Rigid body rotation

  • View
    253

  • Download
    1

Embed Size (px)

DESCRIPTION

Physics 7C lecture 13. Rigid body rotation. Tuesday November 12, 8:00 AM – 9:20 AM Engineering Hall 1200. Applying a torque. Torque: τ = r × F. Torque and angular acceleration for a rigid body. The rotational analog of Newton ’ s second law for a rigid body is   z = I z . - PowerPoint PPT Presentation

Text of Rigid body rotation

PowerPoint Presentation

Rigid body rotationPhysics 7C lecture 13Tuesday November 12, 8:00 AM 9:20 AMEngineering Hall 1200

Copyright 2012 Pearson Education Inc.Applying a torqueTorque: = r F

Copyright 2012 Pearson Education Inc.Torque and angular acceleration for a rigid bodyThe rotational analog of Newtons second law for a rigid body is z =Iz.

Copyright 2012 Pearson Education Inc.Another unwinding cableWhat is the angular acceleration, assuming uniform disk?

Copyright 2012 Pearson Education Inc.Another unwinding cable

Copyright 2012 Pearson Education Inc.A. m2g = T2 = T1 B. m2g > T2 = T1 C. m2g > T2 > T1 D. m2g = T2 > T1 E. none of the aboveQ10.5A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 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 T1 (the tension in the horizontal part of the string), T2 (the tension in the vertical part of the string), and the weight m2g of the object?

Copyright 2012 Pearson Education Inc.6Answer: CA. m2g = T2 = T1 B. m2g > T2 = T1 C. m2g > T2 > T1 D. m2g = T2 > T1 E. none of the aboveA10.5

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 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 T1 (the tension in the horizontal part of the string), T2 (the tension in the vertical part of the string), and the weight m2g of the object?

Copyright 2012 Pearson Education Inc.

Rigid body rotation about a moving axisThe 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 Mvcm2 + 1/2 Icm2.Copyright 2012 Pearson Education Inc.

Rigid body rotation about a moving axisWhats the center-of-masss motion?

Whats the motion around center-of-mass?Copyright 2012 Pearson Education Inc.Rolling without slippingThe condition for rolling without slipping is vcm = R.

Copyright 2012 Pearson Education Inc.A yo-yoFind the angular speed of yo-yo.

Copyright 2012 Pearson Education Inc.A yo-yoFind the angular speed of yo-yo.

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 decideQ10.6

Copyright 2012 Pearson Education Inc.13Answer: CA 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 decideA10.6

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?

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?

Copyright 2012 Pearson Education Inc.Acceleration of a yo-yoWe have translation and rotation, so we use Newtons second law for the acceleration of the center of mass and the rotational analog of Newtons second law for the angular acceleration about the center of mass.Find the acceleration.

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

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 aboveE. The answer depends on whether the ball rolls without slipping.Q10.7

Copyright 2012 Pearson Education Inc.19Answer: CA 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 aboveE. The answer depends on whether the ball rolls without slipping.

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

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

Copyright 2012 Pearson Education Inc.Rolling without slippingThe condition for rolling without slipping is vcm = R.

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.Q10.8F

Copyright 2012 Pearson Education Inc.24Answer: AA 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

FCopyright 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.Q10.10F

2012 Pearson Education, Inc.26Answer: AA 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 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.Q10.9F

Copyright 2012 Pearson Education Inc.28Answer: AA 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

FCopyright 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,Q10.9F

Copyright 2012 Pearson Education Inc.30Answer: AWork 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 = zz.

Copyright 2012 Pearson Education Inc.Work and power in rotational motion P = zz.

Copyright 2012 Pearson Education Inc.

Recommended

View more >