Question

Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed $\omega$. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.

Answer 1

From initial frame of reference when we see, the (man-wheel) system, we can find that the wheel moving with a speed of and the man with $(w+\frac{v}{R})$ after the man has started walking.

$({\omega }^1=$ angular velocity after walking)

$(\omega$= angular velocity of the wheel before walking.)

Since^^Ei = 0

extended torque = 0

Therefore,

$(I+m{R}^2)\omega =I{\omega }^1+m{R}^2({\omega }^1+\frac{v}{R})$

$\Rightarrow (I+m{R}^2)\omega =I{\omega }^1+m{R}^2({\omega }^1+\frac{v}{R})$

$\Rightarrow (I+m{R}^2)\omega -MvR$

$=(I+m{R}^2){\omega }^1$

$\Rightarrow {\omega }^1=\frac{(I+M{R}^2)\omega -MvR}{(I+M{R}^2)}$