Question

For a wheel rotating at a rate of $60\text{rpm}$ about an axis passing through its centre, the torque required to stop wheel's rotation in 2 minute will be (Given : MOI of wheel about its axis is $4{\text{kg m}}^2)$

• A. $\frac{\pi }{15}\text{N.m}$
• B. $\frac{2\pi }{15}\text{N.m}$
• C. $\frac{\pi }{30}\text{N.m}$
• D. $\frac{\pi }{60}\text{N.m}$

Answer 1

The correct option is A $\frac{\pi }{15}\text{N.m}$

Given, $I=4{\text{kg.m}}^2$
Let the torque required to stop wheel's rotation be $\tau$
$\Rightarrow \tau =I\alpha ......\left(i\right)$
$\alpha \to$ magnitude of angular acceleration
$\Delta t=2\text{min}=120\text{sec}$
From kinematic relationship
$\alpha =\frac{w_f-w_i}{\Delta t}=\frac{0-2\pi }{120}$
$\Rightarrow \alpha =\frac{-\pi }{60}{\text{rad/s}}^2$
Putting in Eq. ($i$), magnitude of torque:
$\Rightarrow \tau =I\alpha =\left(4\right)\left(\frac{\pi }{60}\right)$
$\therefore \tau =\frac{\pi }{15}\text{N.m}$