The small particle of mass m = 3.8 kg and its restraining cord of length r = 0.46 m are spinning with an angular velocity? = 1.6 rad/s on the horizontal surface of a smooth disk, shown in section. as the force f is slowly relaxed, r increases and? changes. when r = 0.63 m, what is the angular velocity

Use conservation of angular momentum. The initial momentum is L=Iω, where I is the moment of inertia and ω=1.6rad/s. The moment of inertia of a particle of mass m at distance r from the axis of rotation is I = mr². So in the first case it's

I=3.8 (0.46²) = 0.804 kgm²

L=0.804 * 1.6 = 1.2865 kgm²/s

In the second case the angular momentum will be the same, but I will increase to I' and ω will decrease to ω'. We seek to find ω'.

L=1.2865 kgm²/s = I'ω'

I'=3.8 (0.63²) = 1.508 kgm²

ω' = L/I' = 1.2865/1.508 = 0.853rad/s