Question

A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of $\frac{2L}{3}$ from the hinge so that the rod swings to the right. The horizontal reaction at the hinge is

• A. -P
• B. 0
• C. $\frac{P}{3}$
• D. $\frac{2P}{3}$

Answer 1

The correct option is B

0

at this moment
$\omega =0$
but $\alpha \ne 0\text{[start of rotation]}$

$\overrightarrow{a_{CM}}=\overrightarrow{a_r}+\overrightarrow{a_t}$

(because centre of mass is in circular motion at a radius $\frac{L}{2})$

$\overrightarrow{a_r}=0[\therefore \omega =0]$

${\overrightarrow{a}}_{CM}=\overrightarrow{a_t}=\frac{L}{2}\alpha \left(\hat{i}\right)$

Torque, $T=I\alpha \text{(about hinge)}$

Taking moment about A,

$P\times \frac{2L}{3}=\frac{M{L}^2}{3}\alpha$

$\alpha =\frac{2P}{ML}$

${\overrightarrow{a}}_{CM}=\frac{L}{2}\times \frac{2P}{ML}\hat{i}$

${\overrightarrow{a}}_{CM}=\frac{P}{M}\hat{i}$

Let R be the reaction in horizontal direction at hinge:
[In horizontal direction]

$(\Sigma \overrightarrow{f}{)}_{ext}=M{\overrightarrow{a}}_{CM}$

$P+R=M{a}_{CM}$

$P+R=M\times \frac{P}{M}$

$P+R=P$
$R=0$