A string is wrapped around the rim of a wheel of moment of inertia $0.40 \mathrm{kgm}^2$ and radius $10…

A string is wrapped around the rim of a wheel of moment of inertia $0.40 \mathrm{kgm}^2$ and radius $10 \mathrm{~cm}$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40 \mathrm{~N}$. The angular velocity of the wheel after $10 \mathrm{~s}$ is $x \mathrm{rad} / \mathrm{s}$, where $x$ is _______

Solution

$\begin{aligned} & \tau=\mathrm{FR}=\mathrm{I} \alpha \Rightarrow 40 \times 0.1=0.4 \alpha \\ & \alpha=10 \mathrm{rad} / \mathrm{s}^2 \\ & \mathrm{~W}_{\mathrm{f}}=10 \times 10=100 \mathrm{rad} / \mathrm{s}\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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