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A thin rod of mass $m$ and length $2 l$ is made to rotate about an axis passing through its centre and perpendicular to it. Its angular velocity changes from 0 to $\omega$ in time $t$. What is the torque acting on the rod?
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Verified Answer
The correct answer is:
$\frac{m l^2 \omega}{3 t}$
Given that, mass of rod $=m$
Length of $\operatorname{rod}, L=2 l$
Initial and final angular velocities are 0 and $\omega$ respectively,
Time taken $=t$
We know that, torque acting on body having moment of inertia ( $(I)$ and angular acceleration $(\alpha)$ is
where,
$$
\tau=I \alpha
$$
$$
\begin{aligned}
I & =\frac{1}{12} M L^2=\frac{1}{12} m(2 l)^2 \\
& =\frac{1}{3} m l^2
\end{aligned}
$$
and $\quad \alpha=\frac{\omega_f-\omega_i}{t}=\frac{\omega-0}{t}=\frac{\omega}{t}$
Now, substituting the values, we get
$$
\tau=\frac{1}{3} m l^2 \frac{\omega}{t}
$$
Length of $\operatorname{rod}, L=2 l$
Initial and final angular velocities are 0 and $\omega$ respectively,
Time taken $=t$
We know that, torque acting on body having moment of inertia ( $(I)$ and angular acceleration $(\alpha)$ is
where,
$$
\tau=I \alpha
$$
$$
\begin{aligned}
I & =\frac{1}{12} M L^2=\frac{1}{12} m(2 l)^2 \\
& =\frac{1}{3} m l^2
\end{aligned}
$$
and $\quad \alpha=\frac{\omega_f-\omega_i}{t}=\frac{\omega-0}{t}=\frac{\omega}{t}$
Now, substituting the values, we get
$$
\tau=\frac{1}{3} m l^2 \frac{\omega}{t}
$$
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