Question

A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 255 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-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?

Answer 1

Answer:

7 rpm = 0.73 rad/s

Step-by-step explanation:

R = Radius of merry-go-round = 1.8 m

$I_M$= Moment of inertia of merry-go-round = 255 kg m²

$I_C$= Moment of inertia of child

ω = 9 rev/min

m = Mass of child = 24 kg

From the conservation of angular momentum

$I\omega=I'\omega '\\\Rightarrow I\omega=(I_M+I_C)\omega'\\\Rightarrow \omega'= \frac{I\omega}{(I_M+I_C)}\\\Rightarrow \omega'=\frac{I\omega}{(I_M+mR^2)}\\\Rightarrow \omega'=\frac{255\times 9}{(255+24\times 1.8^2)}\\\Rightarrow \omega'=6.9\ rev/min$

∴ New angular speed of the merry-go-round is 7 rpm = $7\times \frac{2\pi}{60}=\mathbf{0.73\ rad/s}$