I don&#x27;t buy perpendicular force = no work

Original post by DrSebWilkes

...

Firstly there's the issue of what causes a mass to set-off on it's (circular) trajectory. Work is done to get a stationary object moving.

Now it's moving. What causes it to undergo circular motion? It's velocity is perpendic to the attractive force between itself and the mass it's orbiting (at all times).

The values of various key quantities need to be right ofc, not everything moves circularly. Even the set-up of just being attracted to one mass is more-or-less unique to space.

The attractive/centripetal force doesn't do any work as 1) The objects KE/speed is always the same 2) It acts perpendic to the object's trajectory.

Your comment on changing direction, yet no gain/loss in KE (no work done on it) was a bit more interesting. But energy is a scalar, it doesn't care about direction, so the curvature of object's trajectory won't matter.

Note Qs like 'why is KE scalar', 'why is work defined as it is, or even exist as a concept' etc will just lead to a philosophical rabbit hole.

(edited 6 years ago)

Reply 3

Original post by DrSebWilkes

I guess it's "okay" I write down the right answers down in exams but I feel almost as if I am lying, even if I know all the big wigs know what they are saying is correct.

I don't get how in space (so no friction) a circular orbit has no work done to it; as I see it the velocity is always changing. sure, every cycle the change is 0; but at each little bit it's different? What annoys me is that I know that by kinetic energy, the speed remains the same (circular orbit) so since energy is scalar, no problem. Damn okay but the velocity is changing!! garghhh see I keep going around in circles?? (pun actually not intended)

Sorry has anyone got any other ways of explaining / proving to me that circular motion (best example is an orbit) doesn't use any energy?

Take an idealised situation where we have a light inextensible string. We tie one end of the string to a ball and the other end to a peg just sticking up from the top of a frictionless surface. We set the ball in circular motion on this frictionless surface. The only force that acts is the perpendicular* tension from the light inextensible string attached to the peg.

The ball will continue to circle around indefinitely since there are no forces acting in its instantaneous direction of motion. Its speed, and hence kinetic energy, remains constant.

The peg also remains stationary.

W = Fd \cos \theta

(where

\theta

is the angle between the force and distance vectors, and

F,d

are magnitudes of force and distance vectors respectively: note that work done is the dot product of the two vectors). Since the peg moves no distance, the force does no work on the peg.

Now let us apply the principle of conservation of energy. If any work were done on the ball, then to where would that energy have gone? Remember, there are no frictional forces in this situation to dissipate heat, and string itself is inextensible meaning it cannot store energy (Hooke's Law

F=kx

, integrating gives

W = \frac{1}{2}kx^2

, but

x=0

since no extension is possible, so no energy stored). So if work were being done on the ball by the perpendicular force, then the energy lost must be manifest in motion of the peg, but this is rooted to the spot. So by this argument, it appears that no work can be done on the ball.

Of course, this is a very idealised situation, but it does not take much effort to extrapolate this to the planets analogy (though true circular orbits are pretty much unheard of in astrophysics, from what I gather)

@Physics Enemy will probably correct me if my intuition is wrong

*Edit

(edited 6 years ago)

Reply 4

Original post by K-Man_PhysCheM

So if work were being done on the ball by the perpendicular force, then the energy lost must be manifest in motion of the peg, but this is rooted to the spot. So by this argument, it appears that no work can be done on the ball.

Only thing I had a qualm with; why would a stationary peg move (let alone have to) when 'losing' energy? Or why it must be rooted, can rotate in sync with the ball.

You gave a nice example with solid theory, but I think Seb was confused about how circular motion starts (and continues) and justification why 'stuff happens' due to a centripetal force and yet no work is done. Hopefully I've helped clarify these areas.

I'll also add here that 'circular motion at constant speed' isn't all that key. You can have acceleration along a trajectory due to a tangential force, and yet zero net work done in any closed loop, if the force is conservative. By extension, all centripetal forces (circular motion) are by default conservative.

(edited 6 years ago)

Reply 5

RogerOxon

21

Original post by DrSebWilkes

Sorry has anyone got any other ways of explaining / proving to me that circular motion (best example is an orbit) doesn't use any energy?

What about satellites? They continue to orbit at a constant speed for years, without any energy input.

For circular motion, if the force towards the centre did any work, where would that energy go? The potential and kinetic energy doesn't change.

