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Period and Tension of a Pendulum

  1. Jan 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A simple pendulum experiment is constructed from a point mass m attached to a pivot by a massless rod of length
    L in a constant gravitational field. The rod is released from an angle θ0 < π/2 at rest and the period of motion is
    found to be T0. Ignore air resistance and friction.
    19. At what angle θg during the swing is the tension in the rod the greatest?
    (A) The tension is the greatest at the point θ_g = θ_0.
    (B) The tension is the greatest at the point θ_g = 0.
    (C) The tension is the greatest at an angle θ_g with 0 < θ_g < θ_0.
    (D) The tension is constant.
    (E) None of the above is true for all values of θ0 with 0 < θ0 < π/2.
    20. What is the maximum value of the tension in the rod?
    (A) mg
    (B) 2mg
    (C) mLθ0_/T_0^2
    (D) mg sin θ_0
    (E) mg(3 − 2 cos θ_0)
    21. The experiment is repeated with a new pendulum with a rod of length 4L, using the same angle θ0, and the period of motion is found to be T. Which of the following statements is correct?
    (A) T = 2T_0 regardless of the value of θ_0.
    (B) T > 2T_0 with T ≈ 2T_0 if θ_0 << 1.
    (C) T < 2T_0 with T ≈ 2T_0 if θ_0 << 1.
    (D) T > 2T_0 for some values of θ_0 and T < 2T_0 for other values of θ_0.
    (E) T_0 and T are undefined because the motion is not periodic unless θ_0 << 1.


    2. Relevant equations


    3. The attempt at a solution
    For 19, I just drew an FBD and realized that the tension in a pendulum is just mgcos(θ) which is maximized when θ=0 so the answer is B.
    For 20, when θ=0 (where the maximum tension is) from the formula the tension is just mg which is (A).
    For 21, I used the formula T=2pi*sqrt(L/g) so quadrupling the length should double the period making it (A).

    Are my answers (and most importantly my justifications) right?
     
  2. jcsd
  3. Jan 21, 2015 #2
    19 and 21 are correct.

    For 20, you'll need to find the velocity of the bob at the lowest point. The tension provides the necessary centripetal force for the semi-circular motion, so consider that as well.

    FC = mv2/r
     
  4. Jan 21, 2015 #3

    haruspex

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    You're overlooking centripetal acceleration. The rod not only has to counter gravity in the direction of the rod shaft, it also needs to provide the centripetal acceleration needed to make the bob travel in an arc. It doesn't change the answer to 19, but
    it does change the answer to 20.
    That formula is only valid for small oscillations. Here θ0 can be up to π/2, which is definitely not small.
     
  5. Jan 21, 2015 #4
    Ok so basically if T-mg=mv^/r then T=mg+mv^/r , but how do I find v^2/r in terms of g and a trig function?

    Also how do you find the period with large oscillations?
     
  6. Jan 21, 2015 #5
    At the initial position, the pendulum will possess potential energy. This will all be converted to kinetic evergy at the lowermost position. Get the value of 'v' using this.
     
    Last edited: Jan 21, 2015
  7. Jan 21, 2015 #6

    haruspex

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    In general, that's a hard problem. Fortunately, all you care about here is how varying the length affects the period, keeping the angle fixed.
    Can you write down the general differential equation for a simple pendulum (not using any small angle approximation)?
     
  8. Jan 21, 2015 #7
    Ok for 20 I did mg(L-Lcos(theta))=.5mv^2 and got v^2=2gL(1-cos(theta)). Substituting that into my "T" equation I got T=mg(3-2cos(theta)). How is that ?
    For 21 774603e9b85eacfc9c3ba8f14df1e0f6.png
     
  9. Jan 21, 2015 #8
    Yes! You got that right, mate..!
     
  10. Jan 21, 2015 #9
    Cool now what about 21?
     
  11. Jan 21, 2015 #10

    haruspex

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  12. Jan 21, 2015 #11
  13. Jan 21, 2015 #12

    haruspex

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    Yes.
    Now here's the clever bit. We want to find out how changing L will change the period, without actually solving the equation.
    Suppose ##\theta = f(t)## is a solution to this differential equation with boundary condition ##f(0) = \theta_0##. Investigate whether ##h(t) = f(At)##, for some constant A, might be a solution to ##\frac {d^2 h}{dt^2} + \frac gL \sin(h(t))## with the same boundary condition.
     
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