Rotational Motion

JEE Main level, numerical answer, medium

144 of 280

Question 144 of 280

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25πNm25 \pi \mathrm{Nm} for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ________ m.

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Answer: 40

Given, m=1 kg\mathrm{m}=1 \mathrm{~kg}
$\begin{aligned}
& \omega_{\mathrm{i}}=1800 \mathrm{rpm}=1800 \times \frac{2 \pi}{60}=60 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\
& \omega_{\mathrm{f}}=2100 \mathrm{rpm}=2100 \times \frac{2 \pi}{60}=70 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\
& \tau_{e x t}=25 \pi \mathrm{Nm} \\
& \mathrm{t}=40 \mathrm{sec}
\end{aligned}$
Using equation of motion
$\begin{aligned}
& \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t} \\
& 70 \pi=60 \pi+\alpha(40) \\
& \alpha=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}^2
\end{aligned}$
Also, τ=Iα\tau=\mathrm{I} \alpha
τ=mR24α\tau=\frac{\mathrm{mR}^2}{4} \alpha
$\begin{aligned}
& 25 \pi=\frac{1 \times \mathrm{R}^2}{4} \times \frac{\pi}{4} \\
& \mathrm{R}=20 \mathrm{~m}
\end{aligned}$
Hence, diameter of disk =2R=2×20=40 m=2 \mathrm{R}=2 \times 20=40 \mathrm{~m}

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