Question

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

Answer 1

Step 1:Given:

$\text{mass}=m$

$\text{length}=2l$

Step 2: Calculate Angular acceleration:

$$\begin{gathered} {\omega }_f={\omega }_o+\alpha t \\ \omega =0+\alpha t \\ \alpha =\frac{\omega }{t} \end{gathered}$$

Here,${\omega }_f$ is final angular velocity, ${\omega }_0$ is initial angular velocity and $\alpha$ is angular acceleration.

Step 3:Calculate Torque:

Formula of moment of inertia is given by having length $l$.

$I=\frac{m{l}^2}{12}$

$$\begin{gathered} T=I\alpha \\ =\frac{m{\left(2l\right)}^2}{12}\times \frac{\omega }{t} \\ =\frac{m{l}^2\omega }{3} \end{gathered}$$

Here,$I$ is moment of inertia.

Hence, Torque is $\frac{m{l}^2\omega }{3}$.