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  • 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 <img alt="\pi :2

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 the, the value of its angular speed will be rad/s.

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Angular momentum, 

\begin{aligned} & \mathrm{L}=\mathrm{I} \omega \\ & \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\ & \text { Energy }=\frac{1}{2} \mathrm{MV}^2+\frac{1}{2} I \omega^2 \\ & \mathrm{E}=\frac{1}{2} M(\omega \mathrm{R})^2+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^2\right) \omega^2 \end{aligned}

\begin{aligned} & =\frac{7}{10} \mathrm{M} \omega^2 \mathrm{R}^2 \\ & \frac{\mathrm{L}}{\mathrm{E}}=\frac{4}{7 \omega}=\frac{\pi}{22} \\ & \omega=\frac{88}{7 \pi}=\frac{88}{7 \times \frac{22}{7}}=4 \mathrm{rad} / \mathrm{s} \end{aligned}

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shivangi.bhatnagar

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