Question

A particle of mass m is attached to an end of a uniform rod of mass $M=2\text{m}$ and length '$l$' which is suspended through its mid point by an inextensible string as shown. Initially, the rod is in horizontal position and at rest. The system is released from this position. Just after the release.

• A. The angular acceleration of the system is $\frac{6g}{5l}$
• B. The angular acceleration of the system is $\frac{2g}{5l}$
• C. The tension in the string is $\frac{12}{5}\text{mg}$
• D. The tension in the string is $\frac{2}{5}\text{mg}$

Answer 1

Answer: (A), (C)

The correct options are
A The angular acceleration of the system is $\frac{6g}{5l}$
C The tension in the string is $\frac{12}{5}\text{mg}$


The FBDs of the rod and the ball are shown. Applying $\tau =I\alpha$ about the C.M. of the rod, we have, $N\left(\frac{l}{2}\right)=\left(\frac{M{l}^2}{12}\right)\alpha ....\left(1\right)$
Writing Newton's second law in the vertical direction on the CM of the rod, we have $T-N-2mg=0$ and writing Newton's II law in the vertical direction on the ball we have,
$mg-N=m\left(\frac{l}{2}\right)\alpha .....\left(2\right)$
From eq. (1) and (2),
$mg-\left(m\frac{l}{2}\right)\alpha =N$
$(mg-\frac{ml}{2}\alpha )\left(\frac{l}{2}\right)=\frac{M{l}^2}{12}\alpha$
$\text{mg}\frac{l}{2}-\frac{m{l}^2}{4}\alpha =\frac{2m{l}^2\alpha }{12}$
$g\frac{l}{2}-\frac{l^2}{4}\alpha =\frac{l^2\alpha }{6}$
$\frac{gl}{2}=l^2\alpha (\frac{1}{6}+\frac{1}{4})$
$\frac{g}{2}=l\alpha \left(\frac{2+3}{12}\right)$
$\alpha =\frac{6g}{5l}$
$\therefore$ Tension will be calculated as
$3\text{mg}-T=\frac{3mg}{5}$
$T=\frac{12}{5}\text{mg}$