- 13 May '05 11:56I haven't seen many applied maths problems on this forum so i thought i would post one for a change. It took me bloody ages to find a satisfactory explaination but when i asked my tutor he solved it in about 2 mins, but hey!

You have a smooth uniform rod resting on a smooth horizontal plane. If you give this rod and impulse "F" at its centre of mass, it shall move off with linear speed "v", without rotation. It's kinetic energy (KE) is "half*m*v^2" (^ = to the power of)

If we then apply the SAME impulse "F", but this time at the end of the rod, it moves off with the same linear velocity "v" but as we applied the impulse at the end it also moves with a angular velocity "w". It's KE is "half*m*v^2 + half*I*w^2" (where I is the moment of inertia and the second term is the rotational KE)

Same rod, same impulse but in the second case the rod has more KE than before, how? - 13 May '05 15:36I don't know the answer, but I like this question.

Could it have something to do with where on the rods you measure the "v"? I was just trying it with my pen. Not very scientific, I know, but it seems the pen moves much further when I hit it in the middle than when I hit it on the side. I'm sure friction has some part to play in that, and of course my shaky hands (thank your lucky stars I'm not a surgeon). But could it be that the "v" measured on the point of impact is the same on both rods, not the "v" measure at the centre of mass? This would ensure that the "v" for the centre of mass of the rotating rod is less than the other rod, effectively partitioning the energy between rotational and translational? - 13 May '05 15:45My first comment is that, at the end of the rod, only infinitely small time impulses can be given with the force still pointing in the same direction. So, to make it feasable, we have to assume an inifinitely high force during an infinitely small time, with the product still the finite value.

Second comment: Impulse is not equal to energy! You would need more energy (force x distance!) on the end of the rod in order to obtain the same 'v'. The difference in energy would be exactly the rotational kinetic energy. - 13 May '05 17:471st - Why would the force from an impact rotate with the rod? Surely itt could continue to point forwards without affecting the nature of the problem.

2nd - That is true, the question is a bit subtle as impulse isn't energy, however if we say that the force applied and the time it is applyed for are constant for both cases then the extra energy still remains. Where does the extra energy come from? - 13 May '05 17:55

Jesus?*Originally posted by nickhawker***1st - Why would the force from an impact rotate with the rod? Surely itt could continue to point forwards without affecting the nature of the problem.**

2nd - That is true, the question is a bit subtle as impulse isn't energy, however if we say that the force applied and the time it is applyed for are constant for both cases then the extra energy still remains. Where does the extra energy come from? - 13 May '05 17:58

1.if the force does not rotate with the rod, it will leave its interaction point, and without re-action there is no action. Meaning the force would not exist*Originally posted by nickhawker***1st - Why would the force from an impact rotate with the rod? Surely itt could continue to point forwards without affecting the nature of the problem.**

2nd - That is true, the question is a bit subtle as impulse isn't energy, however if we say that the force applied and the time it is applyed for are constant for both cases then the extra energy still remains. Where does the extra energy come from?

2.a force, even during very long time periods, does not produce kinetic energy if there is no change in terms of dispkacement (or chemical change or other ways of energy exchange that may ultimately create kinetic energy). For instance, you standing still on the earth for an hour, does not create kinetic energy, despite the impulse of 'your weight' x 'one hour'. - 13 May '05 18:19Newton's second law should explain this one.

F=ma Force=massXacceleration

The point of impact where the impulse is applied moves off at the same (v) and has the 'extra' angular (v), but the total energy transfer measured as velocity must be measured at the center of mass.

In the first case, the force (impulse) is applied to center of mass and thus the whole bar accelerated with uniform velocity.

In the second case, the force is applied to one end, giving that end a greater (v), but leaving the other end of the bar with hardly any.

However, the center of gravity is where the velocity and 'extra' angular velocity must be measured. Totaling the two at this point will keep us from trashing the laws of nature. - 13 May '05 18:20No no, i mean that application point of the force moves with the rod but the direction of the force remains forwards.

I think we are debating cross-purposes here.

Basically how i see it...

In the first case, the force is applied over a certain time, during which the rod is accelerated to speed "v". In this time the rod will cover a certain distance and the force will do a certain work "F*d"

In the second case, the force is applied over the same time but during that time the point of application of the force (the end) moves a greater distance due to the extra rotation thus doing extra work and hence more KE.

I've got another (far harder) problem like this if you like? - 14 May '05 18:12

I suspect that your claim that the rod moves at the same translational velocity is incorrect.*Originally posted by nickhawker***I haven't seen many applied maths problems on this forum so i thought i would post one for a change. It took me bloody ages to find a satisfactory explaination but when i asked my tutor he solved it in about 2 mins, but hey!**

You have a smooth uniform rod resting on a smooth horizontal plane. If you give this rod and impulse "F" at its centre of ma ...[text shortened]... ional KE)

Same rod, same impulse but in the second case the rod has more KE than before, how?

If you whack a rod on a table at the very end, it will spin like crazy, but won't translate much since you missed the center of gravity. To get it translating you want to hit close to the center of gravity. - 14 May '05 20:35 / 2 edits

I don't think using the same force in a different part of the bar can modify the energy, can it? If you swing a hammer with X Joules of energy behind it, you won't create or destroy energy depending where you hit the bar. Nor do I think you'll get more efficiency of energy transfer by hitting the bar on the end.*Originally posted by corp1131***Could it be that the bar is acting as a lever when you push it at the end, ie you have a moment about the CoM, and that pushing it at the CoM produces no such moment. Like using a spanner with a long handle reduces the work, pushing the bar at the end increases its energy.**

Does a hammer with X kinetic energy produce a different force or impulse depending on where on the bar it hits? I kind of doubt that. If this is the case, then the answer is that there is just more energy in the impulse at the end because the hammer is swung harder, and this is where the extra kinetic energy in the bar came from. - 15 May '05 12:09That is true, in reality you would have to swing your hammer harder when hitting at the end in order to produce the same impulse. That's because when you hit it at the end it wil move away from the force faster than at the centre, thus reducting the contact time of the force or the "strengh" of the force "felt" by the bar or whatever as it moves away.

If you hit a rod with a ball at the end and centre, moving at the same speed each time, the ball at the end will hit with a lower impulse as the rod will begin to move away from the ball faster than it would do at the middle. The ball would probably carry on in the same direction. If it hit the middle, the rod would move away slower and with a bigger impulse. The ball would probably stop dead.

So for the case of the impulses being the same, that would mean a "harder blow" as we see it in reality.