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(A) For a disc rolling without slipping,the velocity of any point at a distance $r'$ from the instantaneous axis of rotation $O$ is given by $v = r' \

Vedclass · Accepted Answer

$v_Q > v_C > v_P$

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(A) For a disc rolling without slipping,the velocity of any point at a distance $r'$ from the instantaneous axis of rotation $O$ is given by $v = r' \omega$,where $\omega$ is the angular velocity of the disc.In the given figure,$O$ is the point of contact with the ground,which is the instantaneous center of rotation.The distances of points $P$,$C$,and $Q$ from the instantaneous center $O$ are $r_P$,$r_C$,and $r_Q$ respectively.From the geometry of the disc,it is clear that $r_P Since $v = r' \omega$ and $\omega$ is constant for all points on the disc,the velocity is directly proportional to the distance from the instantaneous center $O$.Therefore,$v_P  v_C > v_P$.

$A$ disc rolls on a horizontal surface (without slipping). $C$ is the center and $Q$ and $P$ are two points at the same distance from $C$. Let $v_P$,$v_Q$,and $v_C$ be the magnitudes of the velocities of points $P$,$Q$,and $C$ respectively. Then:

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