Question

A wheel having moment of inertia $(I=2{\text{kg-m}}^2)$ is rotating about its natural axis. The instantaneous angular velocity of a point on a rotating wheel is given by the equation $\omega \left(t\right)=(2t^2+4)\text{rad/s}$. Find the torque acting on wheel about its axis of rotation at $t=2\text{s}$.

• A. $14\text{Nm}$
• B. $12\text{Nm}$
• C. $10\text{Nm}$
• D. $16\text{Nm}$

Answer 1

The correct option is D $16\text{Nm}$

Given:
$\omega \left(t\right)=2t^2+4$
We know that,
$\alpha =\frac{d\omega }{dt}$
$\therefore \alpha \left(t\right)=\frac{d}{dt}(2t^2+4)=4t$
$\therefore \alpha {|}_{(t=2\text{s})}=4\times 2=8{\text{rad/s}}^2$
Again,
$\tau =I\alpha$
(Given: $I=2{\text{kg-m}}^2$)
$\tau =2\times 8$
$\tau =16\text{Nm}$
$\therefore$ option (d) is correct.