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A hollow cylinder (hoop) is rolling on a horizontal surface at speed $v=3.3 \mathrm{m} / \mathrm{s}$ when it reaches a $15^{\circ}$ incline. (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?

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(a) 4.3$m$(b) $5.2 \mathrm{s}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

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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So here we can say that there are weak and where they were going to assume that there's no dissipated forces and essentially we're simply going to say, use the law of conservation of energy. So for party, we can simply say that the energy at the bottom would be equal to the energy at the top. At the top, we only have. We only have, ah, gravitational potential energy. So we can say that 1/2 and the squared plus 1/2 i omega squared. So at the bottom, we have translational kinetic energy as well as rotational kinetic energy And at the top, we only have a gravitational potential energy and we can then say 1/2 and the squared plus 1/2 m r squared, which would be the moment of inertia multiple by Omega Squared were going to say that Omega equals V over R. Therefore, Omega squared would, of course, equal V squared over r squared and this is equaling again. M g h. Therefore, we can say that H is gonna be equal to of thes squared over G. This is gonna equal 3.3 meters per second squared, divided by 9.80 meters per second squared. Oh my pop, my muse for second quantity squared and then divided by 9.80 meters per second squared. This is giving us a height of 1.111 meters. So we can say that the distance along the plane D would be equal to H divided by sign of fada. So this would be equal to 1.111 meters, divided by sign of 15 degrees, 15 degrees being the angle of the incline. This is equaling 4.3 meters approximately, so that would be your answer for party and then for part B. Ah, we can say that Ah, the time can be doubt tow X equaling the average accelerations of the initial plus B final times t therefore t it's simply being the equal to two times out the X, divided by the final, knowing that the initial is going to be zero because it at the top of the incline there is no kinetic energy. So we can say that T would be equal to two times 4.293 meters. Let's not round into the very end, divided by the final of 3.3 meters per second and we find that tea is gonna be 2.602 seconds. This is the time to go up the plane. Therefore the time to come back down. The plane is the same. So tee total ah would be equal to two times T or two times 2.602 This would equal 5.2 seconds approximately. So this would be the time, essentially, for a round trip. That is the end of the solution. Thank you for watching.

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