- #1

- 69

- 0

## Main Question or Discussion Point

Its been bugging me for a while now-can pseudo forces do work? Is the calculation of amount of work done dependent on our frame of reference?

- Thread starter dpsguy
- Start date

- #1

- 69

- 0

Its been bugging me for a while now-can pseudo forces do work? Is the calculation of amount of work done dependent on our frame of reference?

- #2

- 69

- 0

Cmon!Does nobody know?I've been waiting for 3 days!!

- #3

Doc Al

Mentor

- 44,867

- 1,114

In any case, the work done by a force definitely depends on the frame of reference. (Even without involving noninertial frames.) After all, in a frame in which the point of application doesn't move, the work done is zero.

As for pseudoforces being able to do work, I see no reason why not. (As long as you are in a frame in which those forces exist.)

- #4

- 34

- 0

- #5

- 29

- 0

I have a subsidiary question :

Do you consider that gravity is a pseudo-force ?

Do you consider that gravity is a pseudo-force ?

- #6

- 69

- 0

And if the force does not exist,how can it do work?

As for the work done being dependent on the frame of reference, I have just one question.In a hydroelectric plant water falls from a great height from a dam and causes the blades of a turbine to rotate. Suppose water falls with a terminal vel. v.If an observer falls along with the water,(assuming that he does not die on falling!!),the water is at rest with respect to him and hence its energy does not change.So how would he explain the turning of the blades given that he knows the turbine is fixed to the ground and cannot come up to meet the water. Where would the blades get the energy from if the water does not lose any?

- #7

- 2,985

- 13

As for your water wheel example, your falling with a terminal velocity v. So in your frame of reference you have zero kinetic energy, but the water wheel appears to be moving towards you at a velocity v, so it appears to have the kinetic energy in this frame. hmm, as for gravitational potential energy, I *think* in your moving frame of reference you can call that zero, so the water wheel also has a changing potential energy relative to you. Its like thinking of the situation in opposite. In the wheels frame of reference, the water comes down and hits it. In the waters frame, the wheel comes up and hits it. The only thing im not sure about what I said is that if that were true, the heavy water wheel appearing to move at a velocity v, would have a huge amount of kinetic energy, and it seems to discredit what I wrote, so dont belive it! Someone help me out, now im confused too! :-)

- #8

mukundpa

Homework Helper

- 524

- 3

With application of sudden brakes to a truck, a box appear to slide forwards on the surface of truck.

We call that as the truck is retarding, in the frame fixed with the bus we have to consider a pseudo force in forward direction.

Now my question is that, as the box moves forward against friction, which force is doing work against friction?

- #9

Doc Al

Mentor

- 44,867

- 1,114

The difference between inertial (pseudo) and "real" forces is that real forces have agents (something that exerts the force) while inertial forces do not. Inertial forces are simply due to viewing things from a non-inertial reference frame. There is no reaction to an agentless force!dpsguy said:But do pseudo forcesreallyexist? I mean, a force should have an equal and opposite reaction,no? Now consider a small block placed on top of a large frictionless block. The large block is accelerating.If we analyse the situation from the frame of reference of the small block ,we use a pseudo force in a direction opposite to the acc.,so that we can apply Newtons laws.But where is the reaction to this pseudo force?

And if the force does not exist,how can it do work?

Consider your example of a small block sliding without friction on large block that is accelerating. If you view things from the accelerating frame of the large block, then the inertial force "pushes" the small block along. So the inertial force does work on the small block, increasing its kinetic energy. (Of course, from the

If you view things from the frame of the small block--an inertial frame--then you'll expect to find a real force pushing that large block. (Else why would it accelerate?)

