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Duplication of the Cavendish experiment of 1798

  1. Jul 21, 2007 #1
    In a recent post here about the Cavendish experiment several people suggested that Cavendish measured G. I disagree with this. I believe that the experimental evidence does not support the claim that Cavendish observed the Newtonian force. What better way to prove this than by duplicating the experiment:smile:

    Cavendish experiment itself is the quintessential garage experiment that amateurs can duplicate easily. Cavendish himself was an amateur and he conducted his experiment in [link deleted by Doc Al] his backyard.

    I started a social network called [link deleted by Doc Al] for people who are interested in the Cavendish experiment. I just created it and I don’t yet know how it works. But if you like physics and tinkering then the Cavendish experiment is really a good place to start.

    There are two tests that I want to try in the experiment to prove that Cavendish did not observe the Newtonian force. I welcome comments from readers of Physics forums.

    1. Cavendish experiment assumes an unphysical equilibrium between the linear restoring force of the torsion wire and the inverse square of the Newtonian force.

    We know from elemenary physics that a linear force could never balance an inverse square force. But when you look at the http://www.physics.sfsu.edu/~ggrist/490/Cavrpt/cavrpt.html" [Broken] of the Cavendish experiment you would see that the experiment assumes this unphysical equilibrium when the linear restoring force is equated to the inverse square force. I explained here [link deleted by Doc Al] in detail why this equilibrium will never happen. Please let me know what you think. Do you think a linear force can balance an inverse square force? To me this violates elementary laws of physics. And is this a good proof that Cavendish did not measure G?

    2. I divide the second experiment in two. First I compute the constant G from an experiment where both the attracting weights W and the attracted balls h attached to the pendulum are present. Call this G1.

    In the second experiment I remove the attracted balls h and leave only the attracting weight W. (I keep moment of inertia of the pendulum the same.) Since the equations are independent of h I will again be able to compute a value for G. Call this G2. I claim that

    G1 = G2.

    The fact that I was able to compute the value of G in an experiment where attracted balls were not present will prove that Cavendish computed the density of the Earth from the constants of the pendulum. If we could compute G without the attracted mass we could no longer claim that we have measured “the attraction between two masses W and h.” The mass h is not there. This is a definite proof that Cavendish experiment never measured and could not have measured the Newtonian force. What do you think?

    I would really appreciate comments on these tests. Thanks again. Please join the [link deleted by Doc Al] and let’s duplicate this beautiful experiment and show that the Cavendish never measured the Newtonian occult.
    Last edited by a moderator: May 3, 2017
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  3. Jul 21, 2007 #2


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    It appears from the links you give that the restoring force is linear in theta. However, the gravitational law is inverse square in r. So, how can you make this statement?

    Also, why are you so intent on showing that Cavendish did not measure G? Why not focus on some useful physics?
    Last edited: Jul 21, 2007
  4. Jul 21, 2007 #3
    Good point. But since theta is r/l, where l is the moment arm, the restoring force is proportional to r and the Newtonian force is proportional to inverse square of r. See for instance equation 1.5 in this "ftp://ftp.pasco.com/Support/Documents/English/AP/AP-8215/012-06802b.pdf"[/URL] for cavendish pendulum.

    Can you explain further why you think this statement cannot be made?

    [quote="cristo, post: 1383314"]Also, why are you so intent on showing that Cavendish did not measure G? Why not focus on some useful physics?[/QUOTE]

    I think Cavendish experiment is useful physics as is shown from your comment. I believe that physics is learned best by doing physics. The Cavendish experiment is so simple that it can be tried with smallest of budgets by any amateur and yields fundamental results about nature of force. Thanks for your comment.
    Last edited by a moderator: Apr 22, 2017
  5. Jul 21, 2007 #4


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  6. Jul 21, 2007 #5


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    The Cavendish experiment has been successfully repeated by thousands of professionals and amateurs.

