Question

A wheel is rotating with an angular speed $20\text{rad/sec}$. It is brought to rest by applying constant torque in $5\text{sec}$. If the moment of inertia of the wheel about the axis is $0.1{\text{kgm}}^2$, then the work done by the torque in $2\text{sec}$ is

• A. $30\text{J}$
• B. $12.8\text{J}$
• C. $15\text{J}$
• D. $25.6\text{J}$

Answer 1

The correct option is B $12.8\text{J}$

Given initial angular speed ${\omega }_1=20\text{rad/sec}$.
Let final angular speed after $5\text{sec}$, ${\omega }_2=0$
$t=5\text{sec}$
Using equation of motion for rotational motion,
${\omega }_2={\omega }_1-\alpha t$ (here retardation is taking place)
So, angular retardation
$\alpha =\frac{{\omega }_1-{\omega }_2}{t}=\frac{20}{5}=4{\text{rad/sec}}^2$

Now , angular speed after $2\text{sec}$ is given as
${\omega }_3={\omega }_1-\alpha t=20-4\times 2$
${\omega }_3=12\text{rad/sec}$
By work energy theorem for rotation,
work done by the torque equals to the loss in kinetic energy.
$\therefore$ Work done by torque in $2\text{sec}=$ Loss
in kinetic energy$=\frac{1}{2}I({\omega }_3^2-{\omega }_1^2)$
$=\frac{1}{2}\times 0.1\left[\right(12{)}^2-\left(20{)}^2\right]$
$=-12.8\text{J}$
So, work done by the torque in $2\text{sec}$ is
$-12.8\text{J}.$