Rotation Ques 13
13 Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system (see figure). Because of torsional constant $k$, the restoring torque is $\tau=k \theta$ for angular displacement $\theta$. If the rod is rotated by $\theta_0$ and released, the tension in it when it passes through its mean position will be
( 2019 Main, 9 Jan I)
(a) $\frac{2 k \theta_0^2}{l}$
(b) $\frac{k \theta_0^2}{l}$
(c) $\frac{3 k \theta_0^2}{l}$
(d) $\frac{k \theta_0^2}{2 l}$
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Answer:
Correct Answer: 13.( b )
Solution:
Thus, when it will be released, the system will execute SHM with a time period, $T=2 \pi \sqrt{\frac{I}{k}}$
(Where $I$ is moment of inertia and $k$ is torsional constant) and the angular frequency is given as, $\omega=\sqrt{\frac{k}{I}}$.
If we know look at the top view of the above figure, we have
At some angular displacement ’ $\theta_0$ ‘, at point ’ $A$ ’ the maximum velocity will be $ v_{\max }=\frac{l}{3} \theta_0 \omega=\frac{l}{3} \theta_0 \sqrt{\frac{k}{I}} $
Then, tension in the rod when it passes through mean position will be
$ \begin{aligned} T & =\frac{m \times v_{\max }^2}{\frac{l}{3}}=\frac{m l^2 \theta_0^2 k \times 3}{9 \times l \times I} \\ & =\frac{m l \theta_0^2 k}{3 I} \end{aligned} $
The moment of inertia $I$ at point $O$,
$ \begin{aligned} & =\frac{m}{2}\left(\frac{2 l}{3}\right)^2+m\left(\frac{l}{3}\right)^2=\frac{2 l^2 m}{9}+\frac{m l^2}{9}=\frac{3 m l^2}{9}=\frac{m l^2}{3} \\ \Rightarrow T & =\frac{m l \theta_0^2 k \times 3}{3 \times m l^2}=\frac{\theta_0^2 k}{l}=\frac{k \theta_0^2}{l} \end{aligned} $
BETA
