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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.

Stefan Bischof
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user25368
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2 Answers2

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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.

jkej
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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 $2M(V/2)^2/2 = MV^2/4$.

Mike Dunlavey
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