First slide

Dynamics of rotational motion about fixed axis

Question

A uniform rod of length l and mass m is suspended by two vertical inextensible string as shown in the fig. The tension in the left string when right string snaps, is

Difficult

A

mg/4
B

3mg/4
C

mg/2
D

mg/8

Solution

mg - T = ma.......(1) where, a = Acceleration of centre of mass
$\large \alpha = \frac{\tau }{I} = \frac{{mg\frac{l}{2}}}{{\frac{{m{l^2}}}{3}}} = \frac{{3g}}{{2l}}......(2)$ where α = Angular acceleration
$\large a=\frac l2\alpha......(3)$
solving equations (1), (2) and (3)
we get, $\large T=\frac {mg}{4}$

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