Question

Initially the platform with the child is rotating with an angular speed ${\omega }_1$. If the child starts walking along its periphery in opposite direction with speed $u$ relative to the platform, what will be the new angular speed of the platform.

Answer 1

Let $v_0$-velocity (tangential) of peripheral point w.r.t. ground

v-velocity of boy w.r.t ground

$\therefore u=v+v_0$

Again using conservation of angular momentum

$(I+m{R}^2){\omega }_1=I{\omega }_0+mR(u-v_0)$

$(I+m{R}^2){\omega }_1=I{\omega }_0+mr\mu -m{R}^2{\omega }_0$

$(I+m{R}^2){\omega }_1=(I-m{R}^2){\omega }_0+mRv$

Therefore, ${\omega }_0=\frac{(I+m{R}^2){\omega }_1-mRv}{(I-m{R}^2)}$