight up as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. A10.10 F

28 Copyright © 2012 Pearson Education Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled to the right as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. Q10.9 F

29 Copyright © 2012 Pearson Education Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled to the right as shown, A. the yo-yo rolls to the right. B. the yo-yo rolls to the left. C. the yo-yo remains at rest. D. The answer depends on the magnitude F of the pulling force compared to the magnitude of the friction force. A10.9 F

30 Copyright © 2012 Pearson Education Inc. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled to the right as shown, Q10.9 F

31 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion A tangential force applied to a rotating body does work on it. The total work done on a body by the torque is equal to the change in rotational kinetic energy of the body and the power due to a torque is P =  z  z.

32 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion P =  z  z.


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