# If an rod is hit at its edge with an impulse I, why would it move with the same velocity as if it were hit at its center?

In one of his lectures, Walter Lewin gives the example of a rod which is hit with an impulse on its edge. It starts rotating with angular velocity w, and its center of mass moves with velocity v. Assuming ideal conditions, he then continues to explain that where the object is hit does not matter, the center of mass will always move with velocity v given impulse I. In other words, if the rod was hit right at its center of mass with impulse I, it would still move with velocity v without rotating. I thought that was very unintuitive if we think about energy considerations. My gut instinct told me that the one hit at the center would move with a higher velocity.

If it is truly the case that the center of mass will always move with velocity v, it would seem to me that the one hit at the edge has more energy since it has both rotational and translational kinetic energy (where the latter is equal to the translational kinetic energy of case when it is hit right at its center of mass, since they have the same velocity v). This is despite the fact that the one hit at the edge was acted upon by the same impulse i.e same force as the one hit at the center which only has translational kinetic energy. So how do we reconcile this?

EDIT: Check out video Bullet Block Experiment Result.

• That's because 2nd Newton's law does not depend on the shape of the body being accelerated. – Maxim Umansky Jul 25 '19 at 13:38
• Related video – Aaron Stevens Jul 26 '19 at 3:46
• @AaronStevens thank you that was super helpful – The Ignorant Wanderer Jul 27 '19 at 11:33

To reconcile the difference in energy of the bar, consider the (perhaps microscopic) collision details:

First, the collision at the center of mass. During that (perhaps very short) collision time T, the bar experiences a force F and moves some distance D. The energy transferred is FD, the work done.

Now consider a point away from the center of mass. The bar will move farther than D because it starts moving faster at that point; that was your original observation. But since there’s more distance during the collision, there more force*distance work done: more energy is transferred. That’s what we were trying to reconcile.

You might be concerned that the different collision has a different collision time or force. That’s certainly possible. But as you work through the details, you’ll always find that the “softer” off-CM target moves further and has more work done, hence more energy transferred.

I am not familiar with the lectures, but what he says matches my understanding. The issue is the type of collision the two objects are undergoing.

The short answer is "in any type of collision, momentum is conserved". So regardless of the internal structure of the objects or the precise nature of the situation, the center of mass will move with a velocity

$$\vec{v}=\frac{\vec{I}}{m}$$

where $$I$$ is the impulse and $$m$$ is the mass of the rod. (notice this is a vector equation - the object will move in whatever direction the impulse has).

The same is not true for energy; except in elastic collisions, the energy can be converted from kinetic to other forms which we often don't know how to keep track of. Indeed, if this was an elastic collision, you would have to determine how much work $$W$$ the impulse $$I$$ did and solve

$$\frac{1}{2}I\omega^2 +\frac{1}{2}mv^2=W$$

along with the impulse equation above. Notice that you cannot determine the work done here without knowing more about the force that produced the impulse.

Essentially, rotational problems imply extended objects, which imply internal structure. When objects have internal structure, they typically can store energy in other forms (thermal, sound, deformation, vibrations, etc), and the conservation of energy formula looks more like

$$\frac{1}{2}I\omega^2 +\frac{1}{2}mv^2=W+E_{them}+E_{sound}+E_{vib}+...$$

Without more knowledge of those other terms, we can't use energy conservation for inelastic collisions, which are typically the kinds of collisions we are considering in rotational motion.

EDIT: I guess for clarity; if you hit the rod twice, with the same impulse, the CM will move with the same velocity. If one of those hits is off-center, it's possible that hit had more energy because some of the energy will go into rotational kinetic, as you say. However, because we cannot model the energy transfer of the bar in the first place, you cannot guarantee that is true. You have essentially no control over the energy transfer in these kinds of problems.

• "The same is not true for energy; energy is only conserved in elastic collisions. " should be restated. Energy is conserved in all collisions, but in inelastic collisions a portion of the energy is converted from linear kinetic energy to other kinds of energy. – S. McGrew Jul 25 '19 at 13:49
• @S.McGrew: I tried to thread the needle by stating the thing we often tell students, but clarifying with the equation demonstrating the changing form of energy. I'll rewrite. – levitopher Jul 25 '19 at 16:08
• Ok so let's assume between each case we lose the same amount of energy to the "other forms". Then what? – Aaron Stevens Jul 26 '19 at 3:55
• @AaronStevens: Well, you don't know how much energy you've lost either, so dividing an unknown amount into equal parts doesn't help. However, if you can calculate all the "other forms" of energy, then you can use energy conservation. I would just point out that doing momentum conservation requires rigid objects and mechanics, whereas calculating the energy losses requires electromagnetism, thermodynamics, solid state physics....a much more detailed model, essentially. – levitopher Jul 26 '19 at 12:50
• Between each case I mean between the center collision and end collision. The other answers are fine with talking about it. – Aaron Stevens Jul 26 '19 at 12:59

If it is truly the case that the center of mass will always move with velocity v, it would seem to me that the one hit at the edge has more energy since it has both rotational and translational kinetic energy

You are exactly correct here. The one hit on the end does in fact have more energy than the one hit on center. This means that it requires more work to produce the same impulse off center vs on center. This highlights the fact that energy and momentum are separate* quantities, each independently conserved.

So why does it require more work to provide the same impulse off center? Suppose that the impulse is provided by a force with the same magnitude in both cases. The off center force must travel further than the on center one because as the off center force is applied the rod begins to rotate away from the force, so the force must go further merely in order to “keep up” with the motion of the rod. The same force over a larger distance is a greater amount of work.

Note also, the off center force provides a greater amount of torque since $$\tau=r \times F$$. So a fixed impulse gives a fixed change in linear momentum, but not a fixed change in energy or angular momentum. Impulse, work, and torque are independent quantities.

*Relativity shows that energy and momentum are closely related, but even there they are not the same.

• thank you for your answer. Could you clarify why "the off center force must travel further than the one center one?" Also, since you mention momentum, could you explain how momentum is conserved in both cases? It seems to me that we have the same problem with momentum as with energy (both have the same linear momentum, but one seems to have "extra" angular momentum, despite the fact that both were hit with the same force / impulse) – The Ignorant Wanderer Jul 27 '19 at 11:14
• I am quite confused. Say the rod has a non uniform mass distribution and is very massive at its center compared to its ends. If you hit an end, the rod will spin and, to me, intuitively, it won't "hurt" much. I.e. you won't spend much energy to rotate the rod. However if you hit the center instead, boy will it hurt. Is this reasoning wrong? – thermomagnetic condensed boson Jul 27 '19 at 11:55
• user I have updated the answer to address your concerns. thermo I don’t want to change the scenario as that leads to confusion. But “boy will it hurt” is not a measure of energy. If anything it is more related to pressure. – Dale Jul 27 '19 at 13:23