Question

A playground 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

Answer 1

Answer:

6.4 rpm

Explanation:

$I_{m}$ = 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

$I_{c}$ = 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

$I_{f}$ = $I_{m}$ + $I_{c}$ = 275 + 111.32 = 386.32 kgm²

Total moment of inertia before child jumps is given as

$I_{i}$ = $I_{m}$ = 275 kgm²

$w_{i}$ = initial angular speed = 9 rpm

$w_{f}$ = final angular speed

using conservation of angular momentum

$I_{i}$ $w_{i}$ = $I_{f}$ $w_{f}$

(275) (9) = (386.32) $w_{f}$

$w_{f}$  = 6.4 rpm