Question

A wheel is rotating with an angular speed of $20rad/s$. It is stopped to rest by applying a constant torque in $4s$. If the moment of inertia of the wheel about its axis is $0.20kg/m^2$, then the work done by the torque in two seconds will be

• A. 10 J
• B. 20 J
• C. 30 J
• D. 40 J

Answer 1

Answer: (C)

${\omega }_1=20rad/s,\omega =0,t=4s.$
So angular retardation $\alpha =\frac{{\omega }_1-{\omega }_2}{t}$
$=\frac{20}{4}=5rad/s^2$
Now, angular speed after $2s$
${\omega }_2={\omega }_1-\alpha t=20-5\times 2=10rad/s$
Work done by torque in $2s=$ loss in kinetic energy
$=\frac{1}{2}I\left({\omega }_1{}^2-{\omega }_1{}^2\right)=\frac{1}{2}(0.20)\left((20{)}^2-(10{)}^2\right)$
$=\frac{1}{2}(0.20)(400-100)$
$=\frac{1}{2}(020)\times (300)$
$=\frac{1}{2}\times 60=30$