# Angular Momentum & Moment Of Inertia

1. Jul 1, 2014

### iloafer73

1. The problem statement, all variables and given/known data

Two children on opposite ends of a merry-go-round of radius 1.6 m throw baseballs at the same speed of 30 m/s but in opposite directions as shown. The mass of each baseball is 0.14 kg, and the moment of inertia of the merry-go-round and children combined is 180 kg-m^2. If the merry-go-round is initially at rest, what is the linear speed at which the children are moving after the balls have been thrown?

2. Relevant equations

L = Iω
KE = 1/2 Iω^2
KE = 1/2 mv^2
τ = Iα

3. The attempt at a solution

Since the motion of the merry-go-round, on my assumption, is caused by the torque generated by the two baseballs being thrown, the Kinetic Energy caused by the two balls is equal to the Kinetic Energy of the merry-go-round after the throw. Hence, assuming KE is conserved:

2(1/2mv^2) = 1/2 Iω^2
0.14 x 30 x 30 = 0.5 x 180 x ω^2
ω = 0.837
therefore V= ωr
so V = 1.339 m/s

which turns out to be wrong, the correct answer should be 11.9 cm/s or 0.119 m/s.
Is it something wrong with the formulas I'm using or is it just a mistake in my assumptions.

2. Jul 1, 2014

### Nathanael

Welcome!

Do you have any reasoning to why you think Kinetic Energy should be conserved in this situation?

3. Jul 1, 2014

### iloafer73

Well since the merry-go-round is initially at rest, the way I understand it is that the balls are what cause the thing to start moving, assuming that isn't the kinetic energy released by the children to throw the balls equal to the kinetic energy of the merry go round when it starts to move?

PS - have to say it does sound pretty ludicrous that throwing the balls is what causes it to move, but there doesnt seem to be any other force that causes the torque in the machine

4. Jul 1, 2014

### Nathanael

It probably seems ludicrous because it isn't usually significant enough to overcome friction.

If you imagine throwing a ball in space, though, you might imagine when you throw the ball forward, the ball "throws you back" (you cause an impulse on the ball, and the ball causes an impulse on you). Same thing applies on a merrygoround (except the impulse will be on the entire merrygoround because you're essentially fixed to the the merrygoround via static friction)

Based on your equation ω would be the square root of 1.4 which is greater than 1 so I think you may have a made a simple mistake somewhere in your computing

5. Jul 1, 2014

### Nathanael

But that still doesn't yield the correct answer, so I don't know why I mentioned it (sorry).

Try looking at in terms of impulse instead of kinetic energy

You throw a ball, and the ball pushes back on you for a certain amount of time with some average force. How do you find the torque based off that average force? Then from that torque how do you find the angular acceleration? Then from that acceleration how do you find Δω?

Edit:
(I know you don't know the value of the "average force" (because you don't know the time it takes for the ball to be thrown) but just pretend like you do; it will still work out in the end)

Last edited: Jul 1, 2014
6. Jul 1, 2014

### iloafer73

I implemented that method but still can't find the right solution.
Impulse = ΔP/T = 2MV/T = (2 x 0.14 x 30)/T = 8.4/T
Impulse causes the torque, so Impulse = Iα
8.4/T = 180x Δω/T
8.4/180 = Δω
Δω = 0.0467

Now cause initial ω was zero, we can conclude ω to be 0.0467, and that linear velocity, V = ωr = 0.0467x1.6 = 0.07467 m/s

Which is miles off the correct answer. Is there another error in my calculations, or are my assumptions wrong?

7. Jul 1, 2014

### Nathanael

You're very close now.

I wouldn't call 5cm miles :)

ΔP/T actually equals the average force. (Impulse is just ΔP whereas ΔP/Δt is Force)

So if that is the average force, what is the average torque?

Edit:
(And you already know that $\tau_{avg}=Ia_{avg}=I\frac{Δω}{T}$ where $\tau$ is torque and $T$ is time)

Last edited: Jul 1, 2014
8. Jul 1, 2014

### Vibhor

@iloafer73 : This is a simple problem . Is there any external torque acting on the system (children+merry go round+balls ) ?

9. Jul 2, 2014

### ehild

The torque of the impulse is equal to Iα. What is the torque with respect to the axis of the merry-go round? [/QUOTE]

ehild