Disk is Rotating and asks for Velocity

Question

A solid disk rotates in the horizontal plane at an angular velocity of  rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.15 kg.m². From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer 1

Angular momentum (L) is conserved. So L₁ = L₂ where L₁ is the angular momentum of the plain rotating disc and L₂ is the angular momentum after you put a ring of sand on it.

But L = Iω where I is the moment of inertia and ω is the angular velocity. So, since L₁ = L₂, then I₁ω₁ = I₂ω₂ and solving for ω₂ we get ω₂ = I₁ω₁/I₂

When we add the ring of sand we are adding the moment of inertia of just the ring of sand to that of the plain disk. The moment of inertia of a ring is mr² so:

I₂ = I₁ + mr² = I₁ + (m_sand)(r_sand)² = 0.15 + (0.50)(0.40)² = 0.15 + 0.080.= 0.23 kgm²

So ω₂ = I₁ω₁/I₂ = (0.15)(5.00x10^-2)/(0.23) = 0.03261 rad/s or, with 2 sig figs, 0.033 rad/s = 3.3x10^-2 rad/s