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A wheel having moment of inertia $40 \mathrm{kgm}^2$ about its axis, rotates at $50 \mathrm{rpm}$. The angular retardation required to stop this wheel in $90 \mathrm{~s}$ is rads $^{-2}$.
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Verified Answer
The correct answer is:
$\frac{\pi}{54}$
Given, moment of inertia, $I$
$$
=40 \mathrm{~kg}-\mathrm{m}^2
$$
Initial angular frequency, $\omega_i=50 \mathrm{rpm}$
$$
=\frac{50 \times 2 \pi}{60}=\frac{5 \pi}{3} \mathrm{rads}^{-1}
$$
Final angular velocity, $\omega_f=0 \mathrm{rads}^{-1}$
Time, $t=90 \mathrm{~s}$
Let, $\alpha$ be the angular retardation and since, $\omega_f-\omega_i=\alpha t$
Therefore, $\alpha=\frac{\omega_f-\omega_i}{t}$
$$
=\frac{0-\frac{5 \pi}{3}}{90}=\frac{-\pi}{3 \times 18}=\frac{-\pi}{54} \mathrm{rad} \mathrm{s}^{-2}
$$
Hence, negative sign is for retardation.
$$
=40 \mathrm{~kg}-\mathrm{m}^2
$$
Initial angular frequency, $\omega_i=50 \mathrm{rpm}$
$$
=\frac{50 \times 2 \pi}{60}=\frac{5 \pi}{3} \mathrm{rads}^{-1}
$$
Final angular velocity, $\omega_f=0 \mathrm{rads}^{-1}$
Time, $t=90 \mathrm{~s}$
Let, $\alpha$ be the angular retardation and since, $\omega_f-\omega_i=\alpha t$
Therefore, $\alpha=\frac{\omega_f-\omega_i}{t}$
$$
=\frac{0-\frac{5 \pi}{3}}{90}=\frac{-\pi}{3 \times 18}=\frac{-\pi}{54} \mathrm{rad} \mathrm{s}^{-2}
$$
Hence, negative sign is for retardation.
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