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Two masses m and 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 rod is rotated by $\theta _{0}$ and released, the tension in it when it passes through its mean position will be:

Option 1)
$\frac{3k\theta _{0}\: ^{2}}{l}$

Option 2)
$\frac{2k\theta _{0}\: ^{2}}{l}$
Option 3)
$\frac{k\theta _{0}\: ^{2}}{l}$
Option 4)
$\frac{k\theta _{0}\: ^{2}}{2l}$

Answers (3)

best_answer

Analogue of second law of motion for pure rotation -

$\vec{\tau }=I\, \alpha$

- wherein

Torque equation can be applied only about two point

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

$I = Ml^{2} = \frac{\frac{m^{2}}{2}}{\frac{3m}{2}}l^{2}$

= $\frac{ml^{2}}{3} - (1)$

$\frac{r_{1}}{r_{2}} = \frac{1}{2} \Rightarrow r_{1} = \frac{l}{3} - (2)$

$\Omega = \omega \theta _{0} = Average \ \ velocity$

$T = m\Omega ^{2} r_{1} \rightarrow from (2)$

$= m\Omega ^{2} \frac{l}{3}$

$= m\omega ^{2} \theta_{0}^{2} \frac{l}{3}$

$\omega = \sqrt{\frac{K}{I}} = \sqrt{\frac{3K}{ml^{2}}} \ \ (from (1))$

$\therefore T = m{\frac{3K}{ml^{2}}} \theta _{0}^{2} \frac{l}{3}$

$= \frac{K \theta _{0}^{2} }{l}$

Option 1)

$\frac{3k\theta _{0}\: ^{2}}{l}$

Option 2)

$\frac{2k\theta _{0}\: ^{2}}{l}$

Option 3)

$\frac{k\theta _{0}\: ^{2}}{l}$

Option 4)

$\frac{k\theta _{0}\: ^{2}}{2l}$

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