A long rod being pushed in zero g vacuum

In summary: KE will be the same.In summary, the rod cannot gain rotational kinetic energy and therefore will not rotate no matter how long the rod or how big of a force is applied.
  • #1
Horvath Bela
14
0

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

 
Physics news on Phys.org
  • #2
No, your logic is not sound. Think about momentum instead of energy.
 
  • #3
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.
 
  • #4
Horvath Bela said:
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 
  • #10
Horvath Bela said:
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?
 
  • #11
Horvath Bela said:
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.)
 
  • #12
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.
 
  • #13
The work done in both cases is F*d so they are exacty the same.
 
  • #14
Horvath Bela said:
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.
 
  • #15
Horvath Bela said:
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.)
 
  • #16
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.
 
  • #17
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?
 
  • #18
Horvath Bela said:
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.
 

1. How does the absence of gravity affect the movement of a long rod in a vacuum?

In a zero gravity vacuum environment, there is no force of gravity acting on the long rod. This means that the rod will not experience any weight or resistance, allowing it to move in a straight line with minimal friction or air resistance.

2. What factors influence the speed at which the long rod moves in a zero gravity vacuum?

The speed of the long rod in a zero gravity vacuum is primarily influenced by the initial force applied to push it, as well as the length, weight, and shape of the rod. Other factors such as air resistance and external forces may also play a role.

3. How does the absence of air resistance affect the motion of the long rod in a zero gravity vacuum?

In a zero gravity vacuum, there is no air resistance to slow down or alter the motion of the long rod. This means that the rod will continue moving at a constant speed until acted upon by an external force, such as another object or the end of the vacuum chamber.

4. What happens to the long rod when it reaches the end of the vacuum chamber in zero gravity?

When the long rod reaches the end of the vacuum chamber in a zero gravity environment, it will continue moving in a straight line with the same velocity until it collides with another object or is acted upon by an external force. The lack of air resistance means that the rod will not slow down or change direction upon exiting the chamber.

5. Are there any potential dangers or risks associated with conducting experiments with a long rod being pushed in zero gravity vacuum?

As with any scientific experiment, there are always potential risks and hazards that need to be considered. Some potential risks when working with a long rod in a zero gravity vacuum may include collisions with other objects or damage to equipment. Proper safety measures and precautions should always be taken when conducting experiments in a zero gravity vacuum environment.

Similar threads

Misunderstanding Work-energy theorem and center of mass properties
    • Introductory Physics Homework Help
Replies
7
Views
704
Linear and angular momentum problem: Ball hitting a rod
    • Introductory Physics Homework Help
2
Replies
62
Views
10K
Linear speed of sphere as it passes through lowest point
    • Introductory Physics Homework Help
Replies
2
Views
1K
Rod on a Plane: Calculating Angular and Linear Motion After Impact"
    • Introductory Physics Homework Help
2
Replies
38
Views
2K
Two rods, each with a free and a fixed ball and a spring
    • Introductory Physics Homework Help
Replies
9
Views
836
Freely hinged rods on a table - Linear velocities of CM after the impulse
    • Introductory Physics Homework Help
Replies
11
Views
2K
A bullet hits a rod attached to a pivot at one of its ends....
    • Introductory Physics Homework Help
Replies
10
Views
3K
Mass hanging on a steel wire
    • Introductory Physics Homework Help
Replies
15
Views
325
Hinged rod rotating, falling and hitting a mass
    • Introductory Physics Homework Help
Replies
11
Views
1K
Rod swinging and hitting a ball
    • Introductory Physics Homework Help
Replies
31
Views
3K

Hot Threads

Recent Insights

Back
Top