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A cyclist accelerates from rest at a rate of 1.00 $\mathrm{m} / \mathrm{s}^{2} .$ How fast will a point at the top of the rim of the tire (diameter $=68 \mathrm{cm}$ ) be moving after 2.5 $\mathrm{s} ?[$ Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest - see Fig. $63 . ]$

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$5.0 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

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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A cyclist accelerates from…

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<1> A cyclist accele…

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A bicycle initially at res…

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A typical road bike wheel …

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A motorcyclist who is movi…

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A bicycle with 0.80-m-diam…

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Additional ProblemsA m…

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A bicyclist accelerates at…

So first we can say that the linear velocity is related to the angular velocity bias doctor are or the radius s. So we can say that the velocity of the top relative to the ground would then be equal to the velocity of the top relative to the center, plus the velocity of the center relative to the ground. Uh, this would be equal to velocity V plus V, which would be equal to two V to the right. The velocity of the top relative to the ground would then be equal to to be, ah, equaling two times the initial plus a t. Ah, we know the initials gonna be zero. So this would simply be two times the acceleration times time. Let's solve So two times 1.0 meters per second squared multiplied by 2.5 seconds. And we find that the velocity of the top relative to the ground would be equal to approximately 5.0 meters per second, where the velocity of the center relative to the ground would be approximately half that, or two and 1/2 meters per second. That is the end of the solution. Thank you for watching

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