Reply 6

Original post by Physics Enemy

That was the only thing I had a qualm with; why would a stationary peg move (let alone have to) when 'losing' energy? Or why it has to be rooted, can rotate in sync with the ball.

While having dinner just now I came up with this same thought. I supposed that energy would be transferred from ball to peg (or vice versa) when work is done and so peg's kinetic energy would change, but I suppose the peg need not move relative to the surface and instead the energy may be transferred via the peg to the surface. And indeed I agree that the peg could rotate.

I have a different refutation to my own analogy though: consider projecting the ball directly away from the peg (assume instantaneous acceleration). When ball distance reaches string length, string would become tense and immediately do work on the ball to get it to return directly back to the peg. By my logic, the peg would have to move in this case, but everyday experience shows that a well-rooted peg would not move. That being said, I suppose it would impart an equal force on the Earth, and hence do work on the Earth (though its effect would be negligible due to the Earth's mass).

Alternatively, if no energy is transferred to the peg/surface then the ball would return at equal speed (in opposite direction). By the argument in my previous post: since the kinetic energy of the ball did not change, then no work was done on the ball. But clearly in this case string tension acted along the line of movement, so work must have been done...

Or perhaps I am misunderstanding. Now that I think of it, if the string were inextensible then the change in direction would be instantaneous, so ball would move no distance while force was acting on it, so it seems in this case no work would be done on ball either! But that doesn't seem intuitive, so there's probably an underlying misunderstanding going on here on my part...

Original post by Physics Enemy

You gave a nice example with solid theory, but I think Seb was confused about how circular motion starts (and continues) and justification why 'stuff happens' due to a centripetal force and yet no work is done. Hopefully I've helped clarify these areas.

The more I think about it, the less I seem to understand too... but your post does help clear things up.

Original post by Physics Enemy

I'll also add here that 'circular motion at constant speed' isn't all that key. You can have acceleration along a trajectory due to a tangential force, and yet zero net work done in any closed loop, if the force is conservative. By extension, all centripetal forces (circular motion) are by default conservative.

Ahh, that makes a lot of sense. I hadn't considered those points while answering this question, and the final point is not one I had thought of specifically before since I had associated conservative forces with conservative fields.

Thanks for tagging me in this thread, it has raised some interesting points for discussion.

Spoiler

Reply 7

Original post by K-Man_PhysCheM

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The logic 'a stationary object can move if it loses energy' caused alarm bells, but the stipulation 'must' caused even more. You appeared to get stuck in the same trap as Seb, which I addressed 1st; 'If X, then ...' but you don't account for how motion starts and the initial work done there.

RE your hypotheticals, think you've answered it; a well rooted peg means it can exert a force on the ground, thus do work on / transfer energy to it. Other energy transfers relating to the peg are possible too (in small amounts I'd expect). The peg rotating in sync with the ball may have simplified your example somewhat.

(edited 6 years ago)

Reply 8

OP

Original post by K-Man_PhysCheM

The more I think about it, the less I seem to understand too

/thread xD #feels

Jokes aside though thanks for the help you guys

So Mr. Enemy, I understand how something starts with a given amount of energy (I actually love, for instance, the total energy of an orbit, because it's surprisingly intuitive for being a negative scale)

And in terms of angular momentum, nothing is changing. Yet another reason.

SolC even said, rightly, that because the tangiential motion is effectively 0 at a given point (as it keeps moving around) f*d=0, since d=0. Okay fair enough actually.

There is still one little niggle.

Let's say I have a HALF circle (or any 'cirlce' as long as it is not equal to n complete cirlces)

I start of, an instant before it turns into the circle, with velocity "v". By the time it gets to the other side, the velocity is -v. Ke stays the same BUT the momentum has changed. We had +P, now -P.

That's my niggle!

Reply 9

Original post by DrSebWilkes

Because the tangential motion is effectively 0 at a given point (as it keeps moving around), W = f*d=0, since d=0.

Tangential velocity is constant as the net tangential force is 0, so W = 0 as f = 0.

DrSebWilkes

Let's say I have a half circle (or any 'circle' as long as it's not equal to n complete circles).

I start off, an instant before it turns into the circle, with velocity 'v'. By the time it gets to the other side, the velocity is -v. KE stays the same but the momentum has changed. We had + P, now - P.

Momentum vector is always changing in circular motion, due to the changing direction and hence velocity, caused by the centripetal force.

