Conservation of Angular Momentum of merry-go-round

[nx01]

Homework Statement

A merry-go-round of radius 2m has a moment of inertia 250kg*m^2, and is rotating at 10rpm on a frictionless axle. Facing the axle and initially at rest, a 25kg child hops on the edge of the merry-go-round and manages to hold on. What will be the new angular velocity of the merry-go-round after the child jumps on?

Homework Equations

Li (system) = Lf (system)

The Attempt at a Solution

Let g = merry-go-round and c = child.

Li = Lf, so (IW)g-initial = (IW)g+c-final.

Solving for Ic:

Ic = mr^2 = (25kg)(2m)^2 = 100kg*m^s

Solving for Ig+c:

Ig+c = Ig + Ic = 350kg*m^s

Solving for final angular velocity of system:

W(g+c)-final = (IW)g-initial / Ig+c = [(250kg*m^2)(10rpm)] / 350kg*m^s = 7.14rpm.


My question is, should I be treating the child as a point-mass, allowing me to use I = mr^2 -- or should I approximate its shape in determining I? (Treating it as a cylinder - I = .5mr^2 - I get 8.33rpm.)

Any insight into this will be much appreciated!

 

[tiny-tim]

hi nx01!

… should I be treating the child as a point-mass, allowing me to use I = mr^2

yes

-- or should I approximate its shape in determining I? (Treating it as a cylinder - I = .5mr^2 - I get 8.33rpm.)

no, it would be I = .5mr² + mR²

where r and R are the radius of the cylinder and of the merry-go-round, respectively

 

[nx01]

Thank you!