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(I) Estimate the moment of inertia of a bicycle wheel 67 $\mathrm{cm}$ in diameter. The rim and tire have a combined mass of 1.1 $\mathrm{kg} .$ The mass of the hub can be ignored (why?).

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0.12 $\mathrm{kg} \cdot \mathrm{m}^{2}$close to the axis-negligible inertia

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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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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(I) Estimate the moment of…

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Estimate the moment of in…

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(I) Calculate the moment o…

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Estimate the moment of ine…

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Part AEstimate the mom…

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$\cdot$ The moment of iner…

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A bicycle rim has a diamet…

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A bicycle wheel has a radi…

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A wagon wheel is construc…

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A wagon wheel is construct…

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What is the moment of iner…

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so here all of the significant mass is located at the same distance from the axis of rotation. So we considered the moment of inertia would simply be given by that would be equal to M. R. Squared the mass times the distance from the axis of rotation denoted as R squared, so I would then be equal to 1.1 kilogram that mass multiplied by 1/2 times 0.67 meters quantity squared. This is giving us point 12 kilogram meters squared. So this would be our moment of inertia and again with the hub mess, because the hub mass can be ignored because its distance from the axis of rotation is extremely small. So we can say hub Mass can be ignored because, essentially, for the hub Mass, it's our would be approximately equal to zero meters. So because it's so close to the axis of rotation, it wouldn't add any rotational inertia. So essentially the only rotational inertia would be M R squared and that mass at a distance of 1/2 times 0.67 meters again, that final moment of inertia 0.12 kilogram meters squared. That is the end of the solution. Thank you for watching

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