Rotational Motion

JEE Main level, numerical answer, medium

149 of 280

Question 149 of 280

A string is wrapped around the rim of a wheel of moment of inertia 0.40kgm20.40 \mathrm{kgm}^2 and radius 10 cm10 \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 N40 \mathrm{~N}. The angular velocity of the wheel after 10 s10 \mathrm{~s} is xrad/sx \mathrm{rad} / \mathrm{s}, where xx is _______

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Answer: 100

τ=FR=Iα⇒40×0.1=0.4αα=10rad/s2 Wf=10×10=100rad/s\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}

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