Question

A 2.8-m-diameter merry-go-round with a mass of 210 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 31 kg .

What is the merry-go-round's angular speed, in rpm, after John jumps on?

Express your answer in revolutions per minute.

Answer 1

Angular speed of merry-go-round before:
ω1 = 20 rev/min * (2π rad/rev) * (1min/60s) = 2.084 rad/s

This is a conservation of angular momentum problem.

I ω1 + r (m_J v) = (I + m_J r²) ω2
ω2 = (I ω1 + r (m_J v)) / (I + m_J r²)  

Moment of inertia of solid disc:
I = ½ M r²
I = ½ * 210 kg * (2.8m / 2)²
I = 205.8 kg⋅m²

ω2 = (I ω1 + r (m_J v)) / (I + m_J r²)  
ω2 = (205.8 kg⋅m² * 2.084 rad/s + (2.8 m / 2) * 31 kg * 5.0 m/s)) / (205.8 kg⋅m² +31 kg * (2.8 m / 2)²) =645.8872/266.56
ω2 = 2.4230rad/s
ω2 = 2.4230rad/s * (rev/2π rad) * (60s/min) = 23.15 rev/min = 23.15 rpm