Question

A merry-go-arround has a moment of inertia I = 400kgm2 and radius R = 5m.

Andrea runs with a speed of 4m/s that is tangent to the rim of the stationary merry-
go-round. The mass of Andrea is 40kg.

(A) After Andrea jumps to the edge of the merry-go-around, find the angular speed of the merry-go-around and Andrea.

(B) Next, Andrea walks 2m along the radial direction towards the axis of rotation . What is the new angular speed?

Answer 1

I_m = moment of inertia of merry-go-round = 400 kg m²

R = radius = 5 m

V = speed of andrea before jumping on the merry-go-round

m = mass of andrea

moment of inertia of andrea = I_a = m R² = 40 (5)² = 1000

final angular speed = W

Initial Total angular momentum = final total angular momentum

momentum of andrea + momentum of merry-go-round = angular momentum of andrea and merry go round combined

m V R + 0 = (I_a + I_m) W

40 x 4 x 5 = (1000 + 400) W

W = 0.571 rad/s

B)

r = distance from axis of rotation for andrea = R - 2 = 5 - 2 = 3 m

I_a' = moment of inertia of andrea at the new position = m r² = 40 (3)² = 360 kg m²

W' = new angular speed

Using conservation of angular momentum

(I_a + I_m) W = (I'_a + I_m) W'

(1000 + 400) (0.571) = (360 + 400) W'

W' = 1.053 rad/s