# Calculating final angular velocity using rot. KE and ang. momentum?

#### 21joanna12

1. Homework Statement
There is a point particle on the edge of a merry-go-round. It has mass 15kg and the merry-go-round has mass 235kg. The initial angular momentum is 2 radians per second. If the point particle moves from the outer edge of the merry-go round to the centre, what is the final angular momentum?

I've tried to do this problem using conservation of angular momentum and conservation of rotational kinetic energy and I get two different answers...

2. Homework Equations

moment of inertia of particle, Ip = mr2 and of merry-go-round, Im=mr2/2

rotational kinetic energy =Iω2/2

Angular momentum, L=Iω for particle or

3. The Attempt at a Solution

In both cases, the moment of inertia of the partcle in the centre is zero and therefore its rotational kinetic energy and angular momentum are zero .

First, using conservation of angular momentum,

intial angular momentum of merry go round + of particle = final angular momentum of merry-go-round.

0.5x235xr2x2 + 15xr2x2 = 0.5x235xr2final

235 + 30 = 0.5x235xωfinal

ωfinal= 2.2553.....

Using conservation of rotational kinetic energy,

0.5x15xr2x22 + 0.5x0.5x235xr2x22 = 0.5x0.5x235xωfinal2

Then I get ωfinal= 2.1238....

I have a feeling that the issue may be that the rotational KE of the particle isn't zero even when it is at the centre, although I'm not sure...

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#### BvU

Homework Helper
Your teacher has a difficult way of formulating exercises!
We have to make a few assumptions before we can actually do something:
We assume the 2 radians per second is not the angular momentum but the rotational speed (angular speed) $\omega = {d\theta\over dt}$ (An angular momentum has a different dimension than rad/s)
We assume the merry-go-round (mgr from now on) plus point particle (pp from now on) BOTH rotate with that same $\omega$
We assume that whatever brings the pp to the center is attached to the mgr
We assume there is no engine that is coupled to the mgr: the whole system isn't driven but rotates freely (no external torque).
We assume that the final angular momentum that is being asked for is not the total angular momentum of mgr+pp but the final angular speed of mgr+pp.

A bit much for my taste; did you render the whole thing or did you skip a few bits ?

In the case all assumptions are correct, we may use angular momentum conservation.

There is no reason to expect conservation of energy: something has to do work to bring the pp to the center of the mgr.

#### 21joanna12

Your teacher has a difficult way of formulating exercises!
We have to make a few assumptions before we can actually do something:
We assume the 2 radians per second is not the angular momentum but the rotational speed (angular speed) $\omega = {d\theta\over dt}$ (An angular momentum has a different dimension than rad/s)
We assume the merry-go-round (mgr from now on) plus point particle (pp from now on) BOTH rotate with that same $\omega$
We assume that whatever brings the pp to the center is attached to the mgr
We assume there is no engine that is coupled to the mgr: the whole system isn't driven but rotates freely (no external torque).
We assume that the final angular momentum that is being asked for is not the total angular momentum of mgr+pp but the final angular speed of mgr+pp.

A bit much for my taste; did you render the whole thing or did you skip a few bits ?

In the case all assumptions are correct, we may use angular momentum conservation.

There is no reason to expect conservation of energy: something has to do work to bring the pp to the center of the mgr.

Thank you for your reply! 'Angular momentum' was a typo- I meant angular velocity. I see now why we woould have to use conservation of angular momentum. The only thing I missed out was that the 'particle' was actually a toddler crawling to the centre. I see that kinetic energy would increase here because chemical energy from their muscles is transformed into KE.

Thank you :)

"Calculating final angular velocity using rot. KE and ang. momentum?"

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