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Angular Speed

  • Thread starter Husker70
  • Start date
90
0
1. Homework Statement
A playground merry-go round of Radius R=2.00m has a moment of
inertia I = 250 kg.m^2 and is rotating at 10.0 rev/min about a frictionless
vertical axle. Facing the axle a 25.0kg 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?

2. Homework Equations
I = 250 kg.m^2
10.0 rev/min
25.0kg child
radius = 2.00m

I=r^2m
L = r x p
L = Iw

3. The Attempt at a Solution

I = r^2m
m = I/r^2
m = 250kg.m^2/(2.00m)^2 = 62.5kg

I added the 62.5kg + 25.0kg = 87.5kg

I = r^2m
= (2.00m)^2(87.5kg) = 350kg.m^2

I don't know what to do with this from here.
Thanks for any help,
Kevin
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

rock.freak667

Homework Helper
6,230
31
Use the conservation of angular momentum
 

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