A child of mass 50.0 kg stands 2.00 m from the center of a horizontal circular d

Question

A child of mass 50.0 kg stands 2.00 m from the center of a horizontal circular disc of radius 4.00 m and mass 200 kg. The disc is mounted on a frictionless, vertical axis. Initially both the child and the disc are at rest. (Hint: so what is their combined initial angular momentum?) The child then begins to run in a counter clockwise direction. Shortly after the child starts running, it is observed that the disc attains a steady angular velocity of ? = ?0.125rad/s in a clockwise sense. The moment of inertia of the flat disc is given by: I = 1/2 m R^2

(i) Calculate the angular momentum of the disc (NOT the child) about the central vertical axis when the child has reached its steady (constant) tangential speed.

(ii) Calculate the steady tangential speed at which the child is running  

Explanation / Answer

a) I = ½mr² = ½ * 200kg * (4m)² = 1600 kg·m²
L = I = 1600kg·m² * -0.125rad/s = -200 kg·m²/s

b) momentum is conserved:
200 kg·m²/s = I = mr² = 50kg * (2m)² *
= 0.5 rad/s
v = r = 0.5rad/s * 2m = 1 m/s

I've been thinking about this, and it occurs to me that they might be looking for the tangential speed RELATIVE TO THE DISK, rather than relative to some nearby stationary reference frame. Since the disk is "receding" under the child at 0.125rad/s his velocity relative to the disk is

v = (0.5 + 0.125)rad/s * 2m = 1.25 m/s

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