A man of mass 100kg stands at the rim of a turn table of radius 2m and moment of inertia $4000kg{m}^2$  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 1 m/s relative to the earth

Applying conservation of angular momentum.
 ${\overline{L}}_i=L_f;O=I_1{\overline{\omega }}_1+I_2{\overline{\omega }}_2$
$\Rightarrow {\overline{\omega }}_2=\frac{-I_1{\overline{\omega }}_1}{I_2}=-0.05ra{d}^{-1}$ 
Opposition to direction of motion of momentum.
(c)  ${\overline{\omega }}_m-{\overline{\omega }}_p={\overline{\omega }}_{mp}$
$\left|{\overline{\omega }}_{mp}\right|=\frac{1}{2}-(-\frac{1}{20});\left|{\overline{\omega }}_{mp}\right|=\frac{11}{20}$ 
$t=\frac{2\pi }{\left(\frac{11}{20}\right)}=\frac{40\pi }{11}$ 
$\therefore$ Angle turned by table
$\theta \text{'}=\left({\omega }_p\right)\times L=\frac{1}{20}\times \frac{40\pi }{11}=\frac{2\pi }{11}rad$ 
(d) If the man completes one revolution relative to earth, ${\theta }_m=2\pi ;k=\frac{2\pi }{{\theta }_m}=\frac{2\pi }{0.5}$
During the time, angular displacement of table, 
${\theta }_t=-0.05\left[\frac{2\pi }{0.5}\right];{\theta }_t=-\frac{\pi }{5}\times radians$