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(a) A yo-yo is made of two solid cylindrical disks, each of mass 0.050 $\mathrm{kg}$ and diameter 0.075 $\mathrm{m}$ , joined by a (concentric) thin solid cylindrical hub of mass 0.0050 $\mathrm{kg}$ and diameter 0.010 $\mathrm{m} .$ Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its $1.0-\mathrm{m}$ -long string, if it is released from rest. $(b)$ What fraction of its kinetic energy is rotational?

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(a) .84 $\mathrm{m} / \mathrm{s}$(b) 96$\%$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

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.

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should be the free body diagram for the yo yo and essentially a diagram of the yo yo itself just to, uh, help you in calculating the moment of inertia. So we can say that the, um, velocity of the center of the mass of the yo yo would be equaling the radius of the yo yo times the angular velocity of the yo yo, we're going to say that calculating the moment of inertia, um, this would be equaling the moment of inertia of a disc 1/2 m r squared plus two times 1/2 m r squared. And so we can say that the moment of inertia vehicle to 1/2 times m R squared lower case and being the mass of the inner hub. So this would be, um And then this would be, um uh this lower case R would be the radius of the inner hub. And then, uh, capital R would essentially be the radius of the two outer disks. And so we can say that this would be the moment of inertia. Through the center of mass would be 1/2 m r squared plus m r squared. And so we can calculate this moment of inertia. The center of mass would be 1/2 times 5.0 times 10 to the negative third kilograms multiplied by 5.0 times 10 to the A negative third meters quantity squared plus 5.0 times 10 to the negative second kilograms multiplied by 3.75 meters times 10 to the negative second meters quantity squared and we find that the moment of inertia through the center of mass is equaling 7.38 times, 10 to the negative kilograms meters squared. Now we know that the total mass mm is equaling. The mass of the inner hub was two times the mass of the outer disks and so this is equaling 0.105 kilograms and so we can then apply the conservation of kinetic conservation of mechanical energy for part of the potential energy initial. We'll be equal to the kinetic energy final and we can say the total mass times G Times H would be equaling 2 1/2 times the total mass times the velocity of the center of mass squared, plus 1/2 times the moment of inertia through the center of mass times omega squared, and so this would be equal mass. Total times G times H would be equaling 1/2 times the mass total times the velocity of the center of mass squared. This would be plus 1/2 times the moment of inertia through the center of mass times. The velocity through the center of mass, divided by our quantity squared and so this would be equal to the total mass divided by two plus the moment of inertia through the center of mass, divided by two times are squared quantity multiplied by the velocity of the center of the mass squared. So the velocity of the center of mass will be equaling a square root of the total mass multiplied by G times H, and this would be divided by 1/2 times the total mass, plus the moment of inertia through the center of mass. It's over the radius of the inner hub squared and weaken solved, so this should be for the numerator 0.105 kilograms multiplied by 9.80 meters per second, squared multiplied by 1.0 meters, and this would be divided by 1/2 times the total mass of 0.105 kilograms plus 7.38 times 10 to the negative fifth kilograms meter squared and then this would be divided by five 0.0 times, 10 to the negative third meters quantity squared. Clean this up and we finally obtained that the velocity of the center of mass would be equaling point eight 395 meters per second or we can say approximately a 3950.84 meters per second. This would be our final answer for part A and then four part B. They want us to find the ratio of the rotational kinetic energy divided by the total kinetic energy. And so this would be equaling the rotational energy. And we know that the total kinetic energy is equaling the initial potential energy. So this would be equal in 1/2 times the moment of inertia through the center of mass times V. The velocity of the center of mass divided by the radius of the inner hub quantity squared, divided by the total mass times, G times H, and we can say that this is gonna be equaling the moment of inertia through the center of mass multiplied by the velocity through the center of mass squared with a velocity of the center of mass squared times two. Rather divided by two R squared times the total mass times G times H and we can solve. So this would be equaling 7.38 times 10 to the negative fifth kilograms meters squared multiplied by 0.8395 meters per second. Quantity squared, divided by two times 5.0 times 10 to the negative third meters quantity squared multiplied by 0.105 kilograms multiplied by 9.8 meters per second squared multiplied by 1.0 meters and we find that this is equal in 0.96 So the rotational kinetic energy divided by the total kinetic energy well equal essentially 96%. Most of the energy is in the form of rotational kinetic energy. That is the end of the solution. Thank you

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