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A wheel having moment of inertia $2 \mathrm{~kg}$ $\mathrm{m}^2$ about its vertical axis, rotates at the rate of $60 \mathrm{rpm}$ about the axis. The torque which can stop the wheel's rotation in one minute would be:
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Verified Answer
The correct answer is:
$\frac{\pi 5}{15} \mathrm{~N}-\mathrm{m}$
Given, $\mathrm{I}=2 \mathrm{~kg}-m^2$,
$\omega_0=\frac{60}{60} \times 2 \pi \mathrm{rad} / \mathrm{s}$
$\omega_0$ and $t=60 \mathrm{~s}$
The torque required to stop the wheel's rotation is
$\begin{aligned}
\tau & =\mathrm{I} \propto l\left(\frac{\omega_0-\omega}{t}\right) \\
\tau & =\frac{2 \times 2 \pi \times 60}{60 \times 60} \\
& =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}
\end{aligned}$
$\omega_0=\frac{60}{60} \times 2 \pi \mathrm{rad} / \mathrm{s}$
$\omega_0$ and $t=60 \mathrm{~s}$
The torque required to stop the wheel's rotation is
$\begin{aligned}
\tau & =\mathrm{I} \propto l\left(\frac{\omega_0-\omega}{t}\right) \\
\tau & =\frac{2 \times 2 \pi \times 60}{60 \times 60} \\
& =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}
\end{aligned}$
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