Frictionless Rolling Motion

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Alkatran
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Main Question or Discussion Point

I want to make a simple game, where you have a space ship with thrusters on the left and right. Firing one at a time both spins and propels the ship.

The problem is that, in physics 1000, we always assumed our rolling motion was on a surface, allowing us to relate the linear speed to the rotational speed.

Initially I thought I would just ignore the fact that both types of motion would apply, but I then figured the sum of rotational and linear energy had to stay constant (subtracting out work done, of course).

So I guess it boils down to: if I have an object with moment of inertia I and apply a forward impulse a distance R to the right, how will it react?
 

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  • #2
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i assume the object is free in space.
it really depends on the surface where the impulse is applied and the orientation between them. the force vector component pointing to the center of mass will acceleration the object linearly. while the other component will cause the object the rotate around the center of mass.
 
  • #3
Danger
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And keep in mind that in the absence of friction, there will be no rolling. That's caused by a spherical or cylindrical object with forward momentum along with friction on the top or bottom. Without friction, it will just slide.
 
  • #4
Doc Al
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tim_lou said:
... the force vector component pointing to the center of mass will acceleration the object linearly. while the other component will cause the object the rotate around the center of mass.
Just to be clear: A (net) force applied to an object will accelerate the center of mass according to Newton's 2nd law. The full force provides the acceleration, not just the component pointing to the center of mass.
 
  • #5
Alkatran
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Yes, I know both of those things. I suppose I didn't make it clear I meant in free space, and that the force was being applied like so (o = object, /\ = force, -- = weightless arm):

o--/\


So, given that any net force will accelerate the center of mass, do I really just ignore the fact that it will cause rotation while calculating the linear acceleration (remember this is a computer program)?
 
  • #6
Doc Al
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Alkatran said:
So, given that any net force will accelerate the center of mass, do I really just ignore the fact that it will cause rotation while calculating the linear acceleration (remember this is a computer program)?
That's correct. The same force produces both a linear acceleration of the center of mass and (depending upon the direction of the force) an angular acceleration about the center of mass. Calculate them separately.
 
  • #7
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Doc Al said:
That's correct. The same force produces both a linear acceleration of the center of mass and (depending upon the direction of the force) an angular acceleration about the center of mass. Calculate them separately.
Doesn't that means that applying the same force over the same distance you can get more work done? Does rotational energy not count towards the net energy?

Take this system for example:

Code:
m = 1
I = 1
r = 1
    <
    |
\/--+--/\
    |
    >
A ferris wheel in space, with engines instead of seats. All engines are aimed parrallel to the tangent, and clockwise. To move the ferris wheel in one direction, only the engine currently at the top is ever firing (where top is defined so that we go in the direction we want).

So you're saying this system will have a linear motion identical to one of the same mass that has an engine directly in the center, like so:

Code:
m = 1
  |
--+--<
  |
But the first system will somehow have a higher net energy because of the added rotational energy.
 
  • #8
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Doesn't that means that applying the same force over the same distance you can get more work done? Does rotational energy not count towards the net energy?
Yes, rotational energy does count in the total energy of the system.

You are supplying an impulse not a force over a distance to the rigid body.

It is the same effect as translation and rotation, but your two cases would require different size rockets, because as you correctly pointed out, the rotation and translation require more energy to the system. It would be the sum of the two energies.
 
Last edited:
  • #9
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cyrusabdollahi said:
Yes, rotational energy does count in the total energy of the system.

You are supplying an impulse not a force over a distance.

It is the same effect, but your two cases would require different size rockets, because as you correctly pointed out, the rotation and translation require more energy to the system.
I used force over distance because I was trying to relate it directly to the energy put into the system. What do you mean by different size rockets? Let's assume that each rocket is accounted for in the mass and moment of intertia of the whole system and they all put out the same force F while firing.
 
  • #10
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What I am saying is that if you want to fire an off center rocket, it will cause a rotation and a linear translation.

If you put that a rocket at the COM and fire it, you will only get a linear translation. Because you have no rotation, the energy of the system is purely translational. This means the rocket motor needs to provide less thrust to produce this same linear translation (in terms of energy).

I gota run, get a pro like Doc Al :cool: to explain what I have probably confused you with <sorry>.
 
  • #11
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cyrusabdollahi said:
What I am saying is that if you want to fire an off center rocket, it will cause a rotation and a linear translation.

If you put that a rocket at the COM and fire it, you will only get a linear translation. Because you have no rotation, the energy of the system is purely translational. This means the rocket motor needs to provide less thrust to produce this same linear translation (in terms of energy).

I gota run, get a pro like Doc Al :cool: to explain what I have probably confused you with <sorry>.
I understand what you're saying, except that it's the opposite of what Doc Al is saying. Based on what you've said, if I follow his instructions to calculate them seperately, I will end up with the energy of the system increasing faster than the work done increases, which is impossible.
 
