Consider a mass of arbitrary form that's freely floating in space and stationary in our frame of reference. Can we say that after a force has been applied on one side of the mass, the linear kinetic and rotational energy are equal, in the same way, that the equipartition theorem states that the energy of a gas is equally distributed over all available degrees of freedom for the gas particles?
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Your terminology is confused. Rotational energy is a form of kinetic energy. I think you are distinguishing between rotational and translational KE. Also, torque is applied to the whole body, not one side only. Probably you mean that a force is applied off-centre so that it causes rotation about the CM as well as translation.
Your question is similar to force applied not on the center of mass. The linear momentum imparted to the CM by an impulse is the same whether or not the line of impact passes through the CM. This is counter-intuitive, because it means that an oblique impulse of the same magnitude imparts additional (rotational) KE compared with an impulse which passes through the CM.
The equipartition theorem does not apply here. It is a statistical law which applies to the averages for many interacting molecules or bodies. It does not apply for each individual molecule or body at all times.
sammy gerbil
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- You need force not torque to calculate the translational kinetic energy (energy associated with the linear motion of the center of mass).
- The equipartition of energy theorem has nothing to do here.
- The rotational energy gained depends on the moment of inertia.
Therefore, the energy supplied will never be shared equally between rotational and kinetic components (in certain cases it may but this is just a coincidence).
Yashas
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