Question

A cord of negligible mass is wound round the rim of a flywheel of mass $20kg$ and radius $20cm$. A steady pull of $25N$ is applied on the cord. The work done by the pull when $2m$ of the cord is unwound is, if wheel starts from rest, what is the kinetic energy of the wheel when .$2m$ of the cord is unwound?

• A. $20J$
• B. $25J$
• C. $45J$
• D. $50J$

Answer 1

Answer: (D)

Here, $R=20cm=0.2m,M=20kg$
As $\tau =FR=(25N)(0.2m)=5Nm$
Moment of inertia of flywheel about its axis is
$I=\frac{M{R}^2}{2}$
$=\frac{(20kg)(0.2m{)}^2}{2}$
$=0.4kg{m}^2$
As $\tau =I\alpha$
$\therefore$ Angular acceleration of the wheel,
$\alpha =\frac{\tau }{I}=\frac{5Nm}{0.4kg{m}^2}=12.5rad{s}^{-2}$
Angular displacement of wheel,
$\theta =\frac{\text{Length of unwound string}}{\text{Radius of the wheel}}$
$=\frac{2m}{0.2m}=10rad$
Let $\omega$ be final angular velocity.
As ${\omega }^2={\omega }_0^2+2\alpha \theta$
Since the wheel starts from rest, therefore ${\omega }_0=0$
$\therefore {\omega }^2=2\times (12.5rad{s}^{-2})(10rad)=250ra{d}^2{s}^{-2}$
$\therefore$ Kinetic energy gained, $K=\frac{I}{2}I{\omega }^2$
$K=\frac{1}{2}\times 0.4kg{m}^2\times 250ra{d}^2{s}^{-2}=50J$