Question

A 2-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs around the merry-go-round at5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 31 kg .

Part A

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

Answer 1

Angular speed of merry-go-round before:

ω1 = 20 rev/min * (2π rad/rev) * (1min/60s) = 2.084 rad/s

Conservation of angular momentum

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 * 250 kg * (2.0 m / 2)²

I = 225 kg⋅m²

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

ω2 = (225 kg⋅m² * 2.084 rad/s + (2.0 m / 2) * 31 kg * 5.0 m/s)) / (225 kg⋅m² + 31 kg * (2.0 m / 2)²)

ω2 = 2.437rad/s

ω2 = 2.334rad/s * (rev/2π rad) * (60s/min) = 22.3 rev/min = 23.272 rpm