Question

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

What torque is required to increase its angular velocity about an axis perpendicular to its plane from zero to $120$ rotations/minute in $8$ seconds?

• A. $\frac{\pi }{4}N/m$
• B. $\frac{\pi }{2}N/m$
• C. $\pi N/m$
• D. $2\pi N/m$

Answer 1

Answer: (A)

$\frac{\pi }{4}N/m$

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-
$\tau =\frac{\pi }{4}N/m$