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Understanding angular momentum

  1. Jul 31, 2009 #1

    daniel_i_l

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    I'm trying to wrap my head around angular momentum using the following setup:
    There's a ball on a frictionless table that's connected to a string. The other end of the string is connected to a hole in the table. In the initial position the ball spins around the hole.
    Now, we start pulling the string through the hole and thus shortening the amount that's connected to the ball.
    From conservation of angular momentum the ball is now spinning faster because it's radius is smaller. Now, angular momentum is conserved because the force of the string only pulls the ball towards the center. But in that case, what causes the angular velocity to increase?
    I think that the answer lies in the fact that even though the force is radial, it still causes acceleration in the direction of the spin via the Coriolis acceleration. Assuming that the speed with which the ball moves towards the center is constant [tex]v_{c}[/tex] then the angular acceleration caused by the Coriolis acceleration is [tex] a(r) = -2 \omega v_{c} = -\frac{2 v(r) v_{c}}{r} [/tex]
    This is a differential equation of [tex] v(r) [/tex] whose solution is [tex] v(r) = \frac{v_{0} r_{0}}{r} => v(r)r = v_{0}r_{0}[/tex] where [tex] v_{0} [/tex] and [tex] r_{0} [/tex] are the initial velocity in the direction of the spinning and the radius. This equation is exactly the conservation of angular momentum.
    So apparently, the conservation of angular momentum is just another way of saying that forces at right angles to the velocity still change the velocity because of Coriolis acceleration.
    Is my conclusion correct? If so, how can I solve the differential equation when [tex]v_{c}[/tex] changes with [tex]r[/tex]?
    Thanks
     
    Last edited: Jul 31, 2009
  2. jcsd
  3. Jul 31, 2009 #2
    To understand this first, instead of thinking about the angular velocity think of how the tangential velocity changes.
    The speed of the ball |v| is constant even when you pull the string in closer, however because the ball is now bound to a shorter circumference it must more rapidly change direction.
    Take your finger and trace out a large circle, then at the same speed trace out a smaller circle and you will find that you complete the smaller circle much faster. The reason the angular velocity increases is because it is in radians/degrees which is independent of the actual distance the ball has to travel. So it just sees that the ball completes the circle faster, even though it is moving at the same speed, just a shorter distance. (less time and less distance = same speed).

    The trade-off for making the ball faster is that now you have to hold onto the string tighter because the force required to keep the ball along the new path has increased.
     
    Last edited: Jul 31, 2009
  4. Jul 31, 2009 #3

    rcgldr

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    The force is radial, but not perpendicualr to the spiral path of the ball that occurs when the string is pulled in or released out. There's a non-zero component of force in the direction of the ball, which speeds up (or slows down if tension reduced) the ball.

    Here's a diagram, showing a short line segment perpendicualr to the ball and the radial string.

    hole.jpg

    For the math part, angular momentum is m x r x s, so if r is halfed, then to conserve momentum, it becomes m x (r/2) x (2s). Based on this assumption, then the relation ship between tension and r is:

    [tex]t(r)\ =\ -m\ (s_0)^2\ (r_0)^2\ /\ r^3[/tex]

    So if string is pulled from [itex]r_0[/itex] to [itex]r_0/2[/itex], then work done is:

    [tex]\int _{r_0} ^{r_0/2} (-m\ (s_0)^2\ (r_0)^2\ /\ r^3)\ dr[/tex]

    [tex]\left[ \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ r^2 \right]_{r_0}^{r_0/2}[/tex]

    [tex]\frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (\frac{r_0}{2})^2 - \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2[/tex]

    [tex]\frac{4}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2 - \frac{1}{2}\ m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2[/tex]

    [tex]\frac{3}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2[/tex]

    [tex]\frac{3}{2} m\ (s_0)^2[/tex]

    Original KE:

    [tex]{KE}_0 = \frac{1}{2} m (s_0)^2[/tex]

    KE after work done:

    [tex]{KE}_1= \frac{1}{2} m(s_0)^2 + \frac{3}{2} m\ (s_0)^2 = \frac{4}{2} m\ (s_0)^2[/tex]

    [tex]{KE}_1= \frac{1}{2} m (2\ s_0)^2 [/tex]

    So if the radius is decreased by 1/2, the speed is doubled and the tension increased by a factor of 8.
     
    Last edited: Jul 31, 2009
  5. Jul 31, 2009 #4

    rcgldr

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    If the goal was to keep the speed of the ball constant, this can be accomplished by having the string wind or unwind from a pole. Here it appears that angular momentum isn't being conservered unless you note that the tension of the string on the pole exerts a torque on whatever the pole is attached to (usually the earth), and only by considering what the pole is attached to will angular momentum be conserved. In this case the force is not radial, but it is perpendicular to the path of the ball:

    pole.jpg

    The math demonstrating that the line perpendicular to the path of the ball is the same as the tangent line to the pole is shown here:

    https://www.physicsforums.com/showpost.php?p=1435942&postcount=32
     
    Last edited: Jul 31, 2009
  6. Aug 1, 2009 #5

    daniel_i_l

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    I don't understand how you got this equation for the tension. Are you taking into account the acceleration in both the tangent and radial directions?
    Thanks
     
  7. Aug 8, 2009 #6

    daniel_i_l

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    Can anyone explain that equation please?
    Thanks
     
  8. Aug 8, 2009 #7

    rcgldr

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    Only the radial direction, since that's the direction that the string is being moved so that it peforms work. Movement perpendicular to the string doesn't involve any work performed by the string, so I'm only condering changes in the radial direction.

    To get the equation, start with

    t(r0) = -m (s0)2 / (r0)

    let r = c r0
    then s = (1/c) s0

    t(r) = -m ( (1/c) s0)2 / (c r0)
    t(r) = -m s02 / ( c3 r0)
    t(r) = -m s02 r02 / ( c3 r03)
    t(r) = -m s02 r02 / ( c r0)3
    t(r) = -m s02 r02 / r3
     
    Last edited: Aug 8, 2009
  9. Aug 10, 2009 #8

    daniel_i_l

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    Thank you.
     
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