Silly Rotational Dynamics Question

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Jorpho
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Silly Rotational Dynamics Question

Postby Jorpho » Wed Oct 28, 2015 2:12 pm UTC

I'm sure the answer to this is dead obvious, but I'm not seeing it right now. Maybe it's just been too long.

Say you've got an object moving along at constant velocity on a frictionless plane. You want to make it move in a circle of radius R, so you apply a force perpendicular to its velocity vector, i.e. which constantly changes direction for the duration of the motion. The object's speed and thus kinetic energy remains constant, so no "work" is done on the object. But energy is still being expended, isn't it?

Can this energy expenditure be calculated just by considering the rotation of the object, i.e., since one circular motion will cause the object to rotate 360 degrees? Or is there some other means of calculating the energy expended?

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Re: Silly Rotational Dynamics Question

Postby cyanyoshi » Wed Oct 28, 2015 2:29 pm UTC

In classical mechanics, you don't need to use up any energy to make an object move in a circle. If you place a moving object on a horizontal, frictionless, circular track, then it should go around and around forever because its energy is conserved (ignoring friction, of course). If zero work needs to be performed, then zero energy needs to be added.

In the real world, it usually takes energy to generate a force, but that is just a result of inefficiency. We usually work with spherical cows in a vacuum in physics class! :wink:

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Re: Silly Rotational Dynamics Question

Postby Jorpho » Wed Oct 28, 2015 2:51 pm UTC

cyanyoshi wrote:If you place a moving object on a horizontal, frictionless, circular track, then it should go around and around forever because its energy is conserved (ignoring friction, of course).
But even in the absence of friction, a force is still being exerted on the object by the edge of the track.

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Re: Silly Rotational Dynamics Question

Postby elasto » Wed Oct 28, 2015 3:02 pm UTC

Jorpho wrote:But even in the absence of friction, a force is still being exerted on the object by the edge of the track.

If you place an object on a table, the table exerts a force on the object which balances the force of gravity - hence the object doesn't move. But this force doesn't 'use up' any energy - the table isn't only going to be able to keep the object stationary for a little while until its 'battery' runs out, whereupon the object falls through it; Theoretically the table could hold the object stationary indefinitely.

Likewise an object could follow a circular frictionless track indefinitely without any work being performed or energy expended.

(Now it's true that the edge of the track exerts a force on the object, and the object exerts the opposite force on the track, but neither the track nor the ball has to expend energy to do so. If the track isn't infinitely massive then it would move as a result of the object's outward push, and so follow an oscillating motion of its own. However this motion too would theoretically continue indefinitely.)

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Re: Silly Rotational Dynamics Question

Postby Jorpho » Wed Oct 28, 2015 5:23 pm UTC

elasto wrote:Now it's true that the edge of the track exerts a force on the object, and the object exerts the opposite force on the track, but neither the track nor the ball has to expend energy to do so.
Technically, isn't energy being lost in the form of the chemical bonds between the atoms of the object and the track, which would in theory wear away over enough time?

How about this: let's say the object is a triangle, pointing "North", starting at the top of a circular track and moving clockwise. And let's say the triangle is mounted to the track by a pin going through its center, around which the triangle can freely rotate. Won't the triangle itself start rotating around the pin as the pin moves around the track, and won't it keep rotating if the pin is stopped?

(I'm sure I'm not thinking about this the right way.)

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Re: Silly Rotational Dynamics Question

Postby elliptic » Wed Oct 28, 2015 5:29 pm UTC

Jorpho wrote:But even in the absence of friction, a force is still being exerted on the object by the edge of the track.

To clarify further: energy = force x distance moved *in the direction of the force*.

For an object held in a circular orbit (which is what we're talking about) the force towards the centre is always at right-angles to the orbital motion, so no energy is being expended. That's all there is to it.

The thing that *will* require an input of energy (when capturing the object into orbit from linear motion) is if you want to keep the same side pointing towards the centre as it orbits. That means the object has to rotate once per orbital revolution, and the angular KE for that has to come from somewhere.

How about this: let's say the object is a triangle, pointing "North", starting at the top of a circular track and moving clockwise. And let's say the triangle is mounted to the track by a pin going through its center, around which the triangle can freely rotate. Won't the triangle itself start rotating around the pin as the pin moves around the track, and won't it keep rotating if the pin is stopped?

