Can Inductive Reasoning Explain the Movement of Strings on Spools and Bikes?

In summary,The string does the work.The string does the work.The string does the work.The string does the work.
  • #1
leafy
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TL;DR Summary
These videos seems like static force can do work. Can anyone explains it?


 
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  • #3
Dale said:
The string does the work.
I can understand spring does work, but string doesn't store energy. Work is force x distance, so one end of the string the displacement is zero. The other end of the string, it displaced with the cylinder through space. But what is the source of energy?
 
  • #4
leafy said:
one end of the string the displacement is zero.
No, the displacement is very clearly not zero on both ends. I am not sure why you would think otherwise. It is very clear that the string moves.

leafy said:
But what is the source of energy?
Obviously the thing pulling the string. There is no mystery here.

The interesting thing about these wheels is how the string winds up under tension. That is a simple matter of choosing the size of the axle, but it is interesting anyway. But there is nothing unusual about the work.
 
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  • #5
Let's say I was pulling the string. I felt a tiny displacement. So my work done is the force I pull times the displacement my hand made. On the cylinder end, the same force carry by the string, but the displacement is many times higher. The work done is then force time big displacement. Obviously, the energy doesn't add up.
 
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  • #6
leafy said:
the displacement is many times higher
Why do you think that. The string looks like normal string which is fairly inextensible. The displacement will be approximately the same on both ends.

You are getting fooled by the fact that the string is being wound up on the axle, but that is a perpendicular displacement and not the displacement of interest.

leafy said:
Obviously, the energy doesn't add up.
This is silly. Of course it does.
 
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  • #7
Dale said:
The displacement will be approximately the same on both ends.
Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.
 
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  • #8
leafy said:
Obviously, the energy doesn't add up.
Haha. Nothing is obvious in this world. Add up all the energy components involved (go as deep as you like) and the total will not change. Your mental experiment is leaving out some vital (not obvious) factors.

Any string will stretch with any applied force (however small) so that involves doing work. If the stretch is elastic than mechanical energy of that work will be returned in some way. If there are internal losses, some of the work will be 'lost' and turn up as heating up the string.
 
  • #9
leafy said:
Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.
Nope. The important quantity is not the force moving through space, it is the distance that the material moves at the point of application of the force. This is hard to see if you haven’t worked these types of problems before because the force is applied to different pieces of material as the axle winds up. But with a bit of experience you will see immediately that it is the same distance on both ends.
 
  • #10
Here I drew the the system. The cylinder moved a distance D at a later time. The force (in red) moves the object for a distance D. The weight on the right also displaced a smaller distance d.

work done by the weight = mgd
work done on the cylinder = mgD
AFCF7F1F-966D-4445-9477-449846C698F7.jpeg
 
  • #11
leafy said:
Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.
So this problem is straightforward to work out, though a little tedious. Here is a free-body diagram for the wheel:
problem.jpg

We have the radius of the wheel ##R## and the radius of the axle ##r## with ##R=r+\Delta r##. ##a## and ##v## are the acceleration and velocity of the center of mass, and ##\alpha## is the torque about the center of mass. The wheel (including the axle) has a mass ##m## and a moment of inertia ##I=kmR^2## where ##k## is some number that depends on the exact geometry of the wheel and axle assembly.

We are only interested in horizontal forces so there is the tension ##T## provided by the string and the friction force ##F## from the ground. The wheel rolls without slipping so ##a=R\alpha##. From Newton's 2nd law and Newton's 2nd law for rotation we have $$T-F = ma$$ $$FR-Tr=kmR^2 \alpha$$ Solving we get $$a=\frac{T \Delta r}{m R (1+k)}$$ $$\alpha=\frac{a}{R}$$

The rate of work is given by the power which is the force times the velocity of the material at the point of application of the force. The velocity of the material of the wheel goes from ##0## at the ground to ##v## at the center, so at the point of application of ##T## the velocity is $$v_T=\frac{\Delta r}{R} v$$ so the rate of work done on the wheel by the string is ##T\cdot v_T=T \frac{\Delta r}{R} v##.

Now on the other end of the string, assuming that the string is ideal (flexible, massless, inextensible), the external force is applying tension ##T##. If the string is a length ##L## then the position of the end of the string ##x_S## is equal to the position of the center of the wheel ##x## plus ##L## and minus the amount of thread that has been wound up onto the spool ##r\theta##. So that is ##x_S=x+L-r\theta##. Now, since we know ##a## and ##\alpha## from above we can solve $$x=\frac{a}{2}t^2$$ $$v=at$$ $$\theta = \frac{\alpha}{2}t^2$$ $$\omega = \alpha t$$ and with those we can obtain $$x_S=x-r \theta + L$$ $$x_S=\frac{a}{2}t^2-r\frac{\alpha}{2}t^2 + L$$ $$v_S=at-r\alpha t =\frac{R a}{R}t-\frac{ra}{R} t $$ $$v_S=\frac{\Delta r}{R}at=\frac{\Delta r}{R}v = v_T$$

So the rate of work done on the string by the external force supplying the tension is also ##T\cdot v_T=T \frac{\Delta r}{R} v## and hence energy is conserved at all points in time.

leafy said:
work done on the cylinder = mgD
No, it is not, see above. As I explained in post 9 it is the motion of the material at the point of application of the force that is the relevant quantity. Not the motion of the point of application of the force.
 
