Question

A rod and a block are of same mass. Initially rod is in horizontal position. What will be acceleration of tip of the rod just after the system is released from this position shown in figure?

• A. Zero
• B. $\frac{3g}{4}$
• C. $\frac{3g}{8}$
• D. $\frac{3g}{2}$

Answer 1

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

Let us consider that just after releasing the angular acceleration of rod is $\alpha$ & linear acceleration of block is a, then from constraint theory $a=l\alpha$, where l is the length of rod.



For rod, $T\times l-mg\times \frac{1}{2}=\frac{m{l}^2}{3}\alpha$
For block, mg - T = ma
Solving above equations we get,
$\frac{g}{2}=\frac{4a}{3}$
$\Rightarrow a=\frac{3g}{8}$ which is also the acceleration of free end of the rod.