From the frame of the water/observer, the turbine (and the earth itself!) is rushing towards him with plenty of energy! And the speed of the water with respect to the earth will change when it hits the turbines, of course. If you want to analyze the interaction from the inertial frame moving with speed v, you must stick to it. (Since the water is really accelerating as it falls, this is a messy frame to use.)As for the work done being dependent on the frame of reference, I have just one question.In a hydroelectric plant water falls from a great height from a dam and causes the blades of a turbine to rotate. Suppose water falls with a terminal vel. v.If an observer falls along with the water,(assuming that he does not die on falling!!),the water is at rest with respect to him and hence its energy does not change.So how would he explain the turning of the blades given that he knows the turbine is fixed to the ground and cannot come up to meet the water. Where would the blades get the energy from if the water does not lose any?

You can understand the frame dependence of work and energy more simply by analyzing this problem: An "ideal" car (no internal friction, air resistance) uses an amount of fuel to accelerate from 0 mph to V mph. We know the energy needed is [itex]1/2 m v^2[/itex]. And to accelerate from V to 2V, it takes an additional 3 times that energy (the total energy is [itex]1/2 m (2v)^2 = 2 m v^2[/itex]). But what if you viewed things from a frame moving at speed V? In that frame the second burst of speed only takes the car from 0 to V? Does the amount of fuel needed depend on the frame that you view things from?

- #10

- 29

- 0

Then gravity is a pseudo-force .... because for someone in free fall gravitational force doesn't exist (GR). I'm not sure of me that's why i asked it before but i didnt get any answer :(Inertial forces are simply due to viewing things from a non-inertial reference frame.

That would also eplain why when you put a not-calibrated accelerometer on a table it indicates that the acceleration is pointing

That's because if you put your accelerometer in an elevetaor in space with nothing around it : it indicates 0 (elevator isnt moving). If you want someont in the elevator to feel like if he were on earth, you have to make the elevator go

Last edited:

- #11

Doc Al

Mentor

- 44,867

- 1,114

Sorry for ignoring your question, but I was purposely trying to stick to classical mechanics and avoid GR considerations. While you are right that gravity is a "pseudo-force" in a GR context, you can't simply treat it like an ordinary inertial force (like centrifugal force) within classical mechanics.BioBen said:Then gravity is a pseudo-force .... because for someone in free fall gravitational force doesn't exist (GR). I'm not sure of me that's why i asked it before but i didnt get any answer :(

My GR is quite rusty, unfortunately, but my trusty (but dusty) copy of Adler, Bazin, & Schiffer (Intro to GR) backs me up with this quote: "... it is impossible to have gravitational forces take the same mathematical form as fictitious forces within the framework of classical analytic mechanics". (You need a description within a four-dimensional space. I'll let GR experts fill in more details.)

- #12

- 2,985

- 13

- #13

- 69

- 0

I tried to use a mathematical approach...and got confused.cyrusabdollahi said:

Suppose the mass of falling water is m and that of the turbine is M(what's more reasonable- m>M or M>m?)Also suppose that the height at which water attains terminal velocity is h metres above the turbine.

Firstly, consider the situation from an external,inertial reference frame.

Taking the ground as reference level,PE of water is mgh and its KE is mu (where u=0.5xv^2).So K=mgh+mu where K is the KE with which the turbine rotates.Here work is done by water on the turbine.

Now consider the RF falling at vel. v.Taking the referene level at height h the PE of turbine is -Mgh and its KE is Mu.So K=-Mgh+Mu.Here work is done by turbine on itself.

This means that m=M(u-gh)/(u+gh) which seems to be a special case.If the masses do not satisfy this reln. will the blades not rotate?Or is this some limiting case, or what?

Also,Doc Al, by saying that fuel needed does not depend on the RF, are you not contradicting what you said earlier-that work depends on our reference frame?

- #14

Doc Al

Mentor

- 44,867

- 1,114

Not at all. (I hope you realize that the amount of fuel used cannot depend on the frame of reference!) Calculation of work and kinetic energydpsguy said:Also,Doc Al, by saying that fuel needed does not depend on the RF, are you not contradicting what you said earlier-that work depends on our reference frame?

