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JEE Main 2013 · Rolling Motion

Rolling Motion question from JEE Main 2013

Q81JEE Main 2013Difficult

Rolling Motion

A hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it ceases to slip?

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Answer: (c)

Friction acts at the contact point, so angular momentum about the contact point is conserved.
Initial: $I\omega_0 = mr^2\omega_0$ . Final (rolling): $mvr + mr^2\dfrac{v}{r} = 2mvr$ .
$mr^2\omega_0 = 2mvr \Rightarrow v = \dfrac{r\omega_0}{2}$ .
So the answer is (c).

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