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Ball Spinning Horizontally on String

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    There is a ball spinning on a string in a circle in a horizontal plane in a gravitational field. What is the rotational frequency w in terms of the angle of the string to the normal line of the plane of motion, the length of the string, m, and g?

    The original homework was actually something much more complicated with 3 masses and I was supposed to use Lagrangian mechanics to solve, but I got a very strange answer that I did not like, so I tried doing a simpler problem using physics 101, but my answer is the same as what I calculated with the lagrangian if you set the other two masses to 0. Please look at this it is very strange.


    2. Relevant equations
    You can solve this using basic physics


    3. The attempt at a solution
    Obviously if the system is in equilibrium then the mass is moving neither up nor down, so mg = Tcos(theta), where mg is gravity, T is tension, and theta is the angle between the string and a line that goes straight up and down.

    Also we have mV^2/r = ma = Tsin(theta). r must be Lsin(theta) and v is rw. Substituting for T and flipping everything around gives you sqrt(g/L/cos(theta)) = w. This is a very, very strange answer. It is not possible to make w = to 0 with any theta with this equation. This means that if you have a point mass hanging from a string in a gravitational field that it spins around for no god damn reason. Since L is on the bottom of the fraction, it would seem to indicate that a point mass merely existing in a gravitational field rotates infinitely quickly.

    The steps are very simple and I've checked them several times. Can someone either explain to me what I did wrong or why a mass hanging from a string always spins in place?

    If you write it in terms of V instead of w then you get V = sqrt((gL)tan(theta)sin(theta)) = V, which looks more reasonable because V = 0 for theta = 0 and V = infinity for theta = 90 degrees.
     
    Last edited: May 13, 2012
  2. jcsd
  3. May 13, 2012 #2

    tiny-tim

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    Hi ShamelessGit! :smile:

    Yes, V ~ √(tanθ), and ω ~ √(secθ),

    so V -> 0 and ω -> constant.​

    Is this surprising?

    If we gradually take energy out, so that θ -> 0, then V -> 0 but shouldn't the angular momentum stay the same? :wink:
     
  4. May 13, 2012 #3
    Angular momentum is only conserved if there are radial forces applied to an object. I would think that if you wanted to decrease the radius and conserve angular momentum then you'd need to apply an inward force, which would speed the object up rather than slow it down. If (r1)^2*w1 = (0)^2*w2, then wouldn't w have to go to infinity rather than be a constant?
     
  5. May 13, 2012 #4
    It just seems like an affront to my intuition to think that something can continue to spin after all the energy has been taken out of it, or that theta can stay the same after you add energy to it. I guess I'll just trust the math like I always do...
     
  6. May 13, 2012 #5
    Hi,
    There is no meaning to ω for θ=0
    Every calculation done depends on something 'spinning' with a certain stability
    obviously, for θ=0, v=0 and well ω=v/r with v=0 and v=0 is undefined
    Remember how you calculated your other relation to get ω as a function of θ, and that was by substituting for T, which is what when the ball does not spin ?
    So this does not tell you that a hanging ball from a spring tends to spin by itself "because even for θ=0 ω seems defined (as a function of θ)"
    On the other hand, as θ gets small, cos(θ) is 1-θ²/2 so it is 1 for first order variations of θ, which tells you that for small values of θ, the angular velocity (or period of rotation) is stable for small variations of θ, as long as there is any (θ>0)

    You could get the same kind of counter intuitive results for an even simpler pendulum where, for θ small enough the oscillation period is constant, and does not seem to depend on θ at all.
    That does not mean anymore than in the previous case that a ball hanging by a string must oscillate because even for θ=0 we would seem to have T=2π√(l/g) (except it does not mean anything because θ
    must be >0 for T to have any meaning at all). On the other hand it tells you that, were it not for friction, any small push should start a never ending oscillation at a stable and characteristic frequency.

    Maybe you were meaning in your example that it would mean, for θ=0 that the ball spins "on itself" at ω ?
    That would be wrong too because that is not the same ω that you calculate and in fact for your calculations the ball is a point and rotating on itself should be meaningless.

    Cheers...
     
  7. May 13, 2012 #6
    You're right oli4. w doesn't make any sense for theta = 0. Thanks
     
  8. May 13, 2012 #7
    You're welcome, thanks for the interesting question, and keep on halting on those kind of doubts, they re very much insightful !

    Cheers...
     
  9. May 13, 2012 #8

    tiny-tim

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    yup, let's forget angular momentum, it doesn't work :redface:

    try this instead …

    ω = √(g/h) where h (= Lcosθ) is the height of the pivot above the circle

    so suppose we have a pole of height h sticking out of the ground,

    and a string running through the top of the pole with a weight on the end, so that it's a circular pendulum with the weight just skimming over the ground

    so we can make the string any length we like but then we must adjust v to keep the weight just above the ground …

    then the period (and ω) of the pendulum is the same whatever the angle, even in the limit of the angle being zero! :wink:

    (this is exactly the same as the period of an ordinary planar pendulum staying the same in the limit of the speed reducing to zero)
     
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