Question

A uniform rod $AB$ is hinged at end $A$ in horizontal position as shown in figure. The other end is connected to a block through a massless pulley- string arrangement as shown in figure. The acceleration of block just after release from the shown position is

• A. $\frac{6g}{13}$
• B. $\frac{3g}{8}$
• C. $\frac{g}{4}$
• D. $\frac{7g}{13}$

Answer 1

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

Angular acceleration of rod about hinged point $A$ is $\alpha$, and $a$ is the linear acceleration of block.
$\Rightarrow$For string to remain taut, the tangential acceleration of rod's end $B$ should be equal to block's acceleration $a$
$\therefore a=r\alpha$
$\alpha =\frac{a}{L}...\left(i\right)$
$\text{MOI}$ of rod about $A,I=\frac{m{L}^2}{3}$
FBD of block
$mg-T=ma...\left(ii\right)$
FBD of rod

From equation of torque for rod about $A$( Taking anticlockwise sense of rotation as $+ve$):
${\tau }_{ext}=I\alpha$
${\tau }_T+{\tau }_{mg}=I\alpha$
$\Rightarrow +TL-\frac{mgL}{2}=\left(\frac{m{L}^2}{3}\right)\alpha$
Substituting from Eq $\left(i\right)$,
$TL-\frac{mgL}{2}=\frac{m{L}^2}{3}\times \frac{a}{L}$
$\Rightarrow 2T-mg=\frac{2ma}{3}...\left(iii\right)$
Multiplying Eq $\left(ii\right)$ with $2\&$adding with Eq $\left(iii\right)$,
$2mg-mg=\frac{8ma}{3}$
$mg=\frac{8ma}{3}$
$\therefore a=\frac{3g}{8}$