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(B) For a wheel rolling with a uniform velocity $v$ of its center of mass,the acceleration of the center of mass is $a_{CM} = 0$.Each point on the rim

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$\frac{2v^2}{R}$

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(B) For a wheel rolling with a uniform velocity $v$ of its center of mass,the acceleration of the center of mass is $a_{CM} = 0$.Each point on the rim of the wheel has a centripetal acceleration directed towards the center,given by $a_c = \frac{v^2}{R}$.Let the topmost point be $A$ and the bottommost point be $B$.The acceleration of point $A$ is $\vec{a}_A = \frac{v^2}{R}$ (directed downwards towards the center).The acceleration of point $B$ is $\vec{a}_B = \frac{v^2}{R}$ (directed upwards towards the center).The relative acceleration of $A$ with respect to $B$ is $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$.Taking the downward direction as positive,$\vec{a}_A = \frac{v^2}{R}$ and $\vec{a}_B = -\frac{v^2}{R}$.Thus,$\vec{a}_{AB} = \frac{v^2}{R} - (-\frac{v^2}{R}) = \frac{2v^2}{R}$.Therefore,the magnitude of the relative acceleration is $\frac{2v^2}{R}$.

$A$ wheel of radius $R$ rolls on the ground with a uniform velocity $v$. The relative acceleration of the topmost point of the wheel with respect to the bottommost point is

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