# After applying a torque (on one side) of a mass in free space, are the rotational and kinetic energy equal?

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?

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.

• Yeah, I'm not sure where the leap to equipartition theorem came in. Why would this behave just like gas molecules? – JMac Feb 15 '17 at 21:53
• @sammy gerbil- I made an edit to the question. Suppose you attach a pulley with a long rope on it (both with zero mass) to an arbitrary point on the mass. If you apply a constant force (perpendicular to the axis that connects the pulley to the COM) to the rope, why doesn't the equipartition theorem [not for the average of many particles, but for the distribution of (two kinds of kinetic energy in this case) in one mass] hold in this situation? Why doesn't the rope impart equal rotational and linear kinetic energy to the mass? Can you give an example that this isn't true? – descheleschilder Feb 15 '17 at 22:37
• There is no equipartition theorem for this situation. What is the difference between the new situation with the pulley and the old situation without it? Why do you think the rotational and translational KE should be equal in this new situation? – sammy gerbil Feb 15 '17 at 23:30
• @sammy gerbil-I understand the situation now. If instead of a pulley we use a rocket to apply the force on the outside of the mass, the mass starts to rotate, but if we apply this to a rod, the linear motion of the rod will be zero after one rotation due to the symmetry of the situation. After a rotation of 360 degrees, the linear motion is reduced by the same amount as the linear motion is increased after a rotation of 180 degrees. So the mass is only rotating if you turn off the rocket after a rotation of 180 degrees. Which contradicts my assumption. I don't know why I thought of the E.P.T. – descheleschilder Feb 16 '17 at 0:25
• If a constant force is applied at a fixed position and direction relative to the object, the situation is the same as here. – sammy gerbil Feb 16 '17 at 0:43
1. You need force not torque to calculate the translational kinetic energy (energy associated with the linear motion of the center of mass).
2. The equipartition of energy theorem has nothing to do here.
3. 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).

• @YashasSamanaga-I don't see how your conclusion follows from the three mentioned points. – descheleschilder Feb 15 '17 at 22:39
• Rotational kinetic energy depends on the torque applied while translational kinetic energy depends on the force applied. The rotational energy acquired depends on the moment of inertia while the kinetic energy acquired depends on the mass. I can change one parameter keeping the other constant. I can make a system with a given ratio of rotational kinetic energy and translational kinetic energy. – Yashas Feb 16 '17 at 4:37