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Homework Help: Torsion pendulum in Cavendish experiment

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data
    In the Cavendish experiment, the two small balls have mass m each and are connected by a light rigid rod with length L. The two large balls have mass M each and are separated by the same distance L. The torsion constant of the torsion wire is κ.
    b) Put the large balls a small distance away from the small balls, when the system reaches the equilibrium, the rigid rod rotates an angle θ and the distance between the centers of balls is r, as shown in the above figure. Find the expression for the gravitational constant G.
    c)If the small balls are perturbed with small angle from the equilibrium position in (b), will the oscillation be harmonic? If so, find the expression for the period of the oscillation.


    2. Relevant equations
    ∑τ = Iα
    T = 2π/ω

    3. The attempt at a solution
    I solved for part b. But I can't get the answer to part c.

    Part b: G = 2π2Lr2θ/(MT2)

    Part c:
    Let the new angle be θ1
    ∑τ = Iα
    GMmL/(r-0.5θcosθ1)2 - κθ1 = Iα
    α = 2GM/(L(r-0.5θcosθ1)2) - 2κθ1/(mL2)
    I can't make α=-ω2θ1 to find for ω to solve for T.
    Last edited: Apr 18, 2015
  2. jcsd
  3. Apr 18, 2015 #2
    Isn't "r" a function of L and θ?
  4. Apr 18, 2015 #3
    Originally when it is in equilibrium, the distance between M and m is r. Then the small balls are perturbed with small angle from the equilibrium position, so the new distance between M and m is r-0.5θcosθ.
  5. Apr 18, 2015 #4
    That's not how I interpret the diagram which clearly shows r depending on the angle θ.

    The distance between M and m at equilibrium when θ=0° is r=L/√2.
  6. Apr 18, 2015 #5
    In part b it says it rotates at an angle θ when it reaches equilibrium. Then in part c it says the small balls are perturbed with small angle from the equilibrium position in part b. So in part c, the new angle isn't it θ+small angle, which we call it θ1. I'll edit my θ in my workings to θ1. I think it's confusing to use the same θ.
  7. Apr 18, 2015 #6
    OK, I see that now.

    This expression is not correct:

    For one thing the dimensions are wrong.

    Since θ1 is a small angle then a good approximation for the distance between M and m would be:
    r' = r-Lθ1

    Now you are squaring this number r'. How much different will r2 be from r'2?
  8. Apr 19, 2015 #7
    Why is it Lθ and not θL/2? Since the radius is L/2.
  9. Apr 19, 2015 #8
    Yes, you're right, L/2.
  10. Apr 19, 2015 #9
    So the new torque would be:
    2GMm/(r-0.5Lθ1)2(L/2) - κθ1 = Iα
    The answer involves a cosθ and it is not in terms of G. Means I got to use part b answer?
  11. Apr 19, 2015 #10
    No, do you know how to solve differential equations?
  12. Apr 19, 2015 #11
    How do I make it into a differential equation? I think I know how to solve them if it is those basic ones.

    Edit: Means I got to change α into d2θ/dt2? And solve it so that I will get an equation relating θ and T?
    Last edited: Apr 19, 2015
  13. Apr 19, 2015 #12
    Yes, you got it. The only trick is to simplify the equation by assuming (r-0.5Lθ1)2 is approximately r2 for small θ1.
  14. Apr 19, 2015 #13
    It will become 2(GMm/r2 - κθ1) = Lm(d2θ1/dt2). How do you solve this? It is not linear and homogeneous.
  15. Apr 19, 2015 #14
    I think it's linear. Regardless, you only need to come up with the period. So why not assume θ1 is in the form of a simple harmonic equation?
  16. Apr 19, 2015 #15
    I tried re-arranging the equation into the form of α = -ω2θ so that I can use T = 2π/ω. But I can't separate out θ1 away from the other terms.
  17. Apr 19, 2015 #16
    Can you solve the homogenous version of this equation?
  18. Apr 20, 2015 #17
    I'm not sure how though. Maybe I haven't learn how to solve this type of differential equations. Is differential equation the only way to solve it?
  19. Apr 20, 2015 #18
    Afraid so.
  20. Apr 20, 2015 #19
    Can you guide me on using differential equation to solve it?
    The equation I have now is 2(GMm/r2 - κθ1) = Lm(d2θ1/dt2)
  21. Apr 20, 2015 #20
    First do what I said in post # 12.
  22. Apr 20, 2015 #21
    OK, I see that you did that already.
  23. Apr 20, 2015 #22
    Yeah, how do I continue from there? How do I alter the equation so I can integrate them?
  24. Apr 20, 2015 #23


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    The net torque may be written $$\tau = \frac{A}{r^2} - \kappa \theta = \frac{A}{(r_0- b\, \delta \theta)^2} - \kappa (\theta_0 +\delta \theta) $$ where ##A## and ##b## are constants.

    ##r_0## and ##\theta_0## are the equilibrium values of ##r## and ##\theta##, and ##\delta \theta## represents displacement of ##\theta## from equilibrium.

    For small ##\delta \theta## you can linearize this expression to get ##\tau = -B \, \delta \theta## for some constant ##B##.
  25. Apr 20, 2015 #24
    You're right, we forgot the original Kθ term. That will simplify the equation down to:

    GMm/r2 + K(θ+θ1) = Iθ1''

    (GMm/r2 + Kθ) + Kθ1 = Iθ1''

    1 = Iθ1''
  26. Apr 20, 2015 #25
    If we have simple harmonic motion then we can assume that the solution to the differential equation will take the form:

    θ1= A⋅sin(ωt) where A and ω are constants to be determined.
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