If the pin is through the triangle's centre of mass then no, it won't rotate.

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Re: Silly Rotational Dynamics Question

Postby Jorpho » Wed Oct 28, 2015 5:48 pm UTC

elliptic wrote:The thing that *will* require an input of energy (when capturing the object into orbit from linear motion) is if you want to keep the same side pointing towards the centre as it orbits. That means the object has to rotate once per orbital revolution, and the angular KE for that has to come from somewhere.
Right, that's what I was thinking.

elliptic wrote:
How about this: let's say the object is a triangle, pointing "North", starting at the top of a circular track and moving clockwise. And let's say the triangle is mounted to the track by a pin going through its center, around which the triangle can freely rotate. Won't the triangle itself start rotating around the pin as the pin moves around the track, and won't it keep rotating if the pin is stopped?
If the pin is through the triangle's centre of mass then no, it won't rotate.
I stopped too soon. By "won't rotate", that means the triangle will keep pointing "North" (in which case it rotates relative to the pin) as opposed to pointing away from the center of the circle (in which case it rotates relative to "North") ?

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Re: Silly Rotational Dynamics Question

Postby gmalivuk » Thu Oct 29, 2015 1:10 am UTC

It will keep pointing north, yes.

In what sense is this rotating around the pin?
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Re: Silly Rotational Dynamics Question

Postby Jorpho » Thu Oct 29, 2015 2:40 am UTC

...Oh, right, I was thinking that the pin itself must be rotating. But that's not necessarily true either.

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Re: Silly Rotational Dynamics Question

Postby Zamfir » Thu Oct 29, 2015 8:06 am UTC

Jorpho wrote:
elliptic wrote:The thing that *will* require an input of energy (when capturing the object into orbit from linear motion) is if you want to keep the same side pointing towards the centre as it orbits. That means the object has to rotate once per orbital revolution, and the angular KE for that has to come from somewhere.
Right, that's what I was thinking.

Perhaps this was already clear to you, but just to be certain:the energy that Elliptic describes is a one-time amount. You need some kinetic energy for the movement of the entire object around the track, and some extra kinetic energy if the object has to stay pointed at the centre all the time.

Once the object is moving and rotating the way you want, it will keep moving and rotating without further energy input. If we assume away friction, of course.

Suppose we have an object on a linear path with velocity V, with a pointer pointing north. We capture the object and force it to move in circle, with the pointer still pointing north all the time. We don't add any extra energy. Then the object will move round the circle with velocity V (measured at its centre of gravity), and it will keep moving that way.

Option number 2: we force it to move in a circle, and the pointer points to the centre of the circle all the time. We still do not add extra energy. Now some of the original kinetic energy goes into the extra rotation of the object. So the object will move around the circle with a constant velocity that is less than V. Again, after the capture it will stay that way forever without further input.
Technically, isn't energy being lost in the form of the chemical bonds between the atoms of the object and the track, which would in theory wear away over enough time?

"Chemical" is not quite the right word. For the rest, yes, in practice there will be forces on the microscopic level between the object and the track. These will sap energy from the system. However, that's exactly the friction that you assumed away earlier. "Friction" is just a catch-all word for such effects.

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Re: Silly Rotational Dynamics Question

Postby Jorpho » Fri Oct 30, 2015 3:59 am UTC

Zamfir wrote:Suppose we have an object on a linear path with velocity V, with a pointer pointing north. We capture the object and force it to move in circle, with the pointer still pointing north all the time. We don't add any extra energy. Then the object will move round the circle with velocity V (measured at its centre of gravity), and it will keep moving that way.
But at the moment of capture, somewhere there would need to be either some input of energy or some loss of velocity, even if the pointer keeps pointing north, right? Like, if the object was snagged by a hook, suddenly that hook would gain rotational energy, and that would have to come from somewhere.