    Furthermore, the statement about balancing different kinds of forces is wrong; the two occurances of "r" in that statement carry different meanings. And even if they refered to the same "r", there could still exist one or more balance points. For example, [itex]F_1(r)=ar[/itex] and [itex]F_2(r)=b/r^2[/itex] have the same value at [itex]r=(b/a)^{1/3}[/itex]. Also, it's easy to see that this point is a point of stable equilibrium, being a local minimum in the potential energy (contradicting the assertion made in the third link in the OP).

    This thread already qualifies as crackpottery - it should be locked and the crackpot links deleted.
    Last edited: Jul 21, 2007
  7. Jul 21, 2007 #6
    So are you saying that a linear force can balance an inverse square force? In other words, are you saying that

    r = 1/(L - r)^2

    is possible? (L is the distance between the masses.)

    For instance, if L = 3, at r=2 we have 2 = 1. This is unphysical and it means that r can never be equal to 1/(L – r)^2.

    I have a graph http://www.densytics.com/wiki/index.php?title=Cavendish_experiment#Dynamic_equilibrium" [Broken] with actual numbers for the Cavendish experiment showing that if the gravitational force couples to the restoring force, however small the couple is, there will never be balance because the gravity will always be greater than the linear force.

    This shows that the assumption made in the modern derivations of the Cavendish experiment is unphysical. According to the laws of elementary physics and mathematics inverse square force will always be greater than the linear force.
    Last edited by a moderator: May 3, 2017
  8. Jul 21, 2007 #7


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    Er... Why don't you actually PLOT the graph and THEN figure out where both curves intersect, rather than simply plugging in random values? Let L be a constant value that you have there, and THEN solve for r. They intersect at ONE point. That is where you have an equilibrium.

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  9. Jul 21, 2007 #8


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    Yikes! Please learn your middle school math. The cubic equation you've written above has a real root (with L=3) at about r=3.53209
  10. Jul 21, 2007 #9

    Thanks for your comment. I agree that Cavendish experiment has been repeated many times. I am not saying that Cavendish experiment was never duplicated. If you point me to a statement by me that says Cavendish experiment was never duplicated I would apologize to you and immediately correct this stupid mistake.

    The geometry of the Cavendish experiment is very simple. Assume that the distance between the weights is L. Cavendish measured in the actual experiment the excursion e of the pendulum arm. We have been calling this excursion r so let's continue with r.

    The restoring force increases with r. Do you agree?

    As the distance between weights decreases as r increases, the force of gravity increases as the inverse square of the distance between the weights. Am I correct so far?

    The distance between weights is given by L - r. Therefore, the restoring force increases linearly as r. Gravity increases dynamically as 1/(L - r)^2. This is elementary physics. And you can see from the geometry that r is the same line in both r and in 1/(L -r)^2. Please correct me if I am wrong.

    Therefore, r will never be equal to 1/(L - r)^2. Do you agree?

    I am posting these questions to get comments so that I can verify if these tests are worth repeating or just garbage. I would appreciate if you would let me know where the above argument goes wrong. You would save me lots of trouble. Thanks again for your comments.

    Can you explain what a and b are in your comment? I did not understand how they relate to the geometry of the experiment.
  11. Jul 21, 2007 #10

    I may be mistaken but if L = 3 any and r=3.5 as you have it this would be unphysical. The way I understand the geometry r must always be smaller than L. If the excursion r is greater than L we would have a situation where lead balls collide and compress.

    So I would exclude any value of r > L as a physical solution. Does this make sense?
  12. Jul 21, 2007 #11

    Doc Al

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    Please write expressions for the force due to gravity and the restoring force as functions of r. You must include the constants--Gokul simply called them a and b.

    No, but what does this have to do with anything (unless the constants are all unity)? Please reread Gokul's response in post #5. You must set the forces equal to solve for the equilibrium point.
    Last edited: Jul 21, 2007
  13. Jul 21, 2007 #12


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    Even if you wish to pursue this argument, note that r=1/(L-r)^2 is a simple cubic equation with a real root. What this means is that for any value of L that you choose, there will be a value of r for which the above equation is satisfied. This is the value of r for which the forces would be balanced and is the equilibrium point about which the torsional pendulum oscillates. With damping, the pendulum will eventually settle at this point.

