Aplayground merry-go-round of radius r = 2.20 m has a moment of inertia i = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round? rev/min

6.4 rpm

Explanation:

= moment of inertia of merry-go-round = 275 kgm²

m = mass of the child = 23 kg

R = radius of the merry-go-round = 2.20 m

= moment of inertia of child after jumping on merry-go-round = mR² = (23) (2.20)² = 111.32 kgm²

Total moment of inertia after child jumps is given as

= + = 275 + 111.32 = 386.32 kgm²

Total moment of inertia before child jumps is given as

= = 275 kgm²

= initial angular speed = 9 rpm

= final angular speed

using conservation of angular momentum

=

(275) (9) = (386.32)

 = 6.4 rpm

i think it's 35 meters

reserve is the answer