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Question

(A) Let $\vec{\omega}$ be the angular velocity of the sphere and $R$ be its radius. The velocity of any point $P$ on the sphere is given by $\vec{V}_P

Vedclass · Accepted Answer

$|\vec{V}_C - \vec{V}_A| = 2 |\vec{V}_B - \vec{V}_C|$

Vedclass · Answer

(A) Let $\vec{\omega}$ be the angular velocity of the sphere and $R$ be its radius. The velocity of any point $P$ on the sphere is given by $\vec{V}_P = \vec{V}_B + \vec{\omega} 	imes \vec{r}_{BP}$,where $\vec{V}_B$ is the velocity of the center.Since the sphere rolls without slipping,the velocity of the contact point $A$ is zero,i.e.,$\vec{V}_A = 0$.For point $B$ (center),$\vec{V}_B = \vec{V}_B$.For point $C$ (topmost point),$\vec{V}_C = \vec{V}_B + \vec{\omega} 	imes \vec{R}_{BC}$. Since $\vec{V}_B = \vec{\omega} 	imes \vec{R}_{BA}$ (where $\vec{R}_{BA}$ is the vector from $A$ to $B$),we have $\vec{V}_C = 2\vec{V}_B$.Now,let's evaluate the magnitudes:$|\vec{V}_C - \vec{V}_A| = |2\vec{V}_B - 0| = 2|\vec{V}_B|$.$|\vec{V}_B - \vec{V}_C| = |\vec{V}_B - 2\vec{V}_B| = |-\vec{V}_B| = |\vec{V}_B|$.Comparing these,we get $|\vec{V}_C - \vec{V}_A| = 2 |\vec{V}_B - \vec{V}_C|$.

$A$ sphere rolls without slipping on a fixed horizontal surface. In the figure,$A$ is the point of contact,$B$ is the center,and $C$ is the topmost point. Then...

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