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Homework Help: Particle rolling on a circular path

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data

    At the end of the lecture in http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed13.htm [Broken], Prof. Walter Lewin compares the motion of a slider on a circular air track (no friction) with the motion of a ball on a circular path (with friction). Based on the first experiment, the period of the second turns out to be larger than expected. Prof. Lewin explains that the reason has nothing to do with friction, and leaves the question unsettled.

    2. Relevant equations

    3. The attempt at a solution

    I made an attempt at giving an explanation: in the second experiment, the ball may not be considered a particle, since it is rolling. I tried to derive the period of the oscillation taking this into account and got the following:

    T = 2pi * square root[ ((R-r)^2 + 2/5 R^2) / g(R-r) ], (using small angle approximation)

    where R is the radius of the track and r is the radius of the ball.

    If r << R, this is almost equal to

    T = 2pi * sqaure root[ 7/5 * R/g ]

    This exceeds the period of the slider on the air track by a factor of sqroot(7/5) = 1.183.
    Hence, the period may be expected to be larger by 18%.

    Is this correct?

    Thank you very much.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 7, 2010 #2

    collinsmark

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    Hello Diegocas,

    [Edit: Scratch my last comment. Looking at your solution a bit more; yes I think you know the answer to why the rolling ball has a slower period! (You didn't call the reason by name, which I guess is why I overlooked it in your equation.)]
     
    Last edited: May 7, 2010
  4. May 8, 2010 #3
    Thank you! But I have one more question about it.
    In my calculation, I assumed the ball was rolling, not sliding. Why can we make that assumption? That is only possible because of the friction, isn't it?

    I did the following:
    Assume the ball is in a position where the angle from the vertical line is [tex] \theta [/tex].
    Then we must have:

    [tex] m a_{cm} = mg \sin \theta - F [/tex]

    where [tex] F [/tex] is the friction force (which I suppose upwards) and [tex] a_{cm} [/tex] is the accleration of the center of mass.
    In addition, considering the momenta of the forces with respect to the center of mass we get:

    [tex] Fr = I\alpha [/tex]

    where [tex] \alpha [/tex] is the angular acceleration of the sphere and

    [tex] I = \frac{2}{5}mr^2 [/tex]

    is the moment of intertia of the sphere.

    Now, for the ball to be rolling, we must have [tex] a_{cm} = r\alpha [/tex].

    This equation, together with the two previous ones, give the following expression for the friction:

    [tex] F = \frac{2}{7} mg \sin \theta [/tex]

    Since [tex] F [/tex] is a static friction, we must have

    [tex] F \leq \mu_s N = \mu_s mg \cos \theta [/tex]

    Hence, the condition for the ball to be rolling results:

    [tex] \frac{2}{7} \tan \theta \leq \mu_s [/tex]

    Now comes my "argument". Since [tex] \theta [/tex] is less than 20 degrees, we need that

    [tex] \mu_s \geq 0.103 [/tex]

    which is a rather mild condition, isn't it?

    That's why I guess the ball must be rolling.

    Is this correct? Or am I overlooking something?

    Thanks!
     
  5. May 9, 2010 #4

    collinsmark

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    Yes, you can assume the ball is rolling, and not sliding. And yes, friction is the reason the ball keeps rolling.

    But, I don't want you to concentrate too much on the frictional part. Friction is only indirectly related, in-so-far that it keeps the ball rolling (frictional losses are not significantly involved with the longer period). The instructor asked why the period is so much longer in the case with the rolling ball (as opposed to the sliding track, which acts like a swinging pendulum). One student in the classroom suggested that it might be due to friction. But the instructor correctly pointed out that frictional losses more-or-less have nothing to do with it.

    In the case of the rolling track, gravitational potential energy was converted to translational kinetic energy and back and forth, in a way that resembles simple harmonic motion (SHM). Frictional losses were ignored.

    In the case of the rolling ball, frictional losses can also be ignored. Yet in the case of the rolling ball, there is another type of energy involved, which is negligible in the case of the sliding track (but pronounced in the rolling ball case). Although I deleted this statement in my earlier post, this other type of energy/momentum is one of the most fundamental in all of physics, from simple Newtonian mechanics, to magnetism & electrodynamics, to special/general relativity, to the most advanced aspects of quantum mechanics. And it has nothing to do with friction.

    I haven't gone through your math, but I see this other aspect already in there, so I think you already know what I'm talking about, concerning this "other" type of energy/momentum.
     
    Last edited: May 9, 2010
  6. May 9, 2010 #5
    Yes, I know that the rolling ball has not only translational kinetic energy but also rotational kinetic energy.
    The part that puzzles me about the rolling ball is the following: on the one hand, friction may be ignored and it is the additional rotational kinetic energy that accounts for the longer period. On the other hand, it is because of friction that the ball can roll.
     
  7. May 9, 2010 #6

    diazona

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    But there are two kinds of friction, static and kinetic. Static friction is the kind that keeps two surfaces from sliding against each other, and it does no work, whereas kinetic friction acts against sliding motion and does do work. (There's rolling friction too, but that can be lumped under "kinetic")

    When Prof. Lewin says friction has nothing to do with the difference in periods, he's talking about kinetic friction, because a person's first instinct to explain the difference in periods might be to say that kinetic friction is "sucking" energy out of the system as it rolls. That's presumably what the student meant. The friction that keeps the ball rolling is static friction, but since that does no work, it doesn't remove any energy from the ball.

    Basically the idea seems to be that kinetic friction can be ignored, but not static friction.
     
  8. May 9, 2010 #7

    collinsmark

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    Technically, you're right. Friction is indirectly related. It is related in-so-far that friction keeps the ball rolling, but that's it. You can ignore it when it comes to changing forms of energy.

    Putting it another way, frictional losses are what can be ignored. You don't have to worry about any energy being converted to heat when calculating the period. In other words, you don't have to worry about any thermodynamically irreversible processes.

    You can treat the whole thing as a conservative system (in-so-far as estimating the period anyway). In other words, you can approximate that the total kinetic and potential energy is conserved. The key point here is that some of the gravitational potential energy is converted to rotational kinetic energy (as you have pointed out). But the period of the system is a function of the translational kinetic energy. And since there is less translational kinetic energy per total energy in this conservative rolling-ball system, the period is longer. :wink:
     
  9. May 9, 2010 #8
    Thank you very much for both your replies! It's all clear now!
     
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