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Circular Motion and Tension in a string

  1. Jul 31, 2008 #1
    Okay, I have given this a go but its been 2 years since I've done any dynamics so I think I've done something stupid...

    A ball is attached horizontally by a string of length L to a central point C. The mass, m, of the ball is 4.775kg. It is released from rest and allowed to swing downwards. What is the tension in the string (in N) when the ball has fallen through 45 degrees.

    This is what I've done so far:

    [tex]\sum[/tex]F[tex]_{}n[/tex] = ma[tex]_{}n[/tex]
    T- m sin[tex]\alpha[/tex] = m/g (v[tex]^{}2[/tex]/L)
    T = m(sin[tex]\alpha[/tex] + v[tex]^{}2[/tex]/gL)

    [tex]\sum[/tex]F[tex]_{}t[/tex] = ma[tex]_{}t[/tex]

    m cos[tex]\alpha[/tex] = m/g a[tex]_{}t[/tex]
    a[tex]_{}t[/tex] = g cos[tex]\alpha[/tex]

    vdv = a[tex]_{}[/tex]t ds
    ds = L d[tex]\alpha[/tex]
    vdv = a[tex]_{}[/tex]t L [tex]\alpha[/tex]

    vdv = g L cos[tex]\alpha[/tex] d[tex]\alpha[/tex]

    then integrate that equation

    v = [tex]\sqrt{}[/tex]2gl(sin[tex]\alpha[/tex] - sin[tex]\alpha[/tex]0)

    T = m(3 sin [tex]\alpha[/tex] - 2 sin [tex]\alpha[/tex]0)
    = 4.775[3sin(45) - 2sin(0)]
    = 10.13N

    Which is apparently very, very wrong.

    Not sure what I've stuffed up, but help would REALLY be appreciated. Thanks.
  2. jcsd
  3. Aug 1, 2008 #2


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    I think that you have over complicated this question somewhat. Start by writing down the net horizontal and vertical forces acting on the ball.
  4. Aug 3, 2008 #3
    Lets pretend I'm stupid (which I am) and that I don't know anything about dynamics (which I currently don't), can you explain more clearly what you mean? :D

  5. Aug 3, 2008 #4


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    Whilst you seem to heave already put a lot of work in I don't mind doing a bit of work for you ... just this once :wink:.

    I'll admit to making a mistake in my previous post, I should have said radial and tangential components rather that horizontal and vertical. So for the radial components:

    [tex]\sum F_r = T - mg\sin\theta[/tex]

    As you correctly have. Now since the ball is following a circular path and applying Newton's second law we obtain:

    [tex]T - mg\sin\theta = m\frac{v^2}{L}[/tex]

    Now from here, rather than attempting to solve a differential equation, it would be much more straight forward to apply conservation of energy to determine the velocity as a function of theta. Do you follow?

    On a related point, you should note the mistake in going from the first line to the second line in your OP:
    The first term on the LHS should also be divided by a factor of g.
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