Conservation of energy with rotational motion problem

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SUMMARY

The discussion centers on the conservation of energy in a rotational motion problem, specifically addressing the neglect of the rotational kinetic energy associated with the moment of inertia (I_g2) of a person on a ride. It is established that the person does not contribute to rotational kinetic energy when seated upright, as they do not spin around their center of gravity. The key conclusion is that the center of gravity for the bar includes the person, leading to the assumption that I_g2 can be considered negligible in this context.

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  • Understanding of rotational kinetic energy and its formula: 0.5 * I * ω²
  • Familiarity with the concept of moment of inertia (I) in rotational dynamics
  • Knowledge of center of gravity and its significance in rotational motion
  • Basic principles of conservation of energy in physics
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  • Study the implications of moment of inertia in complex systems
  • Learn about the effects of center of gravity on stability in rotational motion
  • Explore examples of rotational kinetic energy in real-world applications
  • Investigate the assumptions made in physics problems involving rotational dynamics
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Students of physics, educators teaching rotational dynamics, and anyone interested in understanding the principles of energy conservation in rotational motion scenarios.

theBEAST
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Homework Statement


Alright so I am confused about why the solution does not include the kinetic energy due to rotational energy from I_g2 as indicated in red in the following image. I don't understand why they would neglect it... I calculated the value and it makes a big difference.

pUlvS.png
 
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The rotational kinetic energy measures energy of the body due to rotation around its center of gravity.

The person is not spinning around his center of gravity (located somewhere in his stomach); presumably he stays upright during the ride.
 
aralbrec said:
The rotational kinetic energy measures energy of the body due to rotation around its center of gravity.

The person is not spinning around his center of gravity (located somewhere in his stomach); presumably he stays upright during the ride.

But the person spins around the pivot? OH is it because the center of gravity for the bar INCLUDES the person on it?

Are we suppose to make an assumption?
 
Last edited:
theBEAST said:
But the person spins around the bar? OH is it because the center of gravity for the bar INCLUDES the person on it?

I just saw the question says 'neglect the size of the passenger' which means IG2 = 0. Notice that IG2 << IG1 anyway.

But just to continue... the person and bar are definitely treated separately here. If the person were rigidly attached to the bar so that he does not move with respect to the bar, you would be right to say the person has rotational kinetic energy 0.5*IG2*w^2. This is because the person would be like he is part of the bar and would posses the same w.

I assumed that this ride is like most rides where the person is seated in a carriage that is pinned to the end of the arm so that the person always stays upright. Then the person does not spin and has no rotational kinetic energy.
 

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