A 50 kg boy is at the edge of a circular platform of radius 4.0 m which is rotating freely

at 0.5 rad s". The moment of inertia of the platform is 1000 kg m'. Determine the new

angular velocity of the platform if the boy walks 1.0 m towards the centre of the

platform

0.62 rad/s

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

where I is moment of inertia and ω is angular velocity.

The total moment of inertia is the sum of the platform's inertia and the boy's.

(I + m r₁²) ω₁ = (I + m r₂²) ω₂

Given I = 1000 kg m², m = 50 kg, r₁ = 4.0 m, r₂ = 3.0 m, and ω₁ = 0.5 rad/s:

(1000 + 50 (4.0) ²) (0.5) = (1000 + 50 (3.0) ²) ω₂

900 = 1450 ω₂

ω₂ = 0.62 rad/s