Question

A wheel is rotating with an angular speed of 20 rad/sec. 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

The correct option is C 30 J

${\omega }_1=20$ rad\sec, ${\omega }_2=0,$ t = 4 sec. So angular retardation $\alpha =\frac{{\omega }_1-{\omega }_2}{t}=\frac{20}{4}=5rad{\sec }^2$
Now angular speed after 2 sec ${\omega }_2={\omega }_1-\alpha t=20-5\times 2=10rad\sec$
Work done by torque in 2 sec = loss in kinetic energy $=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)=\frac{1}{2}\left(0.20\right)\left(\right(20{)}^2-\left(10{)}^2\right)$
$=\frac{1}{2}\times 0.2\times 300=30J.$