The two uniform discs rotate separately on parallel axles. The upper disc (radius a and momentum of inertia ${\mathrm{I}}_1$) is given an angular velocity ${\mathrm{\omega }}_0$ and the lower disc of (radius b and momentum of inertia ${\mathrm{I}}_2$) is at rest. Now the two discs are moved together so that their rims touch. Final angular velocity of the upper disc is

Solution:

The two discs exert equal and opposite forces on each other when in contact. The torque due to these forces changes the angular momentum of each disc. From angular impulse-angular momentum theorem, we have

$\mathrm{fa}∆\mathrm{t} = {\mathrm{I}}_1({\mathrm{\omega }}_0-{\mathrm{\omega }}_1)$----------(i)

and $\mathrm{fb}∆\mathrm{t} = {\mathrm{I}}_2{\mathrm{\omega }}_2$---------(ii)

From eqns. (i) and (ii), we get $\frac{\mathrm{a}}{\mathrm{b}} = \frac{{\mathrm{I}}_1({\mathrm{\omega }}_0-{\mathrm{\omega }}_1)}{{\mathrm{I}}_2{\mathrm{\omega }}_2}$------(iii)
When slipping ceases between the discs, the contact points of the two discs have the same linear velocity,
i.e.,
${\mathrm{a\omega }}_1 = {\mathrm{b\omega }}_2-----\left(\mathrm{iv}\right)$

On substituting ${\mathrm{\omega }}_2$ in eqn. (iii) we get

${\mathrm{\omega }}_1 = \frac{\left({\mathrm{I}}_1{\mathrm{\omega }}_0\right)}{[{\mathrm{I}}_1+({\displaystyle \frac{{\mathrm{a}}^2{\mathrm{I}}_2}{{\mathrm{b}}^2}}\left)\right]}$