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JEE Main 2023 (13 Apr, Shift 1) · Rolling Motion

Rolling Motion question from JEE Main 2023 (13 Apr, Shift 1)

Q40JEE Main 2023 · 13 APR · SHIFT 1Medium

Rolling Motion

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $\pi : 22$ then, the value of its angular speed will be ____________ $\text{rad/s}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$L = I\omega = \frac{2}{5}mR^2\omega$ .
Total KE of rolling sphere: $E = \frac{7}{10}mv^2 = \frac{7}{10}mR^2\omega^2$ .
$\dfrac{L}{E} = \dfrac{4}{7\omega} = \dfrac{\pi}{22} \Rightarrow \omega = \dfrac{88}{7\pi}$
With $\pi = \frac{22}{7}$ : $\omega = 4$ rad/s.

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