Rotational Motion: Energy and Momentum Conservation

1. Oct 12, 2015

Prannoy Mehta

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

A child with mass m is standing at the edge of a merry go round having moment of inertia I, radius R and initial angular velocity x as shown. (The figure shows a disc moving anticlockwise, with the velocity v (Mentioned at the end) pointing upwards to the right most edge of the disc. ) The child jumps off the edge of the merry go round with tangential velocity v, w.r.t. the ground. The new angular velocity of the merry go round is.

2. Relevant equations

Basic Rotational motion equations. Mainly rotational energy, and angular momentum.

3. The attempt at a solution

The conservation of angular momentum yielded the correct result for the problem stated. But when I do this using conservation of energy it does not (I have missed something out, I need to know what). Here is what I have done.

0.5 * m *v^2 + 0.5 * I * y^2 = 0.5 (I + mr^2)*x^2

Taking y as the final angular momentum. The result is something else. The answer according the reference book is given as:

((I + mr^2)*x^2 - mvr)/I

Thank you for the support and all your help.

2. Oct 12, 2015

haruspex

If a child jumps, does the child use energy?

3. Oct 12, 2015

Prannoy Mehta

Yes, he takes a part of the energy of the entire system.

4. Oct 12, 2015

haruspex

I would not call that jumping. Sounds more like falling off. What does a jump involve?

5. Oct 12, 2015

Prannoy Mehta

The velocity with he lands with ? Not even that. I am not sure then. He would a apply a backward force on the ride and then use the force to jump forward. Action reaction. As it is internal forces acting energy conservation and momentum conservation can be applicable. I am not sure what you mean ?

6. Oct 12, 2015

haruspex

Jump up in the air. Think about energy.

7. Oct 13, 2015

Prannoy Mehta

Are you referring to Gravitational Potential energy ? mgh? I can take take that as the velocity just as he leaves. Assuming the disc to be my frame of reference..
I did not get you..

8. Oct 13, 2015

haruspex

If you jump up in the air, where does the energy come from to do that?

9. Oct 14, 2015

Prannoy Mehta

The energy comes from the ground to me ? (Conservation of Energy, I push the ground back, and then the ground pushes me back in front)

10. Oct 14, 2015

haruspex

No, that's not conservation of energy, that's action and reaction being equal and opposite.
You do not get free energy from the ground. Why do you have muscles? Why do you need to eat?

11. Oct 15, 2015

Prannoy Mehta

Muscles do provide energy. But then do we consider that while solving a numerical ?

12. Oct 15, 2015

Prannoy Mehta

So there is practically no way to do energy conservation here ? Taking this to be ideal scenario ?

13. Oct 15, 2015

haruspex

The boy jumped, he didn't fall. You have no idea (a priori) what energy he supplied in doing so.
Instead, you can solve the problem using conservation of angular momentum, then deduce the work done by the boy.

14. Oct 15, 2015

Prannoy Mehta

Yes, thank you :)

15. Oct 15, 2015

Prannoy Mehta

Just another question, if he did manage to fall out of the merry go round. Then we would conserve energy, not momentum.. ?

16. Oct 15, 2015

haruspex

Both would be conserved, but then the given information would not be feasible. The numbers would be different.

17. Oct 15, 2015

Prannoy Mehta

Makes more sense, thank you.