Question

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

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

Answer 1

The correct option is C $30J$

${\omega }_1=20rad/s,\omega =0,t=4s$. So angular retardation

$\alpha =\frac{{\omega }_1-{\omega }_2}{t}=\frac{20}{4}=5rad/s^2$

Now, angular speed after $2s$

${\omega }_2={\omega }_1-\alpha t=10rad/s$

work done by torque in $2s$ $=$ loss in kinetic energy

$=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)=\frac{1}{2}\left(0.20\right)[{\left(20\right)}^2-{\left(10\right)}^2]=\frac{1}{2}\times 60=30$