Question

Uniform rod $AB$ is hinged at the end $A$ in a horizontal position as shown in the figure (the hinge is frictionless, that is, it does not exert any friction force on the rod). The other end of the rod is connected to a block through a massless string as shown. The pulley is smooth and massless. Masses of the block and the rod are same and are equal to ${}^{\prime }{m}^{\prime }$. Acceleration due to gravity is $g$. The tension in the thread, and angular acceleration of the rod just after the release of the block from this position:

• A. $\frac{3mg}{8},\frac{g}{8\ell }$
• B. $\frac{5mg}{8},\frac{3g}{8\ell }$
• C. $\frac{mg}{8},\frac{5g}{8\ell }$
• D. $\frac{7mg}{8},\frac{7g}{8\ell }$

Answer 1

The correct option is B $\frac{5mg}{8},\frac{3g}{8\ell }$

As shown in figure , $T$= Tension of string, $a$=acceleration of block, $N$ = Normal reaction at hinged point.

From the FBD of a block,

$mg-T=ma$ .............1

From FBD of rod,

Taking Moments about hinged point, $T\times l-mg\times \frac{l}{2}=\frac{m{l}^2}{3}\times \alpha$

$\Rightarrow T-mg=\frac{ml}{3}\times \alpha$

Using constraint relation $a=\alpha \times l$ .......2

$\Rightarrow T-mg=\frac{ma}{3}$ .........3

Solving Equation 1 and 2,

$\Rightarrow a=\frac{3g}{8}$, .......4

Using equation 4 in equation 1,

$\Rightarrow T=\frac{5mg}{8}$

From the equation 2 and 4, $\Rightarrow \alpha =\frac{3g}{8l}$