Physics Question - Find the Angular Speed?

Question

A playground carousel is free to rotate about its center on frictionless bearings. The carousel has an angular speed of 3.14 rad/s, a moment of inertia of 125 kg m^2, and a radius of 1.80m. A 40.0 kg person, standing still next to the carousel, jumps onto it very close to the outer edge. Find the resulting angular speed of the arousel and person.

Answer 1

Angular momentum (L) must be conserved here, so L(before) = L(after).  L = I * ω.

L(before) = I * ω1 = (125 kg*m²) * (3.14 rad/s) = ...

L(after) = [I(c) + I(p)] * ω2

I(c) is for a disk (the carousel):  I(c) = (1/2)*m*r²

I(p) is for a ring (the person's path):  I(p) = m*r²

Find mass of carousel by rearranging I=(1/2)*m*r²:

m1 = [2 * I(c)] / r² = ...

Find person's moment of inertia (using radius of carousel):

I(p) = m2 * r² = ...

Next, use conservation of angular momentum to find ω2:

L(before) = L(after)

[(1/2)*m1*r²] * ω1 =  [(1/2)*(m1)*r² + (m2)*r²] * ω2

{the r²'s cancel}

[(1/2)*m1] * ω1 =  [(1/2)*(m1) + (m2)] * ω2

{multiply by 2 to remove the 1/2's}

m1 * ω1 = (m1 + 2*m2) * ω2

{rearrange and solve for ω2}

ω2 =  (m1 * ω1) / (m1 + 2*m2) = ...