Easy physics question?

Question

B. Angular Momentum

3. A playground merry-go-round has a radius of 3.0m and a moment of intertia I = 600 kg . m2. When the merry-go-round is at rest, a 20 kg child runs at 5.0 m/s along a line tangent to the rim, and jumps on. What is the angular velocity of the merry-go-round after the child jumps on?

Answer 1

Angular momentum about the axis of rotation of the merry-go-round is conserved. So the angular momentum of the merry-go-round with the child riding on it is equal to angular momentum  of the merry-ground and of the child before the jump.

L' = L_mgr + L_c

Initially the merry-go-round is at rest and has no angular momentum. Therefore:

L ' = L_c

The angular momentum vector of the child about the axis of rotation of the merry-go-round is given by the cross product of a point vector point from axis to the child and its linear momentum vector:  

< L_c > = <r> × <m_c·v> = m_c · ( <r> × <v> )

Its magnitude is

L_c = m_c·r ·v·sin(φ)

where φ is the angle between <r> and <v>. Because the child moves tangentially to the rim at the instant before the jump, the vectors are perpendicular to each other. So:

L_c = m_c·r ·v

= 20kg · 3m · 5m/s = 300kgm²/s

The magnitude of the angular momentum can also be written as product of moment of inertia and angular velocity:

L' = J'·ω

The moments of inertia of child and merry-go-round sum up to the moment of inertia of the whole system:

J' = J_mgr + m_c·r²

(assuming that child lands directly at the rim)

= 600kgm² + 20kg · (3m)² = 780kgm²

So the angular velocity after the jump is:

ω = L' /J'

= 300kgm²/s / 780kgm²

= 0.3846s⁻¹