Question

A wheel is rotating with an angular speed of $20 \,rad/s$. It is stopped to rest by applying a constant torque in $4s$. If the moment of inertia of the wheel about its axis is $0.20\, kg/m^2$, 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}=20\, rad / s , \omega=0, t=4 \,s . $
So angular retardation $\alpha=\frac{\omega_{1}-\omega_{2}}{t}$
$=\frac{20}{4}=5 \,rad / s ^{2}$
Now, angular speed after $2\, s$
$\omega_{2}=\omega_{1}-\alpha t=20-5 \times 2=10\, rad / s$
Work done by torque in $2 \,s =$ loss in kinetic energy
$=\frac{1}{2} I\left(\omega_{1}{ }^{2}-\omega_{1}{ }^{2}\right)=\frac{1}{2}(0.20)\left((20)^{2}-(10)^{2}\right)$
$=\frac{1}{2}(0.20)(400-100)$
$=\frac{1}{2}(020) \times(300)$
$=\frac{1}{2} \times 60=30$