    For example, if you set L=3, then the above equation is satisfied for r=3.53209. Plug this into a calculator and check it for yourself. Choose another value for L, and you'll find a different point of equilibrium.

    PS: For the real Cavendish experiment, you can not simply use this above equation. You need to find and plug in the values of constants a and b, into [itex]ar=b/(L-r)^2[/itex] (where a is the torsion constant for the wire and b=GMm).
    Last edited: Jul 21, 2007
  14. Jul 21, 2007 #13

    The weights are separated by the length L. The excursion r must be smaller than L otherwise the weights will collide. Do you have any value of r which is smaller than L where r = 1/(L - r)^2 is satisfied?

    I have for instance

    L = 3 and r = 2.9 then I get 2.9 = 100
    L = 3 and r = 2.5 then I get 2.5 = 4
    L = 3 and r = 2.0 then I get 2.0 = 1
    L = 3 and r = 0.5 then I get 0.5 = 0.16
  15. Jul 21, 2007 #14
    Thanks for looking at the graph. This is the kind of feedback I was hoping for.

    In that http://www.densytics.com/wiki/index.php?title=Cavendish_experiment#Dynamic_equilibrium" [Broken] I used the original formula from Cavendish’s derivation. Since Cavendish did not know about G and he did not know a unit of force he is not using equations but proportions. Cavendish compared two pendulums in his derivation: a seconds pendulum and his horizontal pendulum.

    The restoring force of the horizantal pendulum is

    f = t^2e/T^2

    where t and T are periods of two pendulums and e is the excursion of the horizantal pendulum. I computed the value of f for the Cavendish’s pendulum for t, e and T. Then I assumed that the Newtonian force couples to the pendulum arm with a force only minutely greater than the restoring force. Once this happens gravity will increase dynamically but the restoring force will increase linearly. So once gravity sets the pendulum arm in motion the restoring force will always be smaller than gravity. By simple mathematics the force increasing as r will always be smaller than the force increasing as 1/r^2. We see this effect in magnetism also.

    What I did was actually plot r and 1/r^2. Once 1/r^2 is greater than r it will always be greater than r and there will not be an equilibrium. Once the attracting weight couples to the pendulum, the arm must and will hit the attracting weight.

    So given this physical situation what kind of graph do you want me to plot?
    Last edited by a moderator: May 3, 2017
  16. Jul 22, 2007 #15
    Thanks for the comment. I think this goes to the heart of the question I am trying to understand.

    There are two constants: the torsion constant k and constant of gravity G.

    If, as supposed by the modern derivation, we equate both forces we get

    kr = G/(l – r)^2

    l = the distance between masses
    r = the excursion of the pendulum arm

    At r = 0 we have two possibilities:

    1. kr > G/(l-r)^2
    In this case the wire is too stiff and gravity cannot move the pendulum arm. In this case G cannot be measured.

    2. kr < G/(l-r)^2
    In this case the torsion is too soft and G pulls the pendulum arm until both weights collide. Again the value of G cannot be found. (This is what happened in Cavendish's first try. He then replaced his wire with a stiffer one.)

    What about an equilibrium? We need a value of kr which is different than 0 and we must have k small enough so that G can set it in motion. But once G sets the arm in motion G/(l-r)^2 will be greater than kr. Once this happens kr will stay always smaller than G/(l – r)^2.

    Therefore, the third possibility where kr and G/(l-r)^2 are equal at some point where the arm is stationary (torsion of the wire balancing gravity) is unphysical because if the dynamic G couples to linear k it will not let the arm go until the two masses collide.

    I would appreciate comments.

    Does the above answer this also?

