A wheel having moment of inertia $40 \mathrm{kgm}^2$ about its axis, rotates at $50 \mathrm{rpm}$. The…
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 $^{-2}$.
$\frac{\pi}{45}$
$\frac{\pi}{30}$
$\frac{\pi}{54}$
$\frac{\pi}{24}$
Solution
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