A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of $2 \mathrm{~rad} / \mathrm{s}^2$. Moment of inertia of the wheel is _____ $\mathrm{kg} \mathrm{m}^2$.
(Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{FR}=\mathrm{I} \alpha \ & \Rightarrow \mathrm{I}=\frac{\mathrm{FR}}{\alpha}=\frac{10 	imes 0.2}{2}=1 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled…

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