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Say you have a simple rigid body in space that is at rest or travelling with only translational motion at a constant speed. Say that the body is something like a rod and it's not rotating. So, at some point, an external force is acted on the rod, just for an instance and it is not acted on some axes that goes through the center of mass . How much of the energy goes in rotation and how much in translational movement and why??

Edit: As for force , I read in Force acting on a simple rigid body in space that , force on com does not divide. I am unable to link it with energy.

Should energy from translation motion be transferred to rotational, since entire force was spent on it? If yes, how do we calculate it?

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Anubhav Goel
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Any force applied to a 2D or 3D object can be decomposed into the same force applied at the CM and a torque about the CM equal to the force times the perpendicular distance from the CM. Impulse is force times time, so this applies for impulses also.

If the force continues to act as the rigid body translates and rotates, the torque might vary. So it is best to consider only impulsive forces which act for an infinitesimal time, during which the rigid body does not move, and which deliver a finite change in linear momentum. Alternatively you can model the force as an instantaneous collision between 2 bodies, and use the coefficient of restitution to indicate how much KE is conserved - ie how elastic the collision is.

If you supply an impulse to the rod, you will have to solve the problem by calculating the changes in momentum - both linear and angular. When you have done that, then you can compare initial and final KE (linear and rotational) to see how the change has been divided between linear and rotational. There is no way of converting impulse into an equivalent amount of kinetic energy.

The impulse on the rod causes an equal change in linear momentum of the CM. Force and momentum are vectors, so you have to take directions into account - eg by drawing a vector triangle.

In addition, if the impulse is not applied through the CM then there is also an impulsive torque (= impulse x perpendicular distance from CM) which causes an equal change in the angular momentum about the CM. Angular momentum is moment of inertia times angular velocity of rotation. It is a (pseudo-)vector so if the impulse and rod do not lie in the same plane, you will have to do a vector calculation for this also.

sammy gerbil
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  • Can you please help with mathematical treatment? – Anubhav Goel Oct 19 '16 at 16:24
  • No. But if you post your solution attempt as an update in your question I will comment on it. – sammy gerbil Oct 19 '16 at 16:26
  • Then can you give me a hint on how do I calculate change in linear momentum of rod? – Anubhav Goel Oct 19 '16 at 16:29
  • Treat the rod as a point particle located at the CM. Sketch a vector triangle with initial and final momentum on 2 sides and impulse on the 3rd side. (A 1D calculation might be all that is needed, depending on circumstances.) Ignore the fact that the impulse is not applied at the CM : this difference is taken care of in the angular momentum calculation, which is similar (ie 1D or 2D depending on circumstances). – sammy gerbil Oct 19 '16 at 16:32
  • Thank you!! But, I don't know this method. I will try learning it, if you can tell it's name? – Anubhav Goel Oct 19 '16 at 16:35
  • I know addition/subttaction of vector. I don't know how how can we treat a rod like a point particle? – Anubhav Goel Oct 19 '16 at 16:38
  • This is basic. See any textbook for discussion of collisions. – sammy gerbil Oct 19 '16 at 16:48