Question

Uniform rod $AB$ is hinged at end $A$ in horizontal position as shown in the figure. The other end is connected to a block of mass $m$ through a massless string as shown. The pulley is smooth and massless. Mass of block and rod is same and is equal to $m$. Then find the acceleration of the block just after release.

• A. $\frac{6g}{13}$
• B. $\frac{g}{4}$
• C. $\frac{3g}{8}$
• D. None

Answer 1

The correct option is C $\frac{3g}{8}$

Writing Newton's second law for the block,
$F_{net}=ma$
$mg-T=ma\dots \dots \left(1\right)$
Writing the equation of torgue on the rod about the hinge,
$\tau =I\alpha$
$Tl-mg\frac{l}{2}=\frac{m{l}^2\alpha }{3}=\frac{m{l}^2}{3}\frac{a}{l}\dots \dots \left(2\right)$

[Since $\alpha =\frac{a}{l}$]
Solving (1) and (2) $\Rightarrow a=\frac{3g}{8}$