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Projectile pendulum

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data
    I need help deriving some formulas, specifically the theoretical speed of a bob at the bottom of a pendulum where the string is horizontal from the lower rod to the bob.

    Background / Introduction
    1. A small weight at the end of a string of length, L, forming a simple pendulum.
    2. A horizontal rod is located a distance, d, directly beneath the pivot point.
    3. The mass is held so that the string is taut and horizontal and then let fall.
    4. What is the minimum distance, d, such that the mass will cause the string to loop over the pin at least once? In theory, we assume the bob is a point mass and the horizontal rod which is a distance d from the fulcrum is infinitesimally thin. However, experimentally, we are looking for the minimum value of d such that the mass hits the lower rod and falls on the other side of it.

    Some Physics principles to keep in mind:
    • Since the only considered forces acting on the mass are tension and weight, and since the tension does no work on the mass (the tension is perpendicular to the displacement), energy of the mass is conserved.
    • Once the string goes slack, the tension disappears and the only force acting on the bob is the force of gravity. Thus, the mass is in projectile motion.
    • At the point of the string going slack, the tension drops to zero, but at that moment there is still a centripetal force.

    Position Meaning
    A Starting position. String is horizontal from the upper rod to the mass.
    B At bottom of arc. String is vertical from the upper rod to the mass.
    C String is horizontal from the lower rod to the mass
    D String starts having slack
    E Mass just barely makes it over the lower rod.





    2. Relevant equations

    the only equation i've been able to derive is at is V=sqrt(g(L-D)

    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2013 #2

    CWatters

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    Think conservation of energy.
     
  4. Dec 6, 2013 #3

    haruspex

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    Consider the point at which the string has started to wind about the rod, the string is still taut, and the string from rod to mass makes an angle theta to the vertical (the downward vertical from the rod, say).
    Using CWatters' hint, what equation can you write for the velocity there? With that and a free body diagram, what equation can you write for the tension in the string?
     
  5. Dec 9, 2013 #4
    okay so 1/2mv^2=mgh???
     
  6. Dec 9, 2013 #5

    haruspex

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    Yes, but what is that in terms of theta, not h?
     
  7. Dec 9, 2013 #6
    1/2mv^2=mg(L-cosθ) ??
     
  8. Dec 9, 2013 #7

    haruspex

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    Not quite. Try that again.
     
  9. Dec 9, 2013 #8
    thank god you found time to respond tonight! I was pulling my hair out!


    Okay I derived v=2sqrt(2gL) for question 1 under Analysis A.

    I'm trying to figure out step 13 under Procedure B.

    1/2mv^2=mg(L-D) ??
     

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  10. Dec 9, 2013 #9

    haruspex

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    It descended height L to its lowest point, but has started to swing back up. The angle the string makes to the vertical is theta. How far below its starting point is it now?
    [It's mid-afternoon here, and I was out all morning.]
     
  11. Dec 9, 2013 #10
    But I'm trying to find the theoretical speed of the bob at position C (string is horizontal from the lower rod to the mass) and the theoretical angle at position D (string starts having slack)
     
  12. Dec 9, 2013 #11

    haruspex

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    To answer D you will have to grapple with the general question of the speed when the string is at angle theta. But OK, let's just do C first.
    The mass descended distance L to reach the lowest point; now it has risen to be level with the pin. The pin is distance d below the starting point. How far is the mass below the starting point?
     
  13. Dec 9, 2013 #12
    L-d ??
     
  14. Dec 10, 2013 #13

    haruspex

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    No. It's really simple. Draw a diagram. Or concentrate on these words I wrote:
     
  15. Dec 10, 2013 #14
    I don't see how it's not L-D. I'm sorry maybe I'm not visualizing it right. I attached a picture that looks like our experiment.
     

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  16. Dec 10, 2013 #15

    haruspex

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    Now I see that picture I realise we might have some confusion about labels. The picture uses D for the distance from pin to the top of the string and d for another distance later, while the OP uses D for a position of the bob. I'll use s instead of D for the distance from the pin to the top of the string, ok?
    You have the following points already defined:
    A = start position (top right in picture)
    B = lowest position (distance L below A)
    C = leftmost position (level with pin)
    Let me define two more:
    O = top of string.
    P = pin.
    O is at the same height as A
    P is distance s below O
    C is same height as P.

    How far is C below P?
    How far is C below O?
    How far is C below A?
    All the information you need to answer those three questions is in the three statements in bold.
     
  17. Dec 10, 2013 #16
    C is same height as P so it's 0 below P
    C is (O-P)
    C is s below A

    I just left our school's tutoring center and I just wasted another 2 hours because they weren't able to figure it out either.
     
  18. Dec 10, 2013 #17

    haruspex

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    Yes! So now you can get the velocity at point C.
    But, as I posted before, this doesn't help much. You need the velocity when the bob is at some later position D. The diagram has the string at angle theta above the line PC, so let's go with that. How far is D below A?
     
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