- #15

- 69

- 0

Sorry,Doc Al, but I don't see your point.Suppose x l of fuel are used in accelerating from 0 to V. Then 3x l are used up in going from V to 2V.If our RF is moving at a vel. V,then the increase in KE is only 0.5m(V^2). But we know that 3x l of fuel have been used. How would the observer explain this anomaly?Doc Al said:Not at all. (I hope you realize that the amount of fuel used cannot depend on the frame of reference!) Calculation of work and kinetic energydoesdepend on frame of reference. Why don't you work through that example with the car and see for yourself.

Also, in the turbine example,the final KE of the turbine would be the same irrespective of the RF(obviously?). Hence whether we work in the RF of water or the turbine,the work done on the turbine will be same(see mathematical approach).Please correct me if I am wrong.

- #16

Chi Meson

Science Advisor

Homework Helper

- 1,789

- 10

In a rotating reference frame( a turntable for example) , the centrifugal force will appear to be the force that pulls a ball toward the outside of the circle. But this force will not accelerate the ball. The increase in speed of the ball comes from the tangential force (by friction) from the floor. The rotating observer will see the "coriolis effect" as the ball drifts to the left or right as it progresses to the outer edge of the turntable. The ficticious centrifugal force does not do the work even in the rotating frame of reference.

But the inertial reaction of coming to a quick stop (in a car): in the car's frame of reference, it does appear that the inertial force causes the passenger to accelerate forward into the windshield. As Doc Al mentioned, this only "works" if you stay in the accelerated reference frame, but since you can't (you just crashed your car), you always must be able to identify the object that really did the work (in this case the windshield).

- #17

Doc Al

Mentor

- 44,867

- 1,114

That's the problem I askeddpsguy said:Sorry,Doc Al, but I don't see your point.Suppose x l of fuel are used in accelerating from 0 to V. Then 3x l are used up in going from V to 2V.If our RF is moving at a vel. V,then the increase in KE is only 0.5m(V^2). But we know that 3x l of fuel have been used. How would the observer explain this anomaly?

How can kinetic energy be independent of RF? Speed (obviously) depends on the reference frame.Also, in the turbine example,the final KE of the turbine would be the same irrespective of the RF(obviously?).

When I get some time later I will post comments on your water/turbine example. But it's a similar issue as with the car.Hence whether we work in the RF of water or the turbine,the work done on the turbine will be same(see mathematical approach).Please correct me if I am wrong.

- #18

- 69

- 0

I gave it a shot...and missed.Doc Al said:That's the problem I askedyouto figure out! To do that, you must consider that the work done on the car (by friction of the road) is different in different frames. And the fact that work is done on the earth. Give it a shot.

How can kinetic energy be independent of RF? Speed (obviously) depends on the reference frame.

When I get some time later I will post comments on your water/turbine example. But it's a similar issue as with the car.

I tried to take it one frame at a time.What I was trying to prove was that fuel needed during the second burst of speed is three times that needed during the first in the

But I had a problem with the moving reference frame.Suppose the car and the RF are at the same point at t=0. If we assume RF to be at rest, the car seems to go

- #19

- 69

- 0

Say the electricity produced in the plant is proportional to how fast the turbine rotates.If the KE of turbine is dependent on the RF isn't the electricity produced too?

- #20

jcsd

Science Advisor

Gold Member

- 2,090

- 11

In GR gravitational forces are just like a pseudo forces in thta they are frame effects and you can always pick another local refrence frame in which they do not occur.

- #21

Doc Al

Mentor

- 44,867

- 1,114

Thedpsguy said:Say the electricity produced in the plant is proportional to how fast the turbine rotates.If the KE of turbine is dependent on the RF isn't the electricity produced too?

- #22

jtbell

Mentor

- 15,484

- 3,250

You have to take into account the change in KE and momentum of the earth, produced by the reaction force of the tires against the road. In the second reference frame, the earth's KE changes by a larger amount than in the first reference frame.dpsguy said:Sorry,Doc Al, but I don't see your point.Suppose x l of fuel are used in accelerating from 0 to V. Then 3x l are used up in going from V to 2V.If our RF is moving at a vel. V,then the increase in KE is only 0.5m(V^2). But we know that 3x l of fuel have been used. How would the observer explain this anomaly?

