Question

The diameter of a solid disc is $0.5m$ and its mass is $16kg$.

At what rate is the work done by the torque at the end of eighth second?

• A. $\pi W$
• B. ${\pi }^{2}W$
• C. ${\pi }^{3}W$
• D. ${\pi }^{4}W$

Answer 1

The correct option is B ${\pi }^{2}W$

We have torque as $\tau =I\alpha$.
Also the final angular velocity $\omega$ is given as 120 rotations/min or 2 rotations/second or $4\pi$ radians/sec.
Thus we have $\alpha$ as $\frac{4\pi -0}{8}=\frac{\pi }{2}rad/s^2$
And moment of inertia I for solid disc is $\frac{m{r}^2}{2}$
Thus we get torque as
$\tau =(\frac{m{r}^2}{2}\times \frac{\pi }{2})$
Substituting the values we get-
$\tau =\frac{\pi }{4}N/m$

Rate of work done is defined as power.
We have power P as
$P=\tau \omega$
or
$P=\frac{\pi }{4}\times 4\pi$
or
$P={\pi }^{2}W$