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Re: Silly Rotational Dynamics Question

Postby gmalivuk » Fri Oct 30, 2015 4:10 am UTC

Sure, in the real world, but the limit as the mass of the hook goes to zero is zero loss of energy.
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Re: Silly Rotational Dynamics Question

Postby Zamfir » Fri Oct 30, 2015 5:44 am UTC

And you don't need a hook. You could have a p-shaped track, or guide rail or marble run. Object comes in through the leg of the p, ends up circling around the o of the p. There is no fundamental need for other moving parts than the object itself

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Re: Silly Rotational Dynamics Question

Postby Jorpho » Sat Oct 31, 2015 3:53 am UTC

Wouldn't a massless hook or a frictionless track be completely incapable of exerting the force necessary to cause the change in trajectory?

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Re: Silly Rotational Dynamics Question

Postby Twistar » Sat Oct 31, 2015 4:38 am UTC

Just as a note: One convincing reason people hold the misconception that it costs energy to exert a force is the following.

Imagine standing up straight and holding a barbell with both arms. If you do a bicep curl and stop and hold the barbell so that your upper arms are horizontal and supporting its weight you will NOT be able to do hold this position forever*. So going by personal experience it makes sense that something (perhaps chemical energy that has been converted from food) is being used up when a force is exerted. The difference, however, is that when humans are doing a static exercise like that the individual muscle cells in their muscle are rapidly firing/springing/twitching/using up energy and then relaxing and then repeating the process. Not all at once of course. In other words, the muscles cells can only tighten for a short period of time by using up energy. They can't affix themselves into a tightened position and hold it long term.

Summary: Humans actually have to use energy to exert constant forces. Tables and banked friction-free circular tracks do not. This is a source of much misconception


*in the way that a table could up the barbell forever.

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Re: Silly Rotational Dynamics Question

Postby Link » Sat Oct 31, 2015 10:04 am UTC

Jorpho wrote:Wouldn't a massless hook or a frictionless track be completely incapable of exerting the force necessary to cause the change in trajectory?

You could still get collisional forces in those cases. You could also use a magnetic or electric field as your hook.

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Re: Silly Rotational Dynamics Question

Postby gmalivuk » Sat Oct 31, 2015 11:38 am UTC

Jorpho wrote:Wouldn't a massless hook or a frictionless track be completely incapable of exerting the force necessary to cause the change in trajectory?

No.

The amount of tension a hook or string can withstand before breaking is correlated with its mass, but it doesn't come from the mass. Nor does the ability of a wall to keep you from passing through it come from friction.
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Re: Silly Rotational Dynamics Question

Postby Jorpho » Sat Oct 31, 2015 3:20 pm UTC

gmalivuk wrote:The amount of tension a hook or string can withstand before breaking is correlated with its mass, but it doesn't come from the mass. Nor does the ability of a wall to keep you from passing through it come from friction.
Fair enough. I suppose I'm thinking that with zero mass, there would be no transfer of momentum to the hook, as Mr. Elasto alluded to earlier.

Link wrote:You could also use a magnetic or electric field as your hook.
In non-ideal situations, there would still be a corresponding transfer of motion to whatever's creating the field, right?

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Re: Silly Rotational Dynamics Question

Postby Link » Sat Oct 31, 2015 3:59 pm UTC

Jorpho wrote:Fair enough. I suppose I'm thinking that with zero mass, there would be no transfer of momentum to the hook, as Mr. Elasto alluded to earlier.

Link wrote:You could also use a magnetic or electric field as your hook.
In non-ideal situations, there would still be a corresponding transfer of motion to whatever's creating the field, right?

Yes, that's true -- but the fact remains that you don't specifically need a mass to change the direction of motion. In real life there almost always happens to be a mass involved somehow, but real life tends to be messy like that.

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Re: Silly Rotational Dynamics Question

Postby gmalivuk » Sat Oct 31, 2015 6:25 pm UTC

Jorpho wrote:
gmalivuk wrote:The amount of tension a hook or string can withstand before breaking is correlated with its mass, but it doesn't come from the mass. Nor does the ability of a wall to keep you from passing through it come from friction.
Fair enough. I suppose I'm thinking that with zero mass, there would be no transfer of momentum to the hook, as Mr. Elasto alluded to earlier.
Obviously the hook would be attached to something on the other end, and the momentum would be transferred to that.

Link wrote:You could also use a magnetic or electric field as your hook.
In non-ideal situations, there would still be a corresponding transfer of motion to whatever's creating the field, right?
In the limit (again), as the field generator gets bigger, the energy transfer corresponding to the necessary momentum transfer shrinks to nothing.
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