Question

Uniform rod $AB$ of mass $3\text{kg}$ and length $2\text{m}$ is hinged at end $A$ in horizontal position as shown in the figure. The other end is connected to a block of mass $1\text{kg}$ through a massless string. The pulley is smooth and massless. Find the angular acceleration of the rod about the hinge just after release. (Take anti-clockwise rotation positive)

• A. $\frac{g}{8}$
• B. $\frac{g}{4}$
• C. $-\frac{g}{8}$
• D. $-\frac{g}{2}$

Answer 1

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


Given: $l=2\text{m},m=1\text{kg},M=3\text{kg}$
F.B.D of block $m$:


$mg-T=ma$ ... (i)
F.B.D of rod:


Due to string constraint, acceleration of point $B\left(a_B\right)$ will be equal to $a{\text{m/s}}^2$
We know,
$a_B=\alpha \times 2$ (for pure rotation)
$\Rightarrow \alpha =\frac{a_B}{2}=\frac{a}{2}$
$\Rightarrow a=2\alpha$
Torque equation about point $A$
$l\times T-\frac{l}{2}\times Mg=I\alpha$
$2\times (mg-ma)-\frac{l}{2}\times Mg=\frac{M{l}^2}{3}\times \alpha$
$2(g-a)-1\times 3g=4\alpha$
$2(g-2\alpha )-3g=4\alpha$
$\therefore \alpha =\frac{-g}{8}$
(- ve) sign indicates rod will have angular acceleration in clockwise direction.