Dumb question about inertia and rotational inertia

In summary, a repulsive force between two bodies of equal mass will cause the bodies to move at the same speed despite the body that is spinning.
  • #1
Graca
12
1
Consider two bodies, A and B, of equal mass set at a short distance. Body A is spinning and body B is at rest. Then, through some kind of electromagnetic device, a strong repulsive force is established between them. Will both be displaced at the same speed?
 
Physics news on Phys.org
  • #2
I don't see why not, since angular momentum and linear momentum are different quantities.
 
  • #3
But... Doesn't the spinning body offer resistance to change its axis of rotation?
 
  • #4
If the force is only repulsive, why would there be such a rotational change?
 
  • #5
Let's imagine the spinning body is a cylinder and the repulsive force is applied at one extremity? Then, would there be a rotational change?
 
  • #6
Are you saying that the force is applied in the same direction as the body's axis of rotation? In that case I would say that there's no torque and therefore no change in the rotational state. On the other hand, if it's applied to an extremity in any other direction, then I would say that there is a torque and therefore a change in the rotational state, and that changes everything.
 
  • #7
Like two pipes laying side by side (one of them is spinning) and the repulsive force being applied at the end. Would they get apart at the same speed then?
 
  • #8
Even if the repulsive force is applied at one extremity, its effect on the linear motion of the center of mass is the same. The calculations of the rotational acceleration and the linear acceleration of the center of mass can be done separately.
 
  • #9
But if the force is applied at the end, wouldn't that change the plane of rotation and therefore wouldn't the spinning pipe offer more resistance to movement?
 
  • #10
If they're lying parallel to one another and one is spinning such that its angular momentum points along the direction of the pipe, and both pipes are pushed at end points away from each other, then both pipes will take on different rotational states, and the one that was initially at rest will be spinning faster, while the one that was initially spinning will be spinning in a different direction altogether due to precession. (Now having trouble about the linear acceleration part. For me this is no longer so simple a question!)
 
  • #11
Graca said:
But if the force is applied at the end, wouldn't that change the plane of rotation and therefore wouldn't the spinning pipe offer more resistance to movement?
If the center of rotation is held in place, that will supply a force that opposes the linear motion and the result will be only the rotation. But if there is nothing holding the object in place, the center of mass will move linearly exactly as though the force was applied at the center.
 
  • #12
Notice that the equation F=mA does not specify that the force is applied at the center of mass. The equation is correct even if the force is off-center.
 
  • #13
Yes, two lying parallel pipes, but only one is spinning. The repulsive force being applied between them at the extremity of the pipes. I can imagine the pipes getting apart, but the pipe at rest moving with a greater speed that the pipe that is spinning. Is this correct?
 
  • #14
Graca said:
Yes, two lying parallel pipes, but only one is spinning.
How can one be spinning, but remain parallel? [EDIT: I understand this if the axis of rotation is along the length of the pipe. Thanks @snoopies622 ]
The repulsive force being applied between them at the extremity of the pipes. I can imagine the pipes getting apart, but the pipe at rest moving with a greater speed that the pipe that is spinning. Is this correct?
No. I am not completely sure what you mean, but the spin should not matter when the linear acceleration at the center of mass is calculated.
 
Last edited:
  • #15
If the same force is applied to both pipes for the same distance, then the energy invested in both is the same. I suppose one could therefore calculate the linear accelerations by doing the rotational parts first and then subtracting the energies. For me the rotational part is more intuitive.
 
  • #16
FactChecker said:
How can one be spinning, but remain parallel?

I guess he means that the axis of rotation is along the same direction as the pipe itself.
 
  • Like
Likes FactChecker
  • #17
Let's say each pipe is 1 meter long. They are lying parallel. One pipe is spinning around the direction of its length. At the end of each pipe there is some kind of switchable magnet. So, a repulsive force is switched on. The pipes' ends would get apart, but for the spinning one, that means changing its axis of rotation. Then, would it move away at a different speed than the one not spinning?
 
  • #18
You have stated the question clearly, but the more I think about it, the more I don't know the answer! :) Must go to bed anyway, will sleep on this. Good night and good luck, will check in in the morning. Good question.
 
  • #19
Thanks, good night!
 
  • #20
Graca said:
Let's say each pipe is 1 meter long. They are lying parallel. One pipe is spinning around the direction of its length. At the end of each pipe there is some kind of switchable magnet. So, a repulsive force is switched on. The pipes' ends would get apart, but for the spinning one, that means changing its axis of rotation. Then, would it move away at a different speed than the one not spinning?
No. The spin can give some stability of orientation, but it does not change the linear acceleration at the center of mass. The equation ##F = m A## is beautifully simple. It applies in all cases, regardless of complications.
 
  • #21
I understand that. But let me give another example: two parallel bycicle wheels (only one spinning, the same repulsive force as in the previous example). What means then stability if they move away at the same speed?
 
