A long rod being pushed in zero g vacuum

[Horvath Bela]

Homework Statement

Lets take a long rod of length l and mass m in a zero g vacuum. If a force F acts upon the center of mass of the rod then the rods linear accelaration will be F/m. If the force acts on a distence of d then the work done by the force is F*d. If the force acts off of the center of mass of the rod on the same distence the work done and the liner acceleration of the rod are both the same. That tells us that the rod cannot start to rotate beacuse if it would it would gain rotational kinetic energy and since the linear accelerations are equal in both cases the terminal velocities are the same hence the kinetic energies of the rods must be the same. So if the rod were to gain rotational kinetic energy that would be energy that did not come from the work of the force F and it would violate the law of concervation of energy. My question is the following: does my reasoning hold and no matter how long the rod and how big of a force pushes it off its center of mass there won't be any torque or does the rod gain less linear speed and starts to rotate instead?

Homework Equations

The Attempt at a Solution

 

[PeroK]

No, your logic is not sound. Think about momentum instead of energy.

 

[Horvath Bela]

A yes no answer does not really help me at all. Could you please explain what exactly is wrong with thinking about energy in this though experiment and then why the thought experiment makes sense when considering momentum. Thank you.

 

[PeroK]

A yes no answer does not really help me at all. Could you please explain what exactly is wrong with thinking about energy in this though experiment and then why the thought experiment makes sense when considering momentum. Thank you.

The rod rotates if you push it off centre. So, you need to analyse more carefully what is happening as the force is applied.

The clue about momentum was to make you think about impulse, which is force by time.

 

[Horvath Bela]

Im sorry for nagging you, but I am clueless regarding what happens to the applied force. I have been thinking about this for weeks and I can not resolve the problem. I would be very greatful if you did so please explain in full detail without giving clues to me beacuse I have falied in solving this problem already. Think of me as a five year old who happens to know a bit of physics and math, try to explain like that.

 

[Doc Al]

No matter where the force is applied, the same force produces the same linear acceleration of the center of mass. But an off-center force will also produce rotation.

 

[PeroK]

Force times distance is the work done. If this provides linear and rotational motion, then the linear motion cannot be the same as when the work all goes to linear motion.

Logically, therefore, the linear momentum cannot be the same in both cases.

 

[Horvath Bela]

Yes but if the linear accelerations are different than that contradicts the fact that regardless of the line of action being at or off the center off mass the linear acceleration has to be F/m.

 

[Horvath Bela]

But if both accelerate the same linearly than both has the same terminal velocity so they both have the same kinetic energy, but one rotates so it has additional rotational kinetic energy, but we have done the same work on both objects(F*d) so they should have the same energy. Otherwise its energy from nothing.

 

[PeroK]

Yes but if the linear accelerations are different than that contradicts the fact that regardless of the line of action being at or off the center off mass the linear acceleration has to be F/m.

Who said the acceleration was different?

Perhaps the time during which the force acts is different?

 

[Doc Al]

but we have done the same work on both objects(F*d) so they should have the same energy.

If you push through the center of mass, all the work goes into translational energy. If you push off-center, you'll end up doing more work to maintain that same force. (The additional work goes into rotational KE.)

 

[Horvath Bela]

You did when you said that the linear motion of the two cases cannot be the same. Since if the acceleration is F/m in both cases then the distence d is done in the same time in both cases and if the acceleration and the time is the same the linear terminal velocity has to be equal.

 

[Horvath Bela]

The work done in both cases is F*d so they are exacty the same.

 

[Doc Al]

The work done in both cases is F*d so they are exacty the same.

If you keep the work done the same, then in the off-center case the translational KE increase will be less. Total work and thus total KE will be the same.

 

[Doc Al]

You did when you said that the linear motion of the two cases cannot be the same. Since if the acceleration is F/m in both cases then the distence d is done in the same time in both cases and if the acceleration and the time is the same the linear terminal velocity has to be equal.

No. The acceleration of the center of mass is the same, but the time for which it is applied will not be. (The point of application of the force moves a distance d, but not the center of mass.)

 

[Horvath Bela]

And would the difference between the distences done by the center of mass and the point of contact be calculatable with concrete data e.g the rod is 1m its mass is 1 kg the distence is 1m and the force is 1N.

 

[Horvath Bela]

I have thought some about what you said. You said that the point of contact travels dintence d and the center of mass does not hance the time it was accelerated is less. But that implies that the contact pont always has larger acceleration than center of mass. How can you be sure about that?

 

[PeroK]

I have thought some about what you said. You said that the point of contact travels dintence d and the center of mass does not hance the time it was accelerated is less. But that implies that the contact pont always has larger acceleration than center of mass. How can you be sure about that?

First, you should Imagine your experiments in the two cases being carried out next to each other.

The rod being pushed at its centre moves uniformly. The rod being pushed at one end moves and rotates.

From your own calculations you can see the the motion of the centre of mass is the same in both cases.

There are, therefore, two key differences.

1) The acceleration of the point of contact must be greater in the rotating case - it's clearly moving ahead of the centre of mass.

2) The force acts over a greater distance in the same time in the rotating case, hence does more work in the same time.