Rolling body kinematics

Discussion in 'Physics & Math' started by arfa brane, Nov 26, 2011.

  1. arfa brane call me arf Valued Senior Member

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    If a disk is rolling in a straight line along a plane surface:
    What is the distance travelled by the centre (or axle) of the disk, between points of contact an arbitrary point P on the circumference makes with the plane as it's moving?

    What distance does P travel in the same time?

    If the cycloid made by P is made into a path for a mass, m, by for instance, building a metal track for a steel ball in the shape of an inverted cycloid, what are the equations of motion for m?
     
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  3. AlphaNumeric Fully ionized Moderator

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    Your post isn't entirely clear so I'll see if I can rephrase it. A circular disk is rolling along without slipping. You mark a point P on the edge of the disk and note the times when this point hits the ground. Between two consecutive times of P hitting the ground you want to know how long is the trajectory P has swept out and how long is the trajectory the central point of the disk has swept out?

    The second one is easy. The disk has radius R and in doing 1 rotation it rolls forward 1 perimeter, which is \(2\pi R\). The question of the path length P moves along is less simple.

    The equations in question to define the position are \(x(t) = R(t-\sin t)\) and \(y(t) = R(1-\cos t)\). Thus \(\mathbf{P}(t) = (x(t),y(t))\) is such \(\mathbf{P}(0) = (0,0)\), it starts on the ground level. It next hits the ground at time \(t = 2\pi\).

    Given a parametrisation you get the path length between those times is \(s = \int_{0}^{2\pi} \sqrt{\dot{x}^{2}+\dot{y}^{2}} dt\). So let's compute that

    \(\dot{x}(t) = R(1-\cos t)\) and \(\dot{y}(t) = R\sin t\) so the bit inside the square root is \(\dot{x}^{2} + \dot{y}^{2} = R^{2}\Big( 1 - 2\cos t + \cos^{2}t + \sin^{2}t \Big) = 2 R^{2}(1-\cos t)\)

    We want to take the square root of this and for those who remember their trig identity \(\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 1-2\sin^{2}\theta\) we set \(\theta = \frac{t}{2}\) and get \(1-\cos t = 2\sin^{2}\frac{t}{2}\) and thus

    \(s = \int_{0}^{2\pi}2R \sin^{2}\frac{t}{2} dt = 4R \int_{0}^{\pi}\sin \tau d\tau = 8R\)

    So surprisingly the answer doesn't involve a factor of pi, it's just 8 times the radius.
     
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  5. Tach Banned Banned

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    Is this homework? AN answered the first part (related to cycloid geometry), the second part is a simple problem of dynamics. You know the equation of the trajectory (it is the inverted cycloid). You know the force acting on the steel ball, its gravity \(m \vec{g}\). Decompose the force into two components, one normal to the trajectory (cancelled by the reaction force of the track) and the second one tangent to the trajectory \(\vec{f}=m \vec{g} sin (\alpha)\).
    Note that \(\alpha=\alpha(t)\) is the time varying angle between \(\vec{g}\) and the normal to the cycloid. Form and solve the ODE describing the motion. You will run into the same difficulty as the pendulum equations, the equation is non-linear , so you will have to get creative. You will not be able to benefit from the approximations for small \(\alpha\) in this problem, so the ODE will be tougher.
     
    Last edited: Nov 26, 2011
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  7. arfa brane call me arf Valued Senior Member

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    But we note that since the wheel's displacement between successive points of contact is: \(2\pi R\), and this is related to the length of the cycloid by \(\frac {4} {\pi} 2\pi R\) (because the cycloid's length is \(\frac {4} {\pi}\) times the horizontal displacement of the wheel, doh!).

    Anyway, I was thinking about the fact that a wheel (or a sphere) rolling on a plane surface has a number of forces acting on it--namely its weight, the normal to the surface and lastly friction at the point(s) of contact. A proper kinematical treatment should consider the radius of gyration for the rolling body.

    And a final note about the cycloidal pendulum, viz: a steel ball rolling along a track in the shape of an inverted cycloid: my reference tells me the period is always \(4\pi \sqrt { \frac {R} {g}} \), where R is the radius of the generating circle. That doesn't give you equations of motion, but isn't it a constraint?
     