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  • #12
Dale said:
No, it is not, see above. As I explained in post 9 it is the motion of the material at the point of application of the force that is the relevant quantity. Not the motion of the point of application of the force.
You're only half correct. Consider the case of a block mass moving through a flat friction surface.

Yes, there is the motion of the block at the point of application of the friction force, a relevant quantity.

But, there is also the motion of the point of application of the friction force moving through the flat surface.
 
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  • #13
leafy said:
You're only half correct. Consider the case of a block mass moving through a flat friction surface.

Yes, there is the motion of the block at the point of application of the friction force, a relevant quantity.

But, there is also the motion of the point of application of the friction force moving through the flat surface.
The statement (mechanical power, the rate of mechanical work, is the product of the force and the velocity of the material at the point of application of the force) is completely correct and applies completely for all mechanical forces including friction. However, the sliding block is not relevant to the current scenario. If you would like it please open a new thread on that topic.

For this scenario it is clear that energy is conserved and the work balances out. There is no mystery, it behaves exactly as I described above.
 
  • #14
The only solution to our argument is to let the block drop a height h, after height h, measure the rotional energy of the wheel, the translational energy of the wheel, and the kinetic energy of the weight. If it add up to the drop height potential, then I submit.
 
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  • #15
leafy said:
The only solution to our argument is to let the block drop a height h, after height h, measure the rotional energy of the wheel, the translational energy of the wheel, and the kinetic energy of the weight. If it add up to the drop height potential, then I submit.
By all means, be my guest. But if you are right then energy is not conserved and the wheel will have substantially more energy than was given to the string. Specifically, the input power is ##T \frac{\Delta r}{R}v## and if you are right then you will have an output power of ##Tv## for a factor ##R/\Delta r## gain in energy. This is not a small violation of energy conservation.

In any case, go ahead and do the experiment. It will be a good test of whether or not you are capable of performing good experimental measurements. I predict that if you do it correctly not only will you see nowhere near a gain of ##R/\Delta r## you will actually get a gain slightly less than 1 due to friction and other losses.
 
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  • #16
leafy said:
The only solution to our argument is to let the block drop a height h, after height h, measure the rotional energy of the wheel, the translational energy of the wheel, and the kinetic energy of the weight. If it add up to the drop height potential, then I submit.
Experiments, along these lines, have been carried out for centuries and, within the limits of experimental error (improving all the time) they have affirmed the principle of Energy Conservation. I am not sure why you don't just accept that your suggested experiment would give the 'accepted' answer.

My standard reply to this sort of question is to say that, if you really think you have spotted a mistake in established Physics, then you would need to start at square one and provide actual evidence (plus, of course, fully understanding all the basic Physics that has developed over centuries).

I'm not just saying you should accept what you are told. What I am saying is that you would have to prove that what you've been told is actually wrong.
 
  • #17
Thanks guys. I just don't get where I misunderstood things. Of course I would continue to prove or disprove myself when I have a chance.

My thought is instead of a tension force pulling on the wheel, you can imagine yourself running and pushing on the same point on the axle. Now compare that to just sit and pull on the wheel. Your intuition would say something is off about input/output. Yes, the gain factor would be huge. It's crazy but I have to go with my intuition.

Another situation is imagine you're on a moving train pushing on a mass with force F. If you stick with mechanical power, it would be F x speed of the mass. But if you calculate the energy gain from a absolute frame, it would be F x speed of train + mass. The force acting on the mass has to move through much larger space in absolute frame. That's where my logic about force moving through space.
 
  • #18
leafy said:
I just don't get where I misunderstood things.
I told you. The important thing is not the motion of the point of application of the force, but the motion of the material at the point of application of the force. That is where you misunderstood.

leafy said:
It's crazy but I have to go with my intuition.
Going with your intuition only works if you have good intuition. Good intuition is something that is built over time. If you do something right, then do it right again, and again and again, then you build good intuition. If you have bad intuition (which you do here) then you need to focus on correct practice to tear down your incorrect intuition and then begin building good intuition in its place.

leafy said:
Another situation is imagine you're on a moving train pushing on a mass with force F. If you stick with mechanical power, it would be F x speed of the mass. But if you calculate the energy gain from a absolute frame, it would be F x speed of train + mass. The force acting on the mass has to move through much larger space in absolute frame. That's where my logic about force moving through space.
There is no such thing as absolute space. However, I would indeed recommend that you open up new threads to discuss how these different scenarios work. All of them follow the rule I posted earlier.
 
