Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $${\omega _1}$$ and $${\omega _2}.$$ They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is

Solution :
When no external torque acts on system then, angular momentum of system remains constant.
Angular momentum before contact $$ = {I_1}{\omega _1} + {I_2}{\omega _2}$$
Angular momentum after the discs brought into contact $$ = {I_{{\text{net}}}}\omega = \left( {{I_1} + {I_2}} \right)\omega $$
So, final angular speed of system $$ = \omega $$
$$ = \frac{{{I_1}{\omega _1} + {I_2}{\omega _2}}}{{{I_1} + {I_2}}}$$
Now, to calculate loss of energy, we subtract initial and final energies of system.
⇒ Loss of energy
$$\eqalign{ & = \frac{1}{2}I\omega _1^2 + \frac{1}{2}I\omega _2^2 - \frac{1}{2}\left( {2I} \right){\omega ^2} \cr & = \frac{1}{4}I{\left( {{\omega _1} - {\omega _2}} \right)^2} \cr} $$