By energy conservation
$$\eqalign{ & {\left( {K.E} \right)_i} + {\left( {P.E} \right)_i} = {\left( {K.E} \right)_f} + {\left( {P.E} \right)_f} \cr & {\left( {K.E} \right)_i} = 0,{\left( {P.E} \right)_i} = mgh,{\left( {P.E} \right)_f} = 0 \cr & {\left( {K.E} \right)_f} = \frac{1}{2}I{\omega ^2} + \frac{1}{2}mv_{cm}^2 \cr} $$
Where $$I$$ (moment of inertia) $$ = \frac{1}{2}m{R^2}\left( {{\text{for}}\,{\text{solid}}\,{\text{cylinder}}} \right)$$
$$\eqalign{ & {\text{so}}\,\,mgh = \frac{1}{2}\left( {\frac{1}{2}m{R^2}} \right)\left( {\frac{{v_{cm}^2}}{{{R^2}}}} \right) + \frac{1}{2}mv_{cm}^2 \cr & \Rightarrow {v_{cm}} = \sqrt {\frac{{4gh}}{3}} \cr} $$

Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement.
$$\therefore \omega = \frac{\theta }{t} = \frac{{2\pi }}{T} = 2\pi \nu \,\,\left( {\because T = \frac{1}{\nu }} \right)$$
Number of revolutions made by the engine wheel $$\left( \nu \right) = 90/\min .$$
$$\therefore $$ Angular velocity of the engine wheel
$$\omega = \frac{{2\pi \nu }}{{60}} = \frac{{2\pi \times 90}}{{60}} = 3\pi \,rad/s$$

$$\eqalign{ & {\bf{KEY }}\,{\bf{CONCPET :}} \cr & {\left( {K.E.} \right)_{{\text{rotation}}}}{\text{ = }}\frac{{{L^2}}}{{2I}} \cr & {\text{Here , }}L = {\text{constant}} \cr & \therefore {\left( {K.E.} \right)_{{\text{rotation}}}} \times I = {\text{constant}} \cr & {\text{When }}I{\text{ is doubled, }}K.E{._{{\text{rotation}}}}{\text{ becomes half}}{\text{.}} \cr} $$

Note : When we are giving an angular acceleration to the rod, the bead is also having an instantaneous acceleration $$a = L\alpha .$$   This will happen when a force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen to the extent to which frictional force is capable of holding the bead.
The frictional force here provides the necessary centripetal force. If instantaneous angular velocity is $$\omega $$  then
$$\eqalign{ & mL{\omega ^2} = \mu \left( {ma} \right) \cr & \Rightarrow mL{\omega ^2} = \mu \,mL\alpha \cr & \Rightarrow {\omega ^2} = \mu \alpha \cr} $$
By applying $$ \Rightarrow \omega = {\omega _0} + \alpha t,$$
We get $$\omega = \alpha t$$
$$\eqalign{ & \therefore {\alpha ^2}{t^2} = \mu \alpha \cr & \Rightarrow t = \sqrt {\frac{\mu }{\alpha }} \cr} $$

$$\eqalign{ & F = 20t - 5{t^2} \cr & \therefore \alpha = \frac{{FR}}{I} = 4t - {t^2} \cr & \Rightarrow \frac{{d\omega }}{{dt}} = 4t - {t^2} \cr & \Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)dt} \cr & \Rightarrow \omega = 2{t^2} - \frac{{{t^3}}}{3}\,\left( {{\text{as}}\omega = 0\,{\text{at }}t = 0,\,6s} \right) \cr & \int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - \frac{{{t^3}}}{3}} \right)} dt \cr & \Rightarrow \theta = 36\,{\text{rad}}\,\,\, \Rightarrow n = \frac{{36}}{{2\pi }} < 6 \cr} $$

$$\eqalign{ & \tau \times \Delta t = I\omega \cr & {\text{or}}\,\,\tau \times 60 = \frac{{2 \times 2 \times 60\pi }}{{60}} \Rightarrow \tau = 0.21\,N - m\,\,\left( {\because f = 60\,rpm\,\therefore \omega = 2\pi f = 2\pi \times \frac{{60}}{{60}}} \right) \cr} $$

After collision velocity of $$COM$$  of $$A$$ becomes zero and that of $$B$$ becomes equal to initial velocity of $$COM$$  of $$A.$$ But angular velocity of $$A$$ remains unchanged as the two spheres are smooth.

When the system is released, heavier mass move downward and the lighter one upward. Thus, centre of mass will move towards the heavier mass with acceleration
$$a = \left( {\frac{{3m - m}}{{3m + m}}} \right)g = \frac{g}{2}$$

$$\eqalign{ & {x_{cm}} = \frac{{1 \times 0 + 1 \times PQ + 1 \times PR}}{{1 + 1 + 1}} = \frac{{PQ + PR}}{3} \cr & {\text{and}}\,{y_{cm}} = 0 \cr} $$

As shown in the diagram, the normal reaction of AB on roller will shift towards O. This will lead to tending of the system of cones to turn left.