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Pendulum and its acceleration

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A small sphere of mass m is fastened to a weightless string of length 0.5m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 60 degrees with the vertical.
    a) What is the velocity of the sphere when it passes through the vertical position?
    b) What is the instantaneous acceleration when the pendulum is at its maximum deflection?

    2. Relevant equations



    3. The attempt at a solution
    a) What is the velocity of the sphere when it passes through the vertical position?
    I think that K should be = U because they only change is the pendulum's speed. Is that correct?
    1/2mv2= mgh

    So the m's cancel and you are left with:
    1/2v2 = 9.8(0.43)
    v = 2.9 m/s

    Did I do that correctly?

    b) What is the instantaneous acceleration when the pendulum is at its maximum deflection?
    I am not sure where to begin for this part.
     
  2. jcsd
  3. Oct 3, 2008 #2

    tiny-tim

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    Hi southernbelle! :smile:
    Nope … you used sin60º …

    but that's the sideways height :cry:

    I think Newton would have wanted you to use the vertical height! :wink:
    Hint: how does centripetal acceleration depend on speed? :smile:
     
  4. Oct 3, 2008 #3
    The acceleration is usually the derivative of the speed, correct? But I don't think it would be = 0.
     
  5. Oct 3, 2008 #4

    Gib Z

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    Acceleration is the derivative of displacement, yes, although its relation to linear velocity/speed is given by

    [tex]a_c = \frac{v^2}{r}[/tex] which you've hopefully seen before.
     
  6. Oct 3, 2008 #5

    Doc Al

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    Staff: Mentor

    Acceleration is the derivative of the velocity, a vector.

    Note that the acceleration has two components: A radial (or centripetal) component and a tangential component. Consider them both.
     
  7. Oct 3, 2008 #6
    Could I use
    a= gcos60?

    Although I guess I can use the length of the string as r in the equation
    ac = v2/r
     
  8. Oct 3, 2008 #7

    LowlyPion

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    Didn't they ask you for a at the point of maximum deflection? What is its v at that point?
     
  9. Oct 5, 2008 #8
    I got 3.1 for the v.
    I did:

    1/2mv2 = mgh
    v2=2gh
    v2 = 2 [9.8(0.5)]
    v=3.1m/s

    Is that correct? Should I have used 0.5 for the height?

    If that is right, then the acceleration would equal 3.1/0.5 = 6.2 m/s2
     
  10. Oct 5, 2008 #9

    LowlyPion

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    But what is the height? Wasn't the deflection only 60 degrees?
     
  11. Oct 5, 2008 #10
    Yes, the deflection is 60 degrees.

    I used 0.5 as the height because that is the length of the string.
     
  12. Oct 6, 2008 #11

    tiny-tim

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    difference in height

    Hi southernbelle! :smile:

    The height, h, in the formula PE = mgh,

    is the difference in height between the initial and final position.

    ('cos you're using conservation of energy, so it's the change that matters :wink:)

    So in this case it's the vertical amount by which the sphere is higher …

    at the bottom position it's at 0.5 below the hinge, and at 60º to the vertical it's at … ? :smile:
     
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