Question

A child m = 30 kg runs at vi = 5 m/s tangentially to a small playground merry-go-round, aiming at the edge with idea of jumping on and spinning around. The merry-go-round has mass 80 kg, radius 1m, and can be modeled as a thin solid disk.

1. What is the angular speed of the merry-go-round after the child jumps on?

2. After the merry-go-round turns through 270◦ , the child jumps off the merry-go-round radially and tumbles away at vf = 0.75 m/s. What is the angular speed of the merry-go-round after the child jumps off?

3. What force was applied to allow the child to change his or her linear momentum (comparing before and after interacting with the merry-go-round)?

Answer 1

M = mass of merry-go-round = 80 kg

R = radius of merry-go-round = 1 m

Moment of inertia of merry-go-round

I = (0.5) M R2 = (0.5) (80) (1)2 = 40 kg m2

m = mass of the child = 30 kg

vi = tangentially velocity of the child before jumping = 5 m/s

w = angular speed of merry-go-round

using conservation of angular momentum

m vi R = (I + m R2) w

(30) (5) (1) = (40 + (30) (1)2) w

w = 2.143 rad/s

2.

vf = final velocity of child after the jump = 0.75 m/s

w = angular velocity of the combination of child and merry-go-round before the jump = 2.143 rad/s

w' = final angular velocity of merry-go-round

Using conservation of momentum

(I + m R2) w = I w' + m vf R

(40 + (30) (1)2) (2.143) = (40) w' + (30) (0.75) (1)

w' = 3.2 rad/s

3.

= angular displacement = 270 deg = 3/2 = 3 (3.14)/2 = 4.71 rad

w = 2.143 rad/s

t = time taken

using the equation

= w t

4.71 = (2.143) t

t = 2.2 s

using impulse-change in momentum equation

F t = m (vf - vi)

F (2.2) = (30) (0.75 - 5)

F = - 58 N

magnitude of force : 58 N