    Thanks again to everyone helping with this.
  17. Jul 22, 2007 #16

    Doc Al

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    Sorry, but I really don't think you understand what's going on.
    The purpose of setting those forces equal (you forgot the masses!) is to find the new equilibrium point of the torsion pendulum. Without gravity, the equilibrium point is at r = 0; with gravity, the equilibrium point will be where the force of gravity just balances the restoring force of the pendulum.

    r = 0 is the equilibrium point in the absence of gravity.

    The two possibilities are: (1) the wire is so stiff that the change in equilibrium point is too small to observe accurately, or (2) the wire is so soft that the masses collide.

    I really don't know what you're talking about when you speak of "dynamic G coupling to linear k".

    Here's how it actually works: When you include the gravitational attraction of the masses, the equilibrium point shifts. For small motions about that new equilibrium point, the torsion pendulum executes simple harmonic motion--just like a pendulum. (To show this, write the expression for the total force and see the dependence of the restoring force on distance from the new equilibrium point. You'll find that for small oscillations it will be linear--just like a pendulum, which is what it is.) The hard part is getting the thing to settle down.

    Do you seriously think that this experiment has never really been done? (Despite the fact that it's a standard undergrad demo in hundreds of classrooms.)
  18. Jul 23, 2007 #17
    Doc Al,

    Many thanks for your comments. I am glad that at least there is some agreement. I think I got the constants right :smile:

    I really like Cavendish experiment and I think it is a good way to learn physics.

    I think that, reading this, I may be using the "equilibrium point," in a wrong sense. It seems to me that your "equilibrium point" refers to what Cavendish called "zero point." This zero point is where the pendulum arm would come to rest if there is no force.

    But the zero point is not the equilibrium point I was talking about. The way I understand it the equilibrium point is where the the pendulum arm is stationary at its extreme position. If e is the excursion of the pendulum arm (the distance from zero point), the equilibrium should occur at maximum value of e. This is the only place where the pendulum arm is stationary (in equilibrium with gravity). This is where the two forces are equal, not at zero point. At zero point, the pendulum arm is moving fastest, it is not in equilibrium.

    http://www.flickr.com/photos/73067828@N00/872351308/" [Broken] I have in mind

    Before I write further, can you comment on this?

    Thanks again for the help.
    Last edited by a moderator: May 3, 2017
  19. Jul 23, 2007 #18

    Doc Al

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    Nope, wrong again. When I said "equilibrium point" that's exactly what I meant. Please reread what I (and others) have already written on this.

    (I'm afraid I'll have to remove the rather dubious links you've seen fit to include in your initial post--they violate our guidelines.)
  20. Jul 23, 2007 #19
    I am sorry I don't understand this. You wrote that

    r = 0 is the equilibrium point in the absence of gravity.

    If there is no gravity what is it that balances the restoring force and puts it in equilibrium?

    I would appreciate more details or references.
    Last edited by a moderator: Jul 23, 2007
  21. Jul 23, 2007 #20

    Doc Al

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    The torsion in the wire provides the restoring force! (See: http://scienceworld.wolfram.com/physics/TorsionalPendulum.html" [Broken])
    Last edited by a moderator: May 3, 2017
  22. Jul 24, 2007 #21
    Ok. Thanks. I looked at the torsion pendulum. I also looked at Pasco's manual for the Cavendish experiment. I believe that the equilibrium you mention is represented by figure 14 in Pasco manual. Is this correct?

    Then, let me use their notation S1 for this equilibrium position but with attracting weights in neutral position. (This is what Cavendish did.)

    Then, with no attracting force, let the arm come to equilibrium at S1.

    Then, bring in the attracting weight. (Position II in the manual.)

    Now let's look at the motion of the arm. Since gravity was able to move the arm it was greater than the torsion force:

    G/(l - r)^2 > kr

    where l is the distance between weights and r the excursion of the pendulum arm from S1.

    From elementary mathematics we know that if G/(1 - r)^2 is greater than kr it will always be greater than kr. As the arm moves kr will increase linearly as distance but gravity will increase as inverse square of distance.

