Calculating Angular Speed of Playground Merry-Go-Round

[Husker70]

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?

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

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

 

[rock.freak667]

Use the conservation of angular momentum

 

[thomate1]

To calculate the new angular speed of the merry-go-round, we can use the conservation of angular momentum equation, which states that the initial angular momentum is equal to the final angular momentum. In this case, the initial angular momentum is given by L = Iω, where I is the moment of inertia and ω is the initial angular speed of the merry-go-round (10.0 rev/min). The final angular momentum is given by L = (I+mR^2)ω, where m is the added mass (25.0 kg) and R is the radius of the merry-go-round (2.00 m). Setting these two equations equal to each other and solving for ω, we get:

Iω = (I+mR^2)ω
I = I+mR^2
ω = I/(I+mR^2)

Plugging in the given values, we get:

ω = 250 kg.m^2/(250 kg.m^2 + 25.0 kg * (2.00 m)^2)
ω = 250/350 = 0.714 rev/min

Therefore, the new angular speed of the merry-go-round is 0.714 rev/min.