Question

A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 265 kg·m2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? rev/min

Name: a playground merry go round of radius r 180 m has a moment of inertia i 265 kgm2 and is rotating at 100 revmin about a frictionless vertical axle facing the axle a 260 kg child hops onto the 63935
Uploaded: 2022-04-14T12:49:04-08:00
Duration: 2 min 30 s
Channel: Prabhu Ramji
Description: a playground merry go round of radius r 180 m has a moment of inertia i 265 kgm2 and is rotating at 100 revmin about a frictionless vertical axle facing the axle a 260 kg child hops onto the 63935

          A playground merry-go-round of
radius R = 1.80 m has a moment of
inertia I = 265 kg·m2 and
is rotating at 10.0 rev/min about a frictionless vertical
axle. Facing the axle, a 26.0 kg child hops onto the
merry-go-round and manages to sit down on its edge. What is the new
angular speed of the merry-go-round?
 rev/min

Added by Jose Angel W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prabhu Ramji

04/14/2022

Step 1

Given: Initial rotational inertia of the merry-go-round, I = 265 kg·m^2 Mass of the child, m = 26.0 kg Radius of the merry-go-round, R = 1.80 m Final rotational inertia, I_final = I_merry-go-round + I_child I_final = 265 kg·m^2 + (26.0 kg)(1.80 m)^2 I_final = 265 Show more…

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A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 265 kg·m2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? rev/min