Question

Two balls of mass $m$ and $4m$ are connected by a rod of length $L$. The mass of the rod is small and can be treated as zero. The size of the balls also can be neglected. We also assume the centre of the rod in hinged, but the rod can rotate about its centre in the vertical plane without friction. What is the gravity induced angular acceleration of the rod when the angle between the rod and the vertical line is $\theta$ as shown.

• A. $\frac{5g}{6L}\sin \theta$
• B. $\frac{g}{6L}\sin \theta$
• C. $\frac{6g}{5L}\sin \theta$
• D. $\frac{g}{3L}\sin \theta$

Answer 1

The correct option is C $\frac{6g}{5L}\sin \theta$

Key Point : The magnitude of torque depends on three quantities, the force applied, the length of the lever arm connecting the axis to the point of force application and the angle between the force vector and the lever arm.
Moment of inertia of the rod and balls system about the centre is
$I=4m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2=\frac{5}{4}m{L}^2$
Net torque on the connected rod of length $L$,
$\tau =4mg\frac{L}{2}\sin \theta -mg\frac{L}{2}\sin \theta =3mg\frac{L}{2}\sin \theta$
We have, $\tau =I\alpha$
So, gravity induced angular acceleration of the rod
$\alpha =\frac{\tau }{I}=\frac{3mg\frac{L}{2}\sin \theta }{\frac{5}{4}m{L}^2}=\frac{6g\sin \theta }{5L}$