Question

An electric motor of power $75W$ rotates a flywheel of moment of inertia $0.36kg-m^2$ at a constant rate of $100rad{s}^{-1}$. 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, $I=0.36kg-m^2,{\omega }_i=100rad/sec$ and
$P=75W$
$\because$ 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 $t=\frac{0.36\times (100{)}^2}{75}=48s$