# Linear and angular momentum

If I push on an object not through the center of mass, we say that the translational effect on the entire object is the same. In other words, we can turn that off-center force into a force through the center of mass and a torque about it.

But if the force was originally through the center of mass, we wouldn't get a rotation.

I can see that in both cases, you are doing the same amount of work because the rotation caused by the force doesn't count towards work. But for the same force at different locations, it seems that one adds more energy to the system due to adding both angular momentum and linear momentum, while the other only adds linear momentum (of the same amount).

Is the linear momentum added to the system really the same regardless of where you push? It seems to me that a push at the center of mass should add more linear momentum due to not adding any angular momentum, but that's apparently not the case.

• What do you mean by "the rotation doesn't do work in the direction of the push"? Rotations never do work. Forces do work. – Brian Moths Apr 20 at 23:07
• Well yeah. I just said it because regardless of where you push, you'll do the same amount of work (because rotation doesn't do work). I just kind of think of it as the rotation caused by the force doesnt count towards its work because it moves in both directions, thus canceling itself out constantly. Either way, it's zero. – Splry00 Apr 20 at 23:17
• physics.stackexchange.com/questions/174135/… – BowlOfRed Apr 20 at 23:41
• Torques can do work. You are mistaken – Aaron Stevens Apr 20 at 23:46
• My wording could be better. A force which causes an object to translate and rotate does no more work than the translational movement it causes. This is because if all the mass is rotating about the center of mass, the same amount of mass is moving forward as backward (which cancels out). But yes, that torque can later be used to do work... but causing it has not done work yet right? Regardless, the post linked to by BowlOfRed answers my question. Thank you. – Splry00 Apr 21 at 0:14

$$W = F \times d$$. You've specified a force, but not the distance it acts over.