A disc is performing pure rolling on a surfac | vedclass

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Question

(C) In pure rolling,the velocity of any point on the disc is the vector sum of the velocity of the center of mass $(V_C)$ and the tangential velocity

Vedclass · Accepted Answer

$V_P > V_C > V_Q$

Vedclass · Answer

(C) In pure rolling,the velocity of any point on the disc is the vector sum of the velocity of the center of mass $(V_C)$ and the tangential velocity due to rotation about the center $(V_{rot} = \omega r)$.Let $r$ be the distance of points $P$ and $Q$ from the center $C$. The velocity of the center is $V_C = \omega R$,where $R$ is the radius of the disc.The velocity of any point at distance $r$ from the center is given by $\vec{V} = \vec{V_C} + \vec{V_{rot}}$.For point $P$,which is in the upper half,the rotational velocity vector $\vec{V_{rot,P}}$ has a component in the same direction as $V_C$. Thus,its magnitude is $V_P = \sqrt{V_C^2 + V_{rot}^2 + 2 V_C V_{rot} \cos 	heta}$,where $	heta$ is the angle between the vectors. Since the components add up,$V_P > V_C$.For point $Q$,which is in the lower half,the rotational velocity vector $\vec{V_{rot,Q}}$ has a component opposing $V_C$. Thus,its magnitude is $V_Q = \sqrt{V_C^2 + V_{rot}^2 - 2 V_C V_{rot} \cos \phi}$. Since the components subtract,$V_Q Therefore,the correct relation is $V_P > V_C > V_Q$.

$A$ disc is performing pure rolling on a surface. The positions of $P$ and $Q$ at an instant are shown in the figure. $C$ is the center of the disc. Which of the following is true for their velocities at the instant when $P$ and $Q$ are at the same distance from the center?

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