  8. Tach Banned Banned

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    There is one way to find out, form the equations of motion and solve them.
    BTW, the above period is only true for small angles, it isn't true in general.
     
  9. prometheus viva voce! Moderator

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    In the OP the question is about deriving the equations of motion, not actually solving them.
     
  10. Tach Banned Banned

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    So, I gave him the steps how to do that.
    Next thing, he asked about calculating the period, the only way to do that is by solving the equations of motion, no?
     
  11. arfa brane call me arf Valued Senior Member

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    That isn't what the reference I have says. It says, "the amplitude of the motion will depend on the point from which the particle is released, but the period will always be \( P\; =\; 4\pi \sqrt {\frac {R} {g}} \).

    It also says, "There is a special arrangement in which the period of a pendulum is independent of the amplitude. This is called a cycloidal pendulum."
     
  12. prometheus viva voce! Moderator

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    This problem is more normally studied as the question: find the curve where the period of oscillation is independent of the amplitude (not the case for a circular pendulum) - you've been given that the solution to this problem is a cycloid so you've got a bit of a head start. It's called the isochrone problem and the method used to attack it is calculus of variations. The simplest way to find the equation of motion of the system is to write down the Lagrangian and hit it with the Euler-Lagrange method. If this is all double Dutch to you, Tach's technique will work, but you will curse your lecturer/teacher for not teaching you Euler-Lagrange sooner in a little while.
     
  13. Tach Banned Banned

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    Cycloidal pendulum is different than what you asked for. See here
     
  14. arfa brane call me arf Valued Senior Member

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    Different how?
    Do you mean a steel ball rolling along a cycloidal track isn't a pendulum?
     
  15. Tach Banned Banned

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    Did you even read the description of the cycloidal pendulum? It is not a ball rolling along the cycloidal track. It has absolutely nothing to do with the ball rolling inside the cycloid track.
     
  16. prometheus viva voce! Moderator

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    A cycloidal pendulum is a pendulum where the bob traces out a cycloid as it oscillates (the length of the string is not a constant), whereas a normal pendulum traces out a circle. I'm struggling to see the relevance if I'm honest, but it's typical of Tach to find an error and point it out, no matter how irrelevant.

    edit: I of course mean an arc of a circle, and a portion of a cycloid, not complete ones.
     
  17. Tach Banned Banned

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    No, see here

    You would if you read arfa brane's scenario vs. the definition of the cycloidal pendulum.
     
  18. prometheus viva voce! Moderator

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    Your objection is completely irrelevant for the last part of the OP, which it is obvious we are talking about since arf has talked about a bead on a rail.
     
  19. prometheus viva voce! Moderator

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    Quote from your link: "the bob of the pendulum also traces a cycloid path."

    Care to try again?
     
  20. arfa brane call me arf Valued Senior Member

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    So your answer is "no, a steel ball rolling along a cycloidal track is not a pendulum".

    But such a system has a period, and an initial amplitude, so what is it if it isn't a pendulum? And why would an undergrad physics text say that it is a pendulum?
     
  21. Tach Banned Banned

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    In a cycloidal pendulum the angle is (severely) constrained, so you can simplify the ODE describing its equations of motion.
    In the case of a ball rolling on a (inverted) cycloid you cannot take advantage of the small angle approximation(s) , so you end up with a more complicated ODE, as explained earlier. The two are very different animals.
    Now, if you include the rotational momentum of the steel bead (as arfa wants to do), the second scenario departs even further from a cycloidal pendulum.
    Is this now clear enough?
     
  22. Tach Banned Banned

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    Sure it can be considered a pendulum. It is NOT a "cycloidal" pendulum, it has DIFFERENT equations of motion, equations that you need to derive.

    See above.
     
  23. prometheus viva voce! Moderator

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    You are inventing objections.

    I don't see how. Assuming the length of the string at rest is sufficient there the only constraint is that the bob must not pass through more than 180 degrees (or else it would be outside a cycloidal path).
    Sure you can use a small angle approximation if it helps. The solution will only be valid for small displacements from equilibrium then, of course.

    I can find no reference to arf wanting to do this.

    One thing is certainly clear, yes.
     

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