And that's fine, momentum is only conserved in the absence of net external force.

(edited 6 years ago)

Reply 10

OP

Original post by Physics Enemy

Tangential velocity is constant as the net tangential force is 0, so W = 0 as f = 0.

Ooh yeah -_- I have also heard that one in my life time.

Original post by Physics Enemy

Momentum vector is always changing in circular motion, due to the changing direction and hence velocity, caused by the centripetal force.

And that's fine, momentum is only conserved in the absence of net external force.

Imagine that you have a source of something that incurs this perpendicular force (to the motion of the particle) and then this source magically turns off / is removed (etc.)

what then? Moreover, what if it was magically whisked away when the momentum vector was opposite; there is no disputing a violation here.

And (okay this is a push so please correct my wrongness if present) it's moved in a circle, which is some distance, so there was a force that acted in accordance to f*d, right? Which is work done?

Sorry to be a burden guys

Reply 11

Original post by DrSebWilkes

Imagine that you have a source of something that incurs this perpendicular force (to the motion of the particle) and then this source magically turns off / is removed (etc.)

what then? Moreover, what if it was magically whisked away when the momentum vector was opposite; there is no disputing a violation here.

Object would continue moving tangentially in a straight line in the direction of its movement the moment it was released.

No violation of momentum conservation, since momentum is conserved only when no force is acting. Clearly there is a centripetal force acting during circular motion, so you expect momentum to change. No work is done, however, since force vector has no component in the direction of the instantaneous "change in position" vector.

When that force is removed, by Newton's First Law the object will remain in motion at constant velocity since no forces would be acting on it.

Reply 12

OP

Original post by K-Man_PhysCheM

Object would continue moving tangentially in a straight line in the direction of its movement the moment it was released.

No violation of momentum conservation, since momentum is conserved only when no force is acting. Clearly there is a centripetal force acting during circular motion, so you expect momentum to change. No work is done, however, since force vector has no component in the direction of the instantaneous "change in position" vector.

When that force is removed, by Newton's First Law the object will remain in motion at constant velocity since no forces would be acting on it.

I get what you're saying but I am not sure you got what I said?

proof of change of momentum.png

velocity is equal (as I said above anyway) so the energy is equalp; no work done.

but the momentum changes and I think it's linkable to say work done is equivilant some momentum change, as you can guage for yourself from my diagram

Reply 13

Original post by DrSebWilkes

I get what you're saying but I am not sure you got what I said?

proof of change of momentum.png

velocity is equal (as I said above anyway) so the energy is equalp; no work done.

but the momentum changes and I think it's linkable to say work done is equivilant some momentum change, as you can guage for yourself from my diagram

Yes, that is what I understood by your post. There is no violation in momentum conservation because an external force acts on the object. EDIT: to clarify, there is no reason why momentum shouldn't change, since momentum is only conserved when no external force acts. Here, an external force acts, so momentum will change.

Why does a change in momentum imply work done?

(edited 6 years ago)

Reply 14

OP

Original post by K-Man_PhysCheM

Yes, that is what I understood by your post. There is no violation in momentum conservation because an external force acts on the object.

Why does a change in momentum imply work done?

Before I answer that, so you're saying if an entirely "still" object within some vast universe exerted a gravitational pull on the particle, and then the object was whisked away magically (or placed so far away "magically" that its effects were negligible) then the initial momentum of the universe was +mv, and then it ends as -mv.

wot ...

xD wow physics 'eh?

Answering your question, well I'm more than bound to be wrong so please do be patient (and correct me! ^_^ ), but since the object has moved a distance and forced acted on it to do so ...and that there was an impulse because we can do momentum after - momentum before etc. etc. I think implies work done?

Reply 15

Original post by DrSebWilkes

Imagine a particle that experiences this perpendic force (to its direction of motion) and then the force is removed. What then?

It travels in a straight (tangential) line at constant velocity and momentum, unless/until acted on by another net external force.

Original post by DrSebWilkes

It moved in a circle, distance 2πr, so there was a force providing W = f*d right?

Repeating 1st few lines of my 1st post: There's a force to get a stationary object moving, does W = f*d on it, providing its KE. After that, centrip/grav force just shapes its trajectory; no work done, constant KE.

