A disc of moment of inertia 4 kgm2 is spinning freely at 3 rad/s. A second disc of moment of inertia 2 kgm2 slides down the …

Q: A disc of moment of inertia 4 kgm2 is spinning freely at 3 rad/s. A second disc of moment of inertia 2 kgm2 slides down the spindle and they rotate together.
(a)What is the angular velocity of the combination ?
(b) what is the change in kinetic energy of the system ?

Sol: (a) Since there are no external torques acting, we may apply the conservation of angular momentum.

$\large I_1 \omega_1 = I_2 \omega_2 $

$\large 4 \times 3 = 6 \omega_2 $

Thus ω2 = 2 rad/s

(b) The kinetic energies before and after the collision are

$\large K_1 = \frac{1}{2}I_1 \omega_1^2 = 18 J$

$\large K_2 = \frac{1}{2}I_2 \omega_2^2 = 12 J$

The change is ∆K = K2 – K1 = -6 J

In order for the two discs to spin together at the same rate, there had to be friction between them. The loss in kinetic energy is converted into thermal energy.