Question

An electric motor of power $75 \,W$ rotates a flywheel of moment of inertia $0.36 \,kg - m ^{2}$ at a constant rate of $100\,rad \,s ^{-1}$. If the power is switched off, the time taken for the wheel to come to rest is

• A. 12 s
• B. 24 s
• C. 36 s
• D. 48 s

Answer 1

Answer: (D)

Given, $I=0.36\, kg - m ^{2}, \omega_{i}=100\, rad / sec$ and
$P=75 \,W$
$\because$ Angular acceleration, $\alpha=\frac{\left(\omega_{f}-\omega_{i}\right)}{t}$
or $\alpha=-\frac{\omega_{i}}{t} \,\,\,\left(\because \omega_{f}=0\right)$
or $t=\frac{\omega_{i}}{\alpha}\,\,\,...(i)$
$\because$ Power of an electric motor, $P=\tau \omega_{i}$
Torque, $\tau=I \alpha$
Where, $I$ is moment of inertia,
$\therefore P=I \alpha \omega_{i}$
$\Rightarrow \alpha=\frac{P}{I \omega_{i}}\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$t=\frac{\omega_{i}}{\left(P / I \omega_{i}\right)}=\frac{I \,\omega_{i}^{2}}{P}$
or $t=\frac{0.36 \times(100)^{2}}{75}=48\, s$