Question

Two balls of masses $m_1=1\text{kg}$ and $m_2=2\text{kg}$ are attached to a massless rod of length $(l=1\text{m})$. Two parallel forces having equal magnitude of $F=1\text{N}$, opposite in direction (couple) are acting on both the balls as shown in the figure. Find the angular acceleration of the system ($\alpha$) about an axis passing through mid-point of the rod and also perpendicular to the plane of the rod. (Take anti-clockwise rotation as +ve)

• A. $\frac{2}{3}{\text{rad/s}}^2$
• B. $\frac{4}{3}{\text{rad/s}}^2$
• C. $\frac{1}{3}{\text{rad/s}}^2$
• D. $\frac{1}{6}{\text{rad/s}}^2$

Answer 1

The correct option is B $\frac{4}{3}{\text{rad/s}}^2$

(Taking anti-clockwise rotation as +ve)
We know that,
Torque due to couple,
$\tau =$ Force $\times$ perpendicular distance between the two forces
$\Rightarrow \tau =F\times l$
$\Rightarrow \tau =1\text{Nm}$

Moment of inertia of system about the given axis,
$I=m_1{r}_1^2+m_2{r}_2^2$
$\Rightarrow I=1\times {\left(\frac{l}{2}\right)}^2+2\times {\left(\frac{l}{2}\right)}^2$
$=1\times {\left(\frac{1}{2}\right)}^2+2\times {\left(\frac{1}{2}\right)}^2=\frac{3}{4}{\text{kgm}}^2$
Now, we also know,
$\tau =I\alpha$
$\therefore \alpha =\frac{\tau }{I}=\frac{1}{\frac{3}{4}}$
$\Rightarrow \alpha =\frac{4}{3}{\text{rad/s}}^2$