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Centripetal force carousel angle

  1. Apr 20, 2013 #1
    Hello everybody,

    i have got a question about the concept of centripetal force in general. Imagine a gondula of a carousel that has just started to rotate. Each gondula moves away from the center of the circular motion and forms an angle with the string.

    What would be the correct physical line of argumentation for explaining the movement away from the center from a coordinate system which is NOT accelerated.

    My thoughts:

    Since the gondula follows a circular motion there must be centripetal force pointing towards the centre. This is the case as long as ω is constant. But right at that moment when the carousel starts moving there must be a force pointing in the opposite diretion of the centripetal force because an observer outside the carousel can see how the gondula forms an angle with the string. But where does it come from?
  2. jcsd
  3. Apr 20, 2013 #2
    I presume you are referring to a so-called "swing ride". I don't think one should make an argument here that has to do with what happens when the ride starts to move.

    What you we can do is show that with constant ω there is only one angle for which the gondola follows circular motion. The radius in this case is not the length of the string, but instead the horizontal component of the length of string. If the gondola is going up or down (assuming the string is taught and has a fixed length) , it follows that the horizontal radius is also changing. So the only way the horizontal radius is not changing is that the gondola is not going up or down. This means the vertical component of the tension in the string must be exactly opposite the force of gravity. There is only one angle where the horizontal tension is what is required for circular motion, and the vertical component of tension exactly opposes the force of gravity.

    If the angle is something other than this angle, then the angle will not be constant in time, even when ω is constant. Finding the force that it experiences to cause this would require some more analysis, but it can be done.

    Attached Files:

    Last edited: Apr 20, 2013
  4. Apr 25, 2013 #3
    Hi Mr X,

    thanks a lot for your time and quick answer but I am afraid I can't see how this answers the question. When ω is constant, there will be only one angle[itex]\varphi[/itex] (see attached picture) but the question is, why does the person on the swing ride moves away from the center in the first place. According to Newton I, there most be a force responsible for that or at least a component of a force in that direction. Since there is noch such force for someone who watches the thing from the outside (non-accelerated coordinate sytem) because the centripetal force ( in this case) is provided by the vector sum of gravitiy (-y direction) and the force of the string/chain. The only thing I can think of is that the process of starting the rotation must be responsible for that because in that case [itex]\omega[/itex] is not constant (the swing ride accelerates) but I can't figure out a way how to calculate this component.

    Attached Files:

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  5. Apr 25, 2013 #4
    I assure you it does not have to with the ride starting. There would be a force even when ω is constant. The equation of motion can be found with Lagrangian mechanics.

    To do this I am going to make θ measured upwards from the horizontal. The length to the center is [itex]\ell[/itex]. The height relative to the center is [itex]z = \ell sin\theta[/itex].

    The velocity squared is
    [itex]\dot{r}^2 = \ell^2 \dot{\theta}^2 + \ell^2 \omega^2 cos^2 \theta[/itex]

    so here is the Lagrangian
    [itex]\mathcal{L} = \frac{1}{2}m\dot{r}^2 - mgz [/itex]
    [itex]\mathcal{L} = \frac{1}{2}m\ell^2 \dot{\theta} + \frac{1}{2}m \ell^2 \omega^2 cos^2 \theta -mg\ell sin\theta[/itex]
    [itex]\frac{\partial \mathcal{L}}{\partial \theta} = -m\ell^2\omega^2cos\theta sin\theta - mg\ell cos\theta [/itex]
    [itex]= \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\theta}} = m\ell^2\ddot{\theta} [/itex]

    so here is the equation of motion

    [itex]\ell^2\ddot{\theta} = -\ell^2\omega^2cos\theta sin\theta - g\ell cos\theta [/itex]

    If you solve for the equilibrium position, [itex]\ddot{\theta} = 0[/itex], you will get the answer as before (except with a minus sign since this theta is measured up/counterclockwise). If you want to get the actual tension that causes this to happen it is some more work but it can be done.
  6. Apr 26, 2013 #5


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    If you watch such a ride start, you will see that there is most definitely some motion associated with start-up. The supports on the carousel start moving and the gondolas stay in place. The resulting angle of the support cables tends to accelerate the gondolas.

    This initial acceleration is in a direction tangent to the rotation, of course.

    1/4 of a rotation later, the motion imparted by that initial acceleration is no longer tangent to the rotation. It is almost entirely outward. In the rotating frame of reference we would see the direction of motion changing and attribute the perceived acceleration to "Coriolis force". In the inertial frame of reference we see straight line motion -- the tangential motion imparted initially has eventually carried the gondola farther away from the center of the carousel.

    Of course, as the gondola's initial motion carries it farther from the center of the carousel on a straight line, the support frame is proceeding in its circular path. The angle of the support cables is changing. The result is continuing tangential acceleration.

    In the rotating frame, Coriolis force is causing the gondola to veer away from the direction of rotation. Meanwhile the support cables are pulling it in against this force toward the equilibrium position. The gondola ends up orbitting around the equilibrium position in the direction of the carousel's rotation until the orbit is ultimately damped and equilibrium is attained.
  7. May 16, 2013 #6
    Jbriggs 444,

    thanks a lot for your last post, it helped me a lot to get a better understanding of the afore mentioned process.
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