    The physics of the pendulum says that if gravity couples to the pendulum arm and sets the arm in motion the torsion force can never exceed the force of gravity. If gravity is coupled to the pendulum arm, the pendulum arm cannot come to rest and reverse its motion to continue oscillating. In physics terms this means that the weights will collides.

    Please let me know if you agree with this reasoning. If not where do you think it is flawed?

  23. Jul 24, 2007 #22

    Doc Al

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    Realize that at S1, r = 0 and thus kr = 0. Of course the gravitational force is greater than zero!

    Elementary mathematics, eh? Let's make this simple: set G = 1 and k = 10 and plug in some numbers.

    At r = 0 : 1 > 0 (of course!)
    At r = 1/4: 1.78 < 2.5 (oops... there goes your theory)
    Ar r = 1/2: 4 < 5 (oops... there goes your theory)

    Don't confuse 1/(1-r)^2 with 1/r^2!

    Once again you falsely claim that this experiment has never been accomplished. (If it's physically impossible that an equilibrium position with r > 0 can be found--as you insist--then that's just what you're claiming.) And yet you reference a standard demo manual for the very experiment that you think violates some principle of physics! Get a grip. :rolleyes:
  24. Jul 30, 2007 #23
    Doc Al,

    Thanks very much for this post. I understand that it is wrong to say that if G/(l-r) is greater than kr it will always be greater than kr. Thanks for correcting this.

    I hope you will help with a couple of other questions.

    When I plot the curve for the difference (gravity - torsion) I see that, given your numbers, they are equal at 16 and 60. After that, the line shoots up and gravity is always greater.

    But that does not mean anything, I think. Because this equation, (the equality of the torsion and gravity) does not describe the motion of the arm. In this scenario the arm keeps moving with uniform motion, not with SHM, and never achieves a stationary equilibrium.

    So the fact that forces are equal does not mean that there is equilibrium.

    Your comments on the above will be appreciated.

    Just to clarify, I am not claiming anything, I am just looking at the Cavendish experiment in order to test if he measured what he said he did. From your comment I see that my first test is not relevant. I think physics has nothing to fear from me :)
  25. Jul 31, 2007 #24

    Doc Al

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    No idea what you mean by "given my numbers" since I had compared 1/(1-r)^2 with 10r. Values of r = 16 or 60 make no sense.

    On what possible basis do you make this claim?

    You are claiming that stationary equilibrium is impossible--a huge, unsubstantiated claim routinely contradicted in labs across the world.
  26. Aug 2, 2007 #25
    Sorry my mistake. I plotted the difference of gravity and torsion. You can see the spreadsheet here.

    By your numbers I mean G=1, k=10 and L=1. Given these numbers there are two points where gravity and torsion are equal. These are r=0.13 and r=0.59. After r=0.59 gravity is always greater than torsion.

    The equation we used to find the equality of kr = G/(l -r)^2 does not describe the motion of the arm. Is this correct? The arm moves with simple harmonic motion not with uniform motion as it is assumed to find the points where the forces are equal.

    For instance, if we remove the weights and if we assume that the arm comes to an equilibrium at r=0.13 as part of its simple harmonic motion this would not be an equilibrim caused by gravity.

    When we bring in the attracting weight, now at each unit of time gravity will add accelearation to the motion of the arm and the arm will no longer be at equilibrium at r=0.13.

    I am trying to understand this mechanism and the equation which describes the actual motion of the arm when the force is present.

    I would think that the arm would move according to some equation like this:

    (simple harmonic motion + acceleration due to gravity)

    Am I on the right track here?

    Let's say I am trying to understand this motion. You already showed that my claim about gravity always being stronger than torsion was wrong. The fact that Cavendish experiment has been repeated in labs across the world does not help me understand the Cavendish experiment and how the arm moves in the presence of the force.

    Personally, this has been very helpful to me. I am learning a lot. And thanks again for commenting.
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