A playground merry-go-round of radius `R = 2.00` has a moment of inertia `I = 250 kg.m^2` and is rotating `10.0 rev//min` about a frictionless, vertical axle. Facing axle, a `25.0-kg` child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is.

Options

A. `5.25 rev//min`
B. `8.45 rev//min`
C. `7.14 rev//min`
D. `3.14 rev//min`

Answer

C

Explanation

( c) From conservation of angular momentum,
`I_i omega_i = I_f omega_f : (250 kg.m^2)(10.0 rev//min)`
=`[250 kg.m^2 + (25.0 kg)(2.0 m)^2] omega_2`
`omega_2 = 7.14 rev//min`.