  • #22
The stability is regarding the orientation. Suppose the rotational momentum vector is very large in one wheel. Then it will take a lot of torque to change its angle much. That can be called stability since it appears to resist a change. But the center of mass of the wheel is a completely different issue. While resisting a rotation, it can still be pushed away as though it was not spinning at all.
 
  • #23
Even if i push far away off the center of mass? Wouldn't that be a rotation of the axis?
 
  • #24
Yes. Suppose you push on one end instead of at the center. That end will move more than the center would, but the center will move exactly according to ##F = mA##. So there will be a move away of the center and also a rotation, with the end you pushed moving away faster, the center moving exactly at ##F=mA##, and the other end moving less. The spin may reduce the amount of rotation due to the force (and make it rotate in a strange direction), but it will not change the linear motion of the center.
 
  • #25
Then, in the case of the two wheels (the spinning one and the not spinning one), with the centers moving according to F=mA , would each of the ends (on the same side) move at the same speed?
 
  • #26
No. The spinning one will resist rotation more than the non-spinning one. The change of the angular momentums will be equal and opposite, but that will appear as a smaller change of orientation in the spinning one.
 
  • #27
Ok, centers of mass will move away at the same speed. The push at the edge will make the spinning one rotate less, the not spinning one rotate more. Thank you.
 
  • Like
Likes FactChecker
  • #28
If I have two pipes floating in space and neither one is rotating, and I give one a shove in its center and another an equal shove at an endpoint - perpendicular to the direction of the pipe - they can't have equal resulting translational velocity because in the second case some of the energy invested goes into rotation instead.
 
  • #29
Pushing on the end with an equal force will cause the end to accelerate more than a push at the center will. Are you sure that you are putting the same amount of energy into the two cases?
 
  • #30
Well, I'm saying that's a premise of my thought experiment. How do I calculate how much of the energy goes into each component of motion? I took a statics and dynamics course many years ago, but there objects were always fixed at a point - not completely free to move.

(perhaps this is worth a new thread altogether)
 
Last edited:
  • #31
I think that the premise of the OP was an equal force, not equal energy. If you apply the same force over the same time on the end as you would at the center of mass, the end will accelerate more and move farther. That means more work is done, so more kinetic energy is put into the motion.

You can calculate the kinetic energy. In all cases, the center of mass reacts according to ##F = mA##. Depending on how long the force is applied, that gives you the amount of linear kinetic energy. The rotational kinetic energy can be calculated using the duration of the force, the torque and the inertial moments.
 
  • #32
FactChecker said:
I think that the premise of the OP was an equal force, not equal energy.
That's why I was thinking of starting a new thread.
FactChecker said:
In all cases, the center of mass reacts according to ##F = mA##.
Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass, although I don't know how to prove it one way or the other.

Edit: Actually now that I think about it, I'll try approaching it in a collision/conservation of momentum framework instead, perhaps that'll make things clear to me.
 
  • #33
snoopies622 said:
Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass
Intuitively, if you push on one end, that end will move more than if you were pushing at the center. But the other end will move less than if you were pushing at the center. The motion at the center will be the average of the motions at the two ends, which is exactly the same as if you had pushed at the center.
 
  • #34
Yes that sounds credible, thanks FactChecker. Will now make up and work on some specific problems to see what happens.
 
  • #35
snoopies622 said:
Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass,...
Applying the same (force x time = impulse) will result in the same linear kinetic energy.
Applying the same (force x distance = work) will result in the same total kinetic energy.
 

Related to Dumb question about inertia and rotational inertia

1. What is inertia?

Inertia is the tendency of an object to resist changes in its state of motion. This means that an object at rest will stay at rest, and an object in motion will continue moving at a constant velocity, unless acted upon by an external force.

2. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotation. It depends on the mass and distribution of mass around an axis of rotation.

3. How is rotational inertia different from linear inertia?

Linear inertia refers to an object's resistance to changes in its linear motion, while rotational inertia refers to an object's resistance to changes in its rotation. They are both forms of inertia, but they apply to different types of motion.

4. How does rotational inertia affect an object's movement?

Rotational inertia affects an object's movement by determining how easily it can be rotated. Objects with a higher rotational inertia will require more torque to change their rotation, while objects with a lower rotational inertia will require less torque.

5. How can rotational inertia be calculated?

Rotational inertia can be calculated using the formula I = mr², where I is the moment of inertia, m is the mass of the object, and r is the distance from the axis of rotation to the mass. This formula assumes that the object has a uniform distribution of mass.

Similar threads

Replies
5
Views
2K
Replies
10
Views
1K
Replies
12
Views
620
Replies
1
Views
1K
Replies
10
Views
1K
Replies
10
Views
1K
Replies
2
Views
2K
  • Mechanics
Replies
30
Views
6K
  • Mechanics
Replies
2
Views
2K
Replies
3
Views
803
Back
Top