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A flywheel having moment of inertia $10 \mathrm{~kg}-\mathrm{m}^2$ is rotating at $50 \mathrm{rad} \mathrm{s}^{-1}$. What amount of work needs to be done in order to bring this flywheel to rest in $10 \mathrm{~s}$ ?
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Verified Answer
The correct answer is:
$12500 \mathrm{~J}$
Given that, moment of inertia, $I=10 \mathrm{~kg} \cdot \mathrm{m}^2$
Initial angular velocity, $\omega_0=50 \mathrm{rad} / \mathrm{s}$
Final angular velocity, $\omega=0$
Time taken to stop wheel, $t=10 \mathrm{~s}$
We know that, work done by torque
= change of rotational $\mathrm{KE}$
$$
\begin{aligned}
W & =\Delta \mathrm{KE}=K_f-K_i \\
W & =\frac{1}{2} I \omega^2-\frac{1}{2} I \omega_0^2 \\
W & =0-\frac{1}{2} \times 10 \times(50)^2 \\
& =-12500 \mathrm{~J}
\end{aligned}
$$
So, amount of work done to bring the flywheel to rest is $12500 \mathrm{~J}$.
Initial angular velocity, $\omega_0=50 \mathrm{rad} / \mathrm{s}$
Final angular velocity, $\omega=0$
Time taken to stop wheel, $t=10 \mathrm{~s}$
We know that, work done by torque
= change of rotational $\mathrm{KE}$
$$
\begin{aligned}
W & =\Delta \mathrm{KE}=K_f-K_i \\
W & =\frac{1}{2} I \omega^2-\frac{1}{2} I \omega_0^2 \\
W & =0-\frac{1}{2} \times 10 \times(50)^2 \\
& =-12500 \mathrm{~J}
\end{aligned}
$$
So, amount of work done to bring the flywheel to rest is $12500 \mathrm{~J}$.
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