- #23

Doc Al

Mentor

- 44,867

- 1,114

Eaxctly! When applying Newton's 2nd law to calculate [itex]F \Delta x[/itex], realize that you must include the "work" done on all the objects involved (Earth and car) and that [itex]\Delta x[/itex] is frame dependent. (For example, in the Earth frame the Earth doesn't move.)jtbell said:You have to take into account the change in KE and momentum of the earth, produced by the reaction force of the tires against the road. In the second reference frame, the earth's KE changes by a larger amount than in the first reference frame.

- #24

jtbell

Mentor

- 15,484

- 3,250

Or simply use conservation of energy and momentum. Consider two masses M and m, at rest next to each other in some inertial reference frame. Their kinetic energies and momenta are both zero. Now add energy Q. This might come from having one of the masses push against the other one, or by exploding a firecracker in between them, or releasing a compressed spring, or whatever.

Now M moves to the left with speed V, and m moves to the right with speed v. Since both M and m are initially stationary, the**change** in their kinetic energies is

[tex]\Delta K_M = \frac{1}{2} MV^2[/tex]

[tex]\Delta K_m = \frac{1}{2} mv^2[/tex]

Now let's look at this in a second reference frame that moves with speed u to the right relative to the first frame. Before we add Q to the system, both M and m are moving to the left with speed u. After we add Q, M is moving to the left with speed V+u and m is moving to the right with speed v-u.

The**change** in kinetic energy for M and m in this frame is

[tex]\Delta K_M = \frac{1}{2} M (V + u)^2 - \frac{1}{2} M u^2

= \frac {1}{2} MV^2 + MVu[/tex]

[tex]\Delta K_m = \frac{1}{2} m (v - u)^2 - \frac{1}{2} m u^2

= \frac {1}{2} mv^2 - mvu[/tex]

So in the second frame, M changes its kinetic energy by a larger amount than in the first frame, whereas m changes its kinetic energy by a smaller amount. Note that MV = mv from conservation of momentum in the first frame, so MVu = mvu.

Now M moves to the left with speed V, and m moves to the right with speed v. Since both M and m are initially stationary, the

[tex]\Delta K_M = \frac{1}{2} MV^2[/tex]

[tex]\Delta K_m = \frac{1}{2} mv^2[/tex]

Now let's look at this in a second reference frame that moves with speed u to the right relative to the first frame. Before we add Q to the system, both M and m are moving to the left with speed u. After we add Q, M is moving to the left with speed V+u and m is moving to the right with speed v-u.

The

[tex]\Delta K_M = \frac{1}{2} M (V + u)^2 - \frac{1}{2} M u^2

= \frac {1}{2} MV^2 + MVu[/tex]

[tex]\Delta K_m = \frac{1}{2} m (v - u)^2 - \frac{1}{2} m u^2

= \frac {1}{2} mv^2 - mvu[/tex]

So in the second frame, M changes its kinetic energy by a larger amount than in the first frame, whereas m changes its kinetic energy by a smaller amount. Note that MV = mv from conservation of momentum in the first frame, so MVu = mvu.

Last edited:

- #25

- 69

- 0

I see 2 reasons why my mathematical approach was wrong:

1.we cannot neglect air resistance(else how would water attain terminal velocity ?)

2.the total energy will not be K in both cases as K is the sum of translational and rotational kinetic energies and translational KE will be different in the two cases, as helpfully pointed out by Doc Al .

But if I take these 2 points into consideration, the maths becomes very messy indeed.Am I still missing out on anything or have I actually chosen a very bad example to understand work done in different reference frames?

BTW, jtbell, I am not sure we can use momentum considerations of one RF in another, as you have done(MV=mv).As Doc Al had said earlier,when you use one RF , stick to it.Or did I get it wrong?Kindly elaborate.

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 39

- Views
- 6K

- Last Post

- Replies
- 12

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 656

- Replies
- 12

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 22K

- Replies
- 1

- Views
- 4K