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A wheel is rotating with an angular speed of $20\,rad/sec$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg-m^2$, then the work done by the torque in two seconds will be .......... $J$
$10$
$20$
$30$
$40$
Solution
${\omega _1} = 20$ $rad/sec$,
${\omega _2} = 0,\,t = 4sec.$
So angular retardation $\alpha = \frac{{{\omega _1} – {\omega _2}}}{t} = \frac{{20}}{4} = 5\ rad/se{c^2}$
Now angular speed after $2\ sec$,
${\omega _2} = {\omega _1} – \alpha t$$ = 20 – 5 \times 2$ = $10\,$$rad/sec$
Work done by torque in $2\ sec$ = loss in kinetic energy = $\frac{1}{2}I\,\left( {\omega _1^2 – \omega _2^2} \right)$$ = \frac{1}{2}(0.20)\,({(20)^2} – {(10)^2})$
$ = \frac{1}{2} \times \,\,0.2 \times 300$= $30\ J.$
