Angular kinetics

Question

A man with a mass of 80 kg runs with a velocity of 3 m/s along a line tangent to a circular platform (m = 160 kg, radius r= 5 m). The platform is initially at rest and it has a moment of inertia equal to 0.5mr2.

a) Given that there is a conservation of angular momentum, calculate the angular velocity of the platform once the runner has jumped on.

b) Then, he walks in the direction of the centre of the circular platform; determine the new angular velocity when he reaches the centre.

Answer 1

Let

M = 80 KG be the mass of the man,

v = 3 m/s be the initial velocity of the man,

m = 160 kg be the mass of the platform,

r = 5 m be the radius of the platform,

I = mr²/2 be the platform's moment of inertia,

ω₁ (to be calculated) be the angular velocity after he jumps on,

ω₂ (to be calculated) be the angular velocity after he reaches the center.

We assume that the man is so thin that any rotation around his own axis

can be ignored when calculating moments of inertia.

Before he jumps on the platform his angular momentum

L = Mvr

This momentum will be preserved in both parts a) and b)

a)

The angular momentum of the man plus the platform equals the initial L:

Mr²ω₁ + Iω₁ = L

ω₁ = L/(Mr² + I)

= Mvr/(Mr² + mr²/2) =

= Mv/(Mr + mr/2) =

80×3/(80×5 + 160×5/2) = 0.3 s^-1

b) After he reaches the center, his radius is 0, so he does not contribute to the angular momentum of the platform.

So the angular momentum of the platform alone is to equal the initial L.

Iω₂ = L

ω₂ = L/I

= Mvr/(mr²/2)

= 2Mv/mr

= 2×80×3/160×5 = 0.6 s^-1