A disc of radius $R$ has a light pole fixed perpendicular to the disc at the circumference which in

Question

A disc of radius $R$ has a light pole fixed perpendicular to the disc at the circumference which in turn has a pendulum of length $R$ attached to its other end as shown in figure. The disc is rotated with a constant angular speed $\omega$. The string is making an angle $30^\circ$ with the rod. Then the angular speed $\omega$ of disc is:

Professor PC Thomas Classes · Answer

Radius of bob's path $r = R + R\sin 30^\circ = 3R/2$. Vertical force balance: $T\cos 30^\circ = mg$. Horizontal centripetal force: $T\sin 30^\circ = m\omega^2(3R/2)$. Dividing the two gives $	an 30^\circ = \frac{\omega^2(3R/2)}{g} \implies \frac{1}{\sqrt{3}} = \frac{3\omega^2 R}{2g} \implies \omega = \left(\frac{2g}{3\sqrt{3}R}ight)^{1/2}$.