A playground merry-go-round of radius 2.5 m has a moment of inertia 207 kg.m^2 a

Question

A playground merry-go-round of radius 2.5 m has a moment of inertia 207 kg.m^2 and is rotating at 8 rev/min about a frictionless vertical axle. Facing the axle, a 33 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? Answer in units of rev/min. A child of mass 51.9 kg sits on the edge of a merry-go-round with radius 2.3 m and moment of inertia 194.931 kg m^2. The merry-go-round rotates with an angular velocity of 1.7 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.943 m from the center. Now what is the angular velocity of the merry-go-round? Answer in units of rad/s.

Explanation / Answer

here,

radius , r = 2.5 m

moment of inertia , I = 207 kg.m^2

initial angular speed , w0 = 8 rev/min

mass of child , m = 33 kg

let the new angular speed be w

using conservation of angular momentum

I * w0 = ( I + m * r^2) * w

207 * 8 = ( 207 + 33 * 2.5^2) * w

w = 4 rev/min

the new angular speed is 4 rev/min

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