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  • #19
Dale said:
I told you. The important thing is not the motion of the point of application of the force, but the motion of the material at the point of application of the force. That is where you misunderstood.
And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.
 
  • #20
leafy said:
And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.
Yes, you said that and that is your misunderstanding. It is always the motion of the material at the point of application that is important for mechanical energy, not the motion of the point of application. Always.

The example is irrelevant to the scenario in the OP, but it is relevant to your misconception. I would recommend opening new threads on those topics.
 
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  • #21
leafy said:
And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.
Have you done a course on Newtonian mechanics? If your intuition was not gained from that sort of background then you need to consider it's possibly wrong. Why are you arguing with @Dale ? He does know what he's talking about and it's no one's job here to nail exactly where you are going wrong. Try to bend to the subject and not expect the subject to bend to your intuition.
 
  • #22
Dale said:
Yes, you said that and that is your misunderstanding. It is always the motion of the material at the point of application that is important for mechanical energy, not the motion of the point of application. Always.
Let's say you're right that the motion of the material is important. How can you be sure that that at time t+delta(t) it will be the same as the time t? The point of material also move upward and at time t+delta(t), or half delta(t), the force already jump to the next point in the arc.
 
  • #23
leafy said:
Let's say you're right that the motion of the material is important. How can you be sure that that at time t+delta(t) it will be the same as the time t? The point of material also move upward and at time t+delta(t), or half delta(t), the force already jump to the next point in the arc.
That is why we use the velocity. That gives us the displacement of the material over an infinitesimal time. This is one of the reasons that we use calculus in Newtonian physics.
 
  • #24
I think there're two methods to replace the tension force on the wheel.

One is you have to station pushers along the path the wheel travel and each time the wheel pass by, each pusher going to push the wheel.

Second is you only need one pusher, but he must chase the wheel to impart a force.

Both yields the same result but one takes more work and energy. Hm...
 
  • #25
leafy said:
one takes more work and energy
Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.
 
  • #26
Dale said:
Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.
oh? Let's say we have 10 pushers station along the path. The wheel translate at 1m/s. The pusher push on material moving at small v(almost zero).
P=Fv x 10

Or we have one pusher moving at 1m/s also push on the material moving at small v.
P=Fv x 10 + F(1m/s)

It is a big difference.
 
  • #27
leafy said:
Or we have one pusher moving at 1m/s also push on the material moving at small v.
P=Fv x 10 + F(1m/s)
You are just randomly throwing numbers here. Physics isn’t a plate of pasta that you throw on the wall and see what sticks.
 
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  • #28
Dale said:
You are just randomly throwing numbers here.
You have to understand that imparting a force while moving cost a lot of energy.
 
  • #29
leafy said:
You have to understand that imparting a force while moving cost a lot of energy.
Not in the scenario you described. (Unless there are some inefficiencies involved)
 
  • #30
If the wheel is moving at 1m/s, imparting a force while moving is equivalent of throwing at rock at greater than 1m/s and transfer energy. Matter of fact, let make the wheel move at 100m/s. You have to throw rock faster than that. Let's say the station pusher throw rock too to move the wheel, they only need to throw rock at approximately 0 m/s to impart the same force.
 
  • #31
leafy said:
If the wheel is moving at 1m/s, imparting a force while moving is equivalent of throwing at rock at greater than 1m/s and transfer energy. Matter of fact, let make the wheel move at 100m/s. You have to throw rock faster than that. Let's say the station pusher throw rock too to move the wheel, they only need to throw rock at approximately 0 m/s to impart the same force.
None of this matters. If something is pushing the wheel in a given spot then the ##\vec v## is fixed to match that velocity regardless of whether it is one thing or many.

If you believe otherwise you will need to rigorously work the math. This pasta on the wall physics is pointless.
 
  • #32
You have no regard for translation speed of the wheel.
 
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  • #33
leafy said:
You have no regard for translation speed of the wheel.
I have regard for math and carefully worked physics.
 
  • #34
It's ok.
 
  • #35
leafy said:
Here I drew the the system. The cylinder moved a distance D at a later time. The force (in red) moves the object for a distance D. The weight on the right also displaced a smaller distance d.

work done by the weight = mgd
work done on the cylinder = mgDView attachment 294549
leafy said:
Let's say I was pulling the string. I felt a tiny displacement. So my work done is the force I pull times the displacement my hand made. On the cylinder end, the same force carry by the string, but the displacement is many times higher. The work done is then force time big displacement. Obviously, the energy doesn't add up.

Using your well drawn diagram, it seems to me that combining D and mg is not correct.
The force doing effective work should be the one applied to the center of mass of the cylinder, which is the effective distance represented by D.

The magnitude of that force is smaller than mg, as many times as the distance of each force to the contact point with the table is different.
If you replace the cylinder with a lever pivoted at the table, you could visualize that you are losing force intensity at D, as you have less and less mechanical advantage as you apply resistance to mg higher and higher.
 

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