  • #12
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(o = object, /\ = force, -- = weightless arm):
With this configuration:

o--/\

it's like rowing a boat with one oar, you'll just go round in circles.

If you have this configuration:

o--o/\

or should I say this

o--o
/\




it'll be more like a canoe.
 
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  • #13
Doc Al
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Alkatran said:
I understand what you're saying, except that it's the opposite of what Doc Al is saying. Based on what you've said, if I follow his instructions to calculate them seperately, I will end up with the energy of the system increasing faster than the work done increases, which is impossible.
Cyrus is saying exactly the same thing as I am (I think). Realize that my comments about force producing a linear acceleration of the cm is instantaneously true. If the force changes, so will the acceleration of the cm. (If your force is a rocket attached to the edge of the object, then the direction of the force continually changes.)

I'll respond to your earlier post shortly.
 
  • #14
Doc Al
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Alkatran said:
Doesn't that means that applying the same force over the same distance you can get more work done?
Not at all. Work is always:
[tex]\int \vec{F}\cdot \,d\vec{s}[/tex]

Does rotational energy not count towards the net energy?
Of course rotational energy counts.
Take this system for example:

Code:
m = 1
I = 1
r = 1
    <
    |
\/--+--/\
    |
    >
A ferris wheel in space, with engines instead of seats. All engines are aimed parrallel to the tangent, and clockwise. To move the ferris wheel in one direction, only the engine currently at the top is ever firing (where top is defined so that we go in the direction we want).

So you're saying this system will have a linear motion identical to one of the same mass that has an engine directly in the center, like so:

Code:
m = 1
  |
--+--<
  |
But the first system will somehow have a higher net energy because of the added rotational energy.
I must not have explained myself very well. (Sorry.) In the first case the object will translate as well as rotate. If I understand your intention, you want to arrange things so that the force is always pointing in the same direction, tangent to the object. If that's what you mean, then the force is constant (same magnitude and direction). Thus the acceleration of the cm will be constant. The system will have both translational and rotational kinetic energy. Note that the point of application of the force moves with a different displacement than the center of mass; the work done by the force equals the change in total kinetic energy.

In the second case the object doesn't rotate, thus the force remains constant. In this case, the acceleration of the cm is constant. All work goes into translational kinetic energy. Note that the displacement of the point of application of the force is identical to the displacement of the cm.

In both cases the work done equals the change in total kinetic energy. The difference, as you apparently realize, is that in the first case you need to move the force through a greater distance to move the cm than you do in the second case. (That added work, of course, becomes the rotational kinetic energy.)

Note that the linear acceleration of the cm is the same in both cases.

(As I said originally: The linear acceleration of the cm is given by the applied force via Newton's 2nd law.)

Let me know if any of this is unclear.
 
  • #15
Alkatran
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Doc Al said:
Not at all. Work is always:
[tex]\int \vec{F}\cdot \,d\vec{s}[/tex]
What kind of integral is that, I've never evaluated one that had a vector. I remember work being equal to force times distance (assuming a constant force in the same direction as the movement).

I think I understand where the difference is coming from. The engine is moving a greater distance than the center of mass in the rotating case, meaning more work can be done.

I'll calculate them seperately, given this information.

so far so good:
 

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  • #16
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Alkatran said:
What kind of integral is that, I've never evaluated one that had a vector. I remember work being equal to force times distance (assuming a constant force in the same direction as the movement).
That integral is just a more general way of defining work that applies in all cases, not just for constant force parallel to the movement. In your case, it's equivalent to F*d, where d is the distance the point of application of the force moves.

I think I understand where the difference is coming from. The engine is moving a greater distance than the center of mass in the rotating case, meaning more work can be done.
Exactly right.
 
  • #17
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cyrusabdollahi said:
Yes, rotational energy does count in the total energy of the system.
this cannot possibly be true.
If it is, then a object rotating while it's center of mass is at rest does not have any energy?
but if you add up all the kinetic energy of individual particles, the energy is NON-ZERO.
 
  • #18
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tim_lou said:
this cannot possibly be true.
If it is, then a object rotating while it's center of mass is at rest does not have any energy?
but if you add up all the kinetic energy of individual particles, the energy is NON-ZERO.
Huh? Reread what you quoted: Cyrus said "Yes, rotational energy does count in the total energy of the system."
 
  • #19
Alkatran
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Doc Al said:
That integral is just a more general way of defining work that applies in all cases, not just for constant force parallel to the movement. In your case, it's equivalent to F*d, where d is the distance the point of application of the force moves.
So the 's' in the integral means displacement, and not time?

Is there a way to mark the thread resolved?
 
  • #20
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sorry about that, didnt read everything before posting.
(was having my lunch while posting)
and yes the linearly acceleration is net force vector divided by m.

:uhh: :devil:
 
  • #21
Doc Al
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Alkatran said:
So the 's' in the integral means displacement, and not time?
Right. Don't know why, but that's standard usage.

Is there a way to mark the thread resolved?
Nope. Just ignore it and it will drift down the screen...
 

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