Physics - Rotation Question with Solution | TestHub

Key PointThe 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,

So, gravity induced angular acceleration of the rod

$\alpha =\frac{\tau }{I}=\frac{3mg\frac{L}{2}\sin \theta }{\frac{5}{4}m{L}^2}=\frac{6g\sin \theta }{5L}$