Question

The diameter of a solid disc is $0.5$ m and its mass is $16$ Kg. What torque will increase its angular velocity from zero to $120$ rotation/minute in $8$ seconds ?

• A. $\frac{\pi }{4}N/m$
• B. $\frac{\pi }{2}N/m$
• C. $\frac{\pi }{3}N/m$
• D. $\pi N/m$

Answer 1

The correct option is A $\frac{\pi }{4}N/m$

Final angular velocity :$\omega =\frac{120\times 2\pi }{60}=4\pi rad/s$.

Thus its acceleration at the end of the 8th second will be $\alpha \frac{4\pi -0}{8}=\frac{\pi }{2}rad/s^2$. Torque is calculated as

$\tau =I\alpha =\frac{m{r}^2}{2}\alpha =\frac{16\left(\right(0.25{)}^2)}{2}\frac{\pi }{2}=\pi /4Nm$