# Energy conservation of a rod in free space

1. Jun 27, 2014

### Volkr16

Hey everyone,

A meager pea brain (me) is in need of some help from you fine physics gentlemen.
This is a concept I've spend a lot of time pondering on (more than I would like to admit).

Imagine a rod of uniform density in free space. If you apply impulse onto the center of mass then it will gain an energy (transverse) of E=1/2mv^2. If you apply an impulse off the center of mass then it will gain an energy of E=1/2mv^2+1/2Iw^2 (transverse+rotational).

Here is what I think I know: Let's assume that this rod gets hit perpendicular to it's length. No matter where it gets hit, the center of mass will have equal transverse motion for equal impulse applied. In other words Ft=mv for center of mass

So let's assume two scenarios, one where the impulse is applied off the center of mass and one where it is on the center of mass. Make the impulse be so, that the velocity of the center of mass will be the same in both cases (From what i stated before i should be able to use the same impulse anywhere on the rod).

My problem lies in the conservation of energy: I know if the impulse is applied off center i will have rotational energy, but i also know that I will have the same translational energy no matter where the rod get hit (perpendicular on the length). Something isn't right here.

Surely my "what i think i know" must be wrong. Could someone please elaborate why?

Volker

2. Jun 27, 2014

### Staff: Mentor

The same impulse means applying a given force for a given time. Depending on where you apply the force, the distance that the point of application travels during that time will be different - and the energy transferred to the rod is equal to the force times the distance, not the force times the time.

3. Jun 27, 2014

### TumblingDice

4. Jun 27, 2014

### Volkr16

Thanks guys, that cleared it up. I searched a lot before posting this, but didn't find that thread - sorry