Question

A wheel having moment of inertia $I=4{\text{kg-m}}^2$ about its natural axis is rotating at the rate of ${\omega }_o=120\text{rpm}$. Find the magnitude of constant torque which can stop the wheel in $3$ minutes.

• A. $4\pi /45\text{N-m}$
• B. $2\pi /45\text{N-m}$
• C. $16\pi /3\text{N-m}$
• D. $160\text{N-m}$

Answer 1

The correct option is A $4\pi /45\text{N-m}$

Given ,
$I=4{\text{kg-m}}^2$
$w_o=120\text{rpm}=120\times \frac{2\pi }{60}\text{rad/s}$
${\omega }_o=4\pi \text{rad/s}$

Let constant angular acceleration required to stop wheel's rotation is $\alpha$
then,
As wheel stops,
$\therefore$ final angular velocity, $\omega =0$
and given: $t=3$ min$=(3\times 60)\text{s}$
$w=w_o+\alpha t$
$\Rightarrow 0=w_o+\alpha (3\times 60)$
$\Rightarrow \alpha =-\frac{4\pi }{3\times 60}{\text{rad/s}}^2$

We know,
Torque $\tau =I\alpha$
$=4\times \frac{-4\pi }{3\times 60}$
$=\frac{-4\pi }{45}\text{N-m}$
Here $-ve$ sign indicates torque is opposing the motion of wheel.
$\therefore |\tau |=\frac{4\pi }{45}\text{N-m}$