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(a) For the yo-yo-like cylinder of Example 19 of "Rota- tional Motion", we saw the downward acceleration of its $\mathrm{CM}$ was $a=\frac{2}{3} g .$ If it starts from rest, what will be the CM velocity after it has fallen a distance $h ?$ (b) Now use conservation of energy to determine the cylinder's $\mathrm{CM}$ velocity after it has fallen a distance $h,$ starting from rest.

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(A) $\sqrt{\frac{4}{3} g h}$(b) $\sqrt{\frac{4}{3} g h}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Simon Fraser University

University of Winnipeg

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

02:48

Cylinder $A$ is moving dow…

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In the cylinder and mass c…

The height of a solid cyli…

03:27

In Fig. $15-59$ , a solid …

01:08

A yo-yo (as shown) falls u…

02:06

A yo-yo can be thought of …

so we can say that the acceleration is found in example 10 19 to be a constant value. A equaling 2/3 of G again, This is from example 10. 19 therefore, accept constant acceleration. Kinnah Matics can be used. We're gonna say downwards as positives and so we can save E Why squared would equal be V Y Final squared equals V Why initial squared plus two times acceleration in the UAE direction times Delta. Why Ah, we know that the initial velocity is gonna be zero. Therefore the Y final would be equal to the square root of two times the acceleration in the right direction times the change in displacement in the UAE direction So the Y final would be equal to the square root of two times 2/3 g times h the height and this would be equal to the square root 4/3 G h equaling the final. So this would be our answer for part A. For part B. However, we're gonna take the zero level for we're gonna take the zero level from gravitational potential energy to be the starting height of the yo yo. So we can say that due to the law of conservation of energy E initial equals E final Ah, we can say that. Then zero equals you final plus que final okay. And we can say that zero equals negative and G h plus 1/2 times the mass times the velocity of the center of mass squared plus 1/2 times the moment of inertia of the yo yo times omega times a center of mass squared So we can say M g h equals 1/2 times the mass tens of velocity of the center of mass squared plus 1/2 times the moment of inertia the yo yo so 1/2 m r squared, multiplied by the velocity of the center of mass divided by our squared Because the yo yo is rolling without slipping we can make this assumption if the yellow was slipping We cannot make this assumption. And then at this point, we're simply going to, uh, solve for the velocity of the center of mass. So the velocity of the center of mass would then be equal to the square roots of 4/3 G. H. This would be our final answer. That is the end of the solution. Thank you for watching

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