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Excercise with conic pendulum

  1. Dec 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A tether ball of mass ##m## is suspended by a rope of length ##L## from the top of a pole. A youngster
    gives it a whack so that it moves with some speed ##v## in a circle of radius ##r = L sin(\theta) < L## around
    the pole.
    a) Find an expression for the tension ##T## in the rope as a function of ##m##, ##g##, and ##\theta##.
    b) Find an expression for the speed ##v## of the ball as a function of ##\theta##.
    Immagine.png

    2. Relevant equations
    Centripetal acceleration
    Newton's Second Law
    Tension
    Gravity

    3. The attempt at a solution
    This is the picture of how it would be with all the forces and with the y-axis upward and the x-axis going right. (I took this picture from another exercise about the conic pendulum in here)
    conicpend.JPG
    Now, we know that there is no actual "Centripetal Force". The centripetal acceleration that the mass has is given by the net force of the other forces in play, in this case the tension ##T## and the gravity ##m g##. We know that there is no motion on the y-axis and only a motion on the x-axis, so we have:
    $$\begin{cases}
    m a_x = T sin \theta \\
    m a_y = T cos \theta - m g = 0
    \end{cases}$$
    We can write ##a_x = a_c = \frac{v^2}{r}##.
    We can then write:
    $$T cos \theta = m g$$
    $$T = \frac{m g}{cos \theta}$$
    And so we have the tension expressed as requested.
    The velocity, though, will be written like this:
    $$v = \sqrt{a_c r}$$
    $$v = \sqrt{a_c L sin \theta}$$
    but we know that ##a_c = \frac{T}{m} sin \theta##, ##a_c = g tan \theta##, and so:
    $$v = \sqrt{L g tan \theta sin \theta}$$
    Am I right? Or I did something wrong?
     
  2. jcsd
  3. Dec 20, 2016 #2

    TSny

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    Homework Helper
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    Your work looks good to me. It's always a good idea to check limiting cases of your answer. For the cases θ = 0 and θ = 90o, does your result for ##v## make sense? Also, does your formula for ##v## give the correct dimensions (or units) for a speed?
     
  4. Dec 20, 2016 #3
    For the dimensions yes.
    If I have ##\theta = 0##, I have ##v = 0##. This is the case where the ball is at the same height as the beginning of the rope. It nullifies because there's no force acting on it?
    If I have ##\theta = 90^o##, I have the tangent that doesn't exist. This is the case where the tether ball is basically at the center and so it doesn't actually move. Or, to better say, that the two forces acting on it nullify themselves.
     
  5. Dec 20, 2016 #4

    TSny

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    Gold Member

    θ = 0 corresponds to the rope being vertical.
    θ = 90o corresponds to the rope being parallel to the ground.
     
  6. Dec 20, 2016 #5
    Oh true! I was looking at the other picture.
    Thank you!
     
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