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When applying a force outside of the center of mass of the body, the body will get both linear and angular momentum. Right?

Does the linear velocity from this force equal to the linear velocity from the same force, which is applied on the center of mass?

$$v_{\text{center}} = v_{\text{non-center}}\quad\text{? (no vectors)}$$

If it does: How is it that the energy applied to the system is equal? In one we have both rotation and linear movement, and in the other just linear movement.

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  • $\begingroup$ A force applies acceleration. An action (force over time) applies momentum. Just to be clear. $\endgroup$
    – John Alexiou
    Commented Jun 3, 2013 at 19:50
  • $\begingroup$ Relevant (but not identical) question: physics.stackexchange.com/q/66446/392 $\endgroup$
    – John Alexiou
    Commented Jun 3, 2013 at 20:05
  • $\begingroup$ applying a force momentarly will also creat momentum when applied not to the center of mass $\endgroup$
    – user25368
    Commented Jun 5, 2013 at 12:50
  • $\begingroup$ That is called an impulse and it is a force over time. An instantaneous finite force will not impart momentum, but acceleration. Maybe this is the source of your confusion. $\endgroup$
    – John Alexiou
    Commented Jun 5, 2013 at 13:38
  • $\begingroup$ Even more relevant question: physics.stackexchange.com/q/43232/392 $\endgroup$
    – John Alexiou
    Commented Dec 17, 2013 at 0:38

2 Answers 2

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If you apply the same force for the same period of time, the linear velocity of the body will be the same in both cases, assuming the body is unconstrained. However, having applied the same force for the same amount of time does not mean that the same amount of energy has been transferred. The energy, or the work done by the force, is the force times the displacement along the direction of the force. This displacement will be greater if the force is not applied through the center of mass.

EDIT: I think the problem is that your intuition tells you that applying a force $F$ to a body for a certain time period $\Delta t$ means that you are transferring energy proportional to $F\cdot\Delta t$. This is not true in the general case. The energy transferred is the work done by the force: $F\cdot d$, where $d$ is the displacement along the direction of the force of the point that the force is applied to. Basically, when you apply the force along the center of mass of the body, the displacement will be smaller, because it corresponds to an equal displacement of the whole body, but when you apply the force further from the center of mass, the displacement will be larger because it is a combination of displacement of the whole body and rotation of the body.

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  • $\begingroup$ The 2 answers contredict eachother. As far as I understand, the second one is right, even thow it is oppesed to the intuition. If you could please elaborate on how is it that the energy is not the same because of the placing of the force $\endgroup$
    – user25368
    Commented Jun 3, 2013 at 17:27
  • $\begingroup$ @user25368 Added elaboration. $\endgroup$
    – jkej
    Commented Jun 3, 2013 at 17:58
  • $\begingroup$ thx! I'm not sure I understand what displacement is exactlly but you did bring me closer to understanding this! $\endgroup$
    – user25368
    Commented Jun 3, 2013 at 18:54
  • $\begingroup$ @user25368 With displacement I just mean the distance that the point that you apply the force to moves during the time that you apply the force. $\endgroup$
    – jkej
    Commented Jun 3, 2013 at 18:58
  • $\begingroup$ I still find it hard to grasp for an instant force $\endgroup$
    – user25368
    Commented Jun 5, 2013 at 12:46
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A simple example would be two baseballs of mass M connected by a 1-meter stiff bar, placed on tees 1 meter apart. You hit one baseball with a bat, but not the other one. This imparts a velocity V to the one you hit, and velocity 0 to the one you didn't hit.

So immediately after hitting, the total momentum is $MV$, and the kinetic energy is $MV^2/2$. If instead you had imparted the same momentum to the center of the bar, the momentum would be divided between the two masses $2MV/2 = MV$ and the energy would be $2M(V/2)^2/2 = MV^2/4$.

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