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Homework Help: Conical Pendulum Problem

  1. Nov 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Conical Pendulum Question

    Mass (m) is attached to the ceiling by a String of Length (s)
    The string makes an angle of ([tex]\theta[/tex]) with the vertical

    Compute the Period of Revolution and the Tension in the string?

    Mass of Bob (m) -- 4.35kg
    Length of String (s) -- 5.50 meters
    [tex]\theta[/tex] -- 64 degrees

    2. Relevant equations

    1/2(mv^2)[tex]/[/tex]radius -- Centripetal Force

    W= mg -- weight

    Not sure what else I need

    3. The attempt at a solution

    I think that the tension in the string would be

    Tcos([tex]\theta[/tex])=mg ---- T=(mg)[tex]/[/tex]cos([tex]\theta[/tex])


    (4.35)(9.81)[tex]/[/tex]cos(64)= ---- 97.35 N

    I have no idea how to get the period though, but I think it has to do with

    Tsin[tex]\theta[/tex]= 1/2(mv^2)[tex]/[/tex]radius
  2. jcsd
  3. Nov 21, 2007 #2

    Doc Al

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    Staff: Mentor

    So far, so good.

    Once you find the speed you can use that to find the period. What's the radius of the bob's path?
  4. Nov 21, 2007 #3
    I think that the radius would be

    string length --- (s) sin (Theta) --- so 5.5sin(64)= 4.94m

    I am not sure the equation to find the period though?

    I have radius, string length, tension in string, and mass of the Bob.
  5. Nov 21, 2007 #4

    Doc Al

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    Staff: Mentor

    The period is just the time the bob takes to make one complete revolution. Find the speed (and figure out the distance).
  6. Nov 21, 2007 #5
    Would I find the speed by using Tsin(theta)= (1/2)(mv^2)/radius ----

    97.35 sin (64) = (1/2)(4.35)(v^2)/(4.94) ----- velocity = 14.097 m/sec ?

    so the distance would be (2)(radius)(pi)= (2)(4.94)(pi)= 31.03m

    I am stuck now on how to find the period. Is there a certain equation that is used to find the period, we just started this topic?
  7. Nov 21, 2007 #6

    Doc Al

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    Staff: Mentor

    How about distance = speed x time?
  8. Nov 21, 2007 #7
    So (2)(pi)(r)=(velocity)(time)


    31.03m=(14.1m/sec)(t) ---- so ---- period= 2.2 sec?
  9. Nov 21, 2007 #8

    Doc Al

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    Staff: Mentor

    Sure. It's just the time required for one revolution. You have the distance and the speed--that's all there is to it.
  10. Nov 21, 2007 #9
    thank you so much for the help, I am going to go back and check all of my work.
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