Angular momentum of a child on a merry go round

Question

At a particular playground there is a merry-go-round that may be thought of as a flat, solid disk of mass 142 kg and radius 1.44 m. A child whose mass is 24.7 kg is running toward the outer rim of the merry-go-round so that his path intersects the rim perpendicular to the radius. The child's speed is 2.5 m/s.

A. What is the moment of inertia of the merry-go-round before the child jumps onto its rim or outer edge?

I_mgr =  kg·m²

B. What will be the moment of inertia of the merry-go-round and boy system after the boy has jumped onto its outer edge?

I_sys =  kg·m²

C. As the child is running toward the merry-go-round, what angular momentum does he have relative to the center of the merry-go-round?

Lchild =  kg·m2

s

D. What will be the final angular speed of the merry-go-round after the child has jumped on? Assume the merry-go-round is initially at rest.

𝜔 =  rad/s

Answer 1

A.     The moment of inertia of merry-go-round in the beginning is

                                                          I_mgr = ½ MR²,

where

M = 142 kg – mass of merry-go-round,

R = 1.44 m – its radius.

Plugging the numbers, we get

                                                                       I _mgr = ½ 142 kg x (1.44m)² = 147 kgm²

B.      The moment of inertia of a system merry-go-round and a child

                                                               I_sys =  I _mgr + I _child

                                  

 Moment of inertia of a child

I _child = mR² ,

Here m = 24.7 kg – mass of a child,

R = 1.44 m – radius of merry-go-round.

Calculation gives 

                                                                          I _child = 24.7 kg x (1.44m)² = 51 kg m²

So the moment of inertia of the system

                                                                            I_sys = 147 kgm² + 51 kgm² = 198 kgm²

                                                             

C.      The angular momentum of the child

                                                                      L_chld = mvR

Let’s plug in the numbers and then we have

                                                                                      L_chld = 24.7 kg x 2.5 m/s x 1.44 m = 89 kgm²/s

D.     Let ω₂ be the final angular speed of the system. We assume that merry-go-round at first was not moving.

The initial angular speed of the child ω₁ = v/R.

Using the law of conservation of angular momentum, we get

                                                                I _child ω₁ = I_sys ω₂

 We can find

                                               ω₂ = (I _child v) / (I_sys R)

Putting numbers, we have

                                             ω₂ = (51 kgm² ·2.5 m/s) / (198 kgm²·1.44 m)

                        

                                                                  ω₂ = 0.45 rad/s