Reply 16

Original post by DrSebWilkes

Before I answer that, so you're saying if an entirely "still" object within some vast universe exerted a gravitational pull on the particle, and then the object was whisked away magically (or placed so far away "magically" that its effects were negligible) then the initial momentum of the universe was +mv, and then it ends as -mv.

"Still" is a relative term. Nothing in the Universe can be still since there is no universal frame of reference: the Michelson-Morley experiment disproved the existence of this "aether" (or universal frame of reference). All motion is relative.

Newton's second law still applies... the Earth is in orbit around the Sun due to the gravitational attraction that provides the centripetal force on the Earth... and by Newton's second law, the Earth exerts a gravitational force of equal magnitude and opposite direction on the Sun.

Going back to your scenario; it depends what you define as your system. I was originally defining the system as just the particle in partial orbit. An external force acts on the system, and so its momentum changes from

+\textbf{p}

to

-\textbf{p}

. No violation since an external force acts on the system (just the half-orbiting particle).

However, the same force that acted on the particle must act on the central object that causes it to orbit. You very rightly state that if we consider the system as the whole Universe, then momentum is conserved since no external forces act on the Universe system (inb4 multiverse theory). It follows that the central thing's momentum would change from

0

to

+2\textbf{p}

*, such that the momentum of the Universe remains at a constant

\textbf{p}

.

If the central object truly did remain stationary, then we would have a violation of conservation of momentum; hence we can deduce that our premise of a stationary central object before AND after must be false, and asking such a question is meaningless.

*(if we choose an inertial frame of reference which is stationary relative to the initial velocity of the central thing).

Original post by DrSebWilkes

xD wow physics 'eh?

Wow physics indeed.

Original post by DrSebWilkes

Answering your question, well I'm more than bound to be wrong so please do be patient (and correct me! ^_^ ), but since the object has moved a distance and forced acted on it to do so ...and that there was an impulse because we can do momentum after - momentum before etc. etc. I think implies work done?

Work done is a scalar, impulse a vector.

By definition, work done is the dot product of the force and change in position vectors, ie

W = \displaystyle \int \textbf{F} \bullet \delta \textbf{x}

This is just mathematical jargon for saying that work done is equal to the force applied multiplied the component of the instantaneous change in position that is parallel to the direction of the force.

Often times, change in momentum DOES imply work is done:

If I project a ball straight at a wall such that there is a completely elastic collision, then kinetic energy is conserved, since the ball bounces back at the same speed. But clearly the wall has applied a force to the ball to change its momentum from

+\textbf{p}

to

-\textbf{p}

.

In this case, work IS done, despite no overall change in kinetic energy, because the force from the wall causes the ball to compress. This transfers the ball's kinetic energy to elastic potential over a very short distance parallel to the direction of the large force. Then the elastic potential energy applies a force on the wall, which replies with a reaction force that again does work on ball (force parallel to motion), and so EPE is transferred back to KE. Work has been done twice! (Momentum is conserved since force of ball on wall applies an impulse on wall/Earth).

Note that in this case the speed of the ball changed too: from initial speed, it decelerated to stationary and then accelerated again. So the ball's kinetic energy changed during the collision, and so work was done on it despite finishing up with the same KE as it started with.

------

However, in the case of circular motion, the force acts perpendicularly to the direction of motion and so does NOT slow down the particle AT ALL. This happens because there is no force component parallel to the direction of motion to cause it to slow down. This is just like in projectile motion, where (in the absence of resistive forces), the ball's horizontal component remains constant because gravity is perpendicular and so has no horizontal component.

Orbits are just a special type of projectile motion, where the object is constantly falling around a central point such that its speed remains constant. Kinetic energy is conserved throughout the whole trajectory, so there are no energy transfers. No work is done on the orbiting particle.

Hopefully I've explained that alright.

Original post by Physics Enemy

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The attractive/centripetal force doesn't do any work as 1) The objects KE/velocity is always the same 2) It acts perpendic to the object's trajectory.

.....

When you say KE/velocity is always the same, I doubt it is right. I can agree that KE is always the same for uniform circular, but the velocity is definitely not the same unless you mean speed.

Original post by Physics Enemy

.....

Note Qs like 'why is KE scalar', 'why is work defined as it is, or even exist as a concept' etc will just lead to a philosophical rabbit hole.

I have a different viewpoint as to the question 'why is KE scalar' will lead to a philosophical rabbit hole. I think it is a valid question. But the question of why we define KE as 1/2mv^2 is a different story.

Reply 18