Rotational Motion: Energy and Momentum Conservation

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SUMMARY

The discussion focuses on the conservation of angular momentum and energy in a scenario involving a child jumping off a merry-go-round. The conservation of angular momentum provides the correct new angular velocity of the system, while attempts to apply conservation of energy yield incorrect results due to overlooked factors. The correct formula for the new angular velocity is derived as ((I + mr^2)*x^2 - mvr)/I. Participants emphasize the importance of distinguishing between internal forces and energy contributions in analyzing the motion.

PREREQUISITES
  • Understanding of basic rotational motion equations
  • Familiarity with angular momentum conservation principles
  • Knowledge of energy conservation concepts in physics
  • Ability to apply mathematical formulas in physics problems
NEXT STEPS
  • Study the principles of conservation of angular momentum in rotational systems
  • Learn about the relationship between linear and angular velocity
  • Explore the implications of internal forces on energy conservation
  • Review examples of energy conservation in non-ideal scenarios
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Physics students, educators, and anyone interested in understanding the dynamics of rotational motion and the principles of energy and momentum conservation.

Prannoy Mehta
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Homework Statement



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.

Homework Equations



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

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)/IThank you for the support and all your help.
 
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If a child jumps, does the child use energy?
 
Yes, he takes a part of the energy of the entire system.
 
Prannoy Mehta said:
Yes, he takes a part of the energy of the entire system.
I would not call that jumping. Sounds more like falling off. What does a jump involve?
 
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 ?
 
Prannoy Mehta said:
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 ?
Jump up in the air. Think about energy.
 
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..
 
Prannoy Mehta said:
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..
If you jump up in the air, where does the energy come from to do that?
 
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
Prannoy Mehta said:
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)
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
Muscles do provide energy. But then do we consider that while solving a numerical ?
 
  • #12
So there is practically no way to do energy conservation here ? Taking this to be ideal scenario ?
 
  • #13
Prannoy Mehta said:
So there is practically no way to do energy conservation here ? Taking this to be ideal scenario ?
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
Yes, thank you :)
 
  • #15
Just another question, if he did manage to fall out of the merry go round. Then we would conserve energy, not momentum.. ?
 
  • #16
Prannoy Mehta said:
Just another question, if he did manage to fall out of the merry go round. Then we would conserve energy, not momentum.. ?
Both would be conserved, but then the given information would not be feasible. The numbers would be different.
 
  • #17
Makes more sense, thank you.
 

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