Question

A light rod of length $l=1\text{m}$ is hinged at its centre and two masses of $m_1=5\text{kg}$ and $m_2=2\text{kg}$ are hung from the ends as shown in figure. Find the initial angular acceleration of the rod assuming that it was horizontal in beginning. (Take $g=10m/s^2$)

• A. $8.5{\text{rad/s}}^2$
• B. $4.4{\text{rad/s}}^2$
• C. $3.4{\text{rad/s}}^2$
• D. $5.4{\text{rad/s}}^2$

Answer 1

The correct option is A $8.5{\text{rad/s}}^2$

Given:
$m_1=5\text{kg}$ and $m_2=2\text{kg}$
$l=1\text{m}$


$\tau =r\times F=I\alpha$
Taking anticlockwise moment as positive,
$\Rightarrow \frac{l}{2}\times m_{1}g-\frac{l}{2}\times m_{2}g=I\alpha$
$\Rightarrow \frac{1}{2}\times 5\times g-\frac{1}{2}\times 2\times g=I\alpha$
$\Rightarrow \frac{3g}{2}=I\alpha ....\left(i\right)$

We know,
$I=M{R}^2=m_1{\left(\frac{l}{2}\right)}^2+m_2{\left(\frac{l}{2}\right)}^2=5{\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^2=\frac{7}{4}{\text{kg-m}}^2....\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$\frac{3}{2}\times 10=\frac{7}{4}\alpha$
$\Rightarrow \alpha =8.5{\text{rad/s}}^2$