Question

A thin rod of mass $m$ and length $2L$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $\omega$ in time $t,$ the torque acting on it is

• A. $\frac{ m \ell^{2} \omega}{12 t }$
• B. $\frac{ m \ell^{2} \omega}{3 t }$
• C. $\frac{ m \ell^{2} \omega}{ t }$
• D. $\frac{4 m \ell^{2} \omega}{3 t }$

Answer 1

Answer: (B)

Since, $\tau=I \alpha$
So, $\tau=\left(\frac{ m (2 \ell)^{2}}{12}\right)\left(\frac{\omega}{ t }\right)$
or $\tau=\frac{ m \times 4 \ell^{2} \times \omega}{12 \times t }$
Or $\tau=\frac{4 m \ell^{2} \omega}{12 t}=\left(\frac{m \ell^{2} \omega}{3 t}\right)$