A man of mass 100kg stands at the rim of a turn table of radius 2m and moment of inertia 4000kg ${\mathrm{m}}^2$ mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of 1m / s relative to the earth .

Applying conservation of angular momentum, ${\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}} \Rightarrow 0={\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2$

$\Rightarrow {\overline{\mathrm{\omega }}}_{\mathrm{i}}=\frac{-{\mathrm{I}}_1{\overline{\mathrm{w}}}_1}{{\mathrm{I}}_2}=-0.05{\mathrm{rads}}^{-1}$ 

Opposite to direction of motion of man

So the angular velocity of the man w.r.t table will be  ${\mathrm{\omega }}_{\mathrm{mt}}=0.5+0.05=0.55\mathrm{rad}/\sec$

So time taken to complete one revolution w.r.t table = $\frac{2\pi }{0.55}$

So the angle rotated by the table meanwhile 

$\mathrm{\theta }\text{'}=\left({\mathrm{\omega }}_{\mathrm{P}}\right)\times \mathrm{t}=\frac{1}{20}\times \frac{2\mathrm{\pi }}{0.55}=\frac{2\mathrm{\pi }}{11}\mathrm{rad}$

If the man completes one revolution relative to earth,

${\mathrm{\theta }}_{\mathrm{m}}=2\mathrm{\pi };\mathrm{t}=\frac{2\mathrm{\pi }}{{\mathrm{\theta }}_{\mathrm{m}}}=\frac{2\mathrm{\pi }}{0.5}$ 

During this time, angular displacement of table,

${\mathrm{\theta }}_{\mathrm{t}}=-0.05\left[\frac{2\mathrm{\pi }}{0.5}\right];{\mathrm{\theta }}_{\mathrm{t}}=-\frac{\mathrm{\pi }}{5}\mathrm{radians}$