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Jerk of the point of contact of a ball rolling on a flat surface

  1. Mar 10, 2004 #1

    I'm working on a kinematics problem of rolling a golf ball on a flat plane (4-position infinitesimally separated solution). The solution sought is to be planar, so we're working in 2D only. This is largely irrelevant though, as I'm facing a rather undergraduate puzzler.

    I need to derive jerk of the so-called instant center of rotation of a circle that rolls (without slip) on a straight line.

    Let's call the ground (straight line) body 1, and the ball (circle) body 2.

    The instant center I on body 1 [itex]I_1[/itex] is moving along the path of rolling. That's easy.
    The instant center I on body 2 [itex]I_2[/itex] is moving along the cycloid. Easy too.

    I mark vectors with underline, like [itex]\underline v[/itex].

    Now, let's look at the instant center as a point in the moving body 2. It's absolute acceleration (ground-relative) is
    [tex]\underline a_I=-\omega^2\cdot \underline r_{I_2/O_2},[/tex]
    where omega is the angular velocity of the rolling ball, and [itex]\underline r_{I_2/O_2}[/itex] is the radius vector from center of the ball to the instant center (point of contact). It's simply the radius of the ball. Fair enough. The sign may be wrong but that's the easiest thing to fix and I'm not concerned about it.

    I forgot to mention that the ball is to be rolling with a constant angular velocity, so obviously [itex]\alpha=\dot\omega=0[/itex]. That's why I left out the zero term [itex]\alpha\times \underline r_{I_2/O_2}[/itex]

    Now, I need the jerk too, which is simply the derivative of acceleration of I vs. time.

    {d\over dt} \underline a_I =
    {d\over dt} \left(-\omega^2\cdot \underline r_I\right) =
    - \left[\left({d\over dt} \omega^2\right)\cdot \underline r_I + \omega^2\cdot\left({d\over dt} \underline r_I\right)\right] =
    - \left(2 \omega \alpha \underline r_I + \omega^2 \underline a_I\right),

    where [itex]\omega[/itex] - ang. vel, [itex]\alpha[/itex] - ang. accel, [itex]\underline r[/itex] - position vector of I, [itex]\underline a[/itex] - acceleration vector of I

    Substituting [itex]\underline a_I[/itex] back into the equation, noting that [itex]\alpha=0[/itex], we get

    {d\over dt} \underline a_I =
    - \omega^2 \left(-\omega^2 \cdot \underline r_I\right) =
    \omega^4 \underline r_I

    Now, jerk of a point is a vector whose elements have SI units of m/s^3 - it's a rate of change of acceleration, so it has the unit of acceleration per time, so m/s^2/s = m/s^3.

    Now, looking at the above derived d/dt a_I it's pretty obvious that the unit would be
    (1/s)^4*m = m/s^4

    My calculus is wrong and I'm too tired to see where did I err. Help :wink:

    Cheers, Kuba
    Last edited by a moderator: Mar 10, 2004
  2. jcsd
  3. Mar 10, 2004 #2


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    Gold Member

    As far as the units go, dr/dt shouldn't = a, it should be a velocity.

    But I think I'm missing something here:

    The instant center is always the point of contact. Doesn't that just move in a straight line along the groundat a constant velocity (and therefore 0 acceleration and 0 jerk)?

    If you're talking about the point that is the instant center on the ball moving in a cycloidal fashion, I see that (but it's not the instant center after that instant, right?). In that case, if you set the origin at the instant center at t = 0, you get the following for the position of that point:

    [tex] x = R(\omega t - \sin \omega t )[/tex]
    [tex] y = R(1 - \cos\omega t) [/tex]


    [tex] \dot x = R(\omega - \omega\cos\omega t) [/tex]
    [tex] \dot y = R\omega\sin\omega t [/tex]


    [tex] \ddot x = R(\omega^2\sin\omega t )[/tex]
    [tex] \ddot y = R\omega^2\cos\omega t [/tex]


    [tex] \frac{d^3x}{dt^3} = R \omega^3\cos\omega t [/tex]
    [tex] \frac{d^3y}{dt^3} = -R\omega^3\sin\omega t [/tex]

    I hope that's useful.
  4. Mar 11, 2004 #3
    Now, take it seriously - fatigue *is* a hideous mode of failure. This statement applies both to mental fatigue and to fatigue of material due to cyclic loading.

    Surely enough [itex]{d\over dt}\underline r[/itex] is the velocity! Stupid me [zz)]

    Thus, correcting myself,
    {d\over dt}\underline a_I
    = -\left[2\omega\alpha \underline r_I + \omega^2 \underline v_I\right]
    = -\omega^2\underline v_I
    = -\omega^2\cdot \omega\times\underline r_I
    = -\omega^3\cdot \hat k \times \underline r_I

    In this problem, the instant center is a point that's instantaneously coincident with point [itex]I_1[/itex] and [itex]I_2[/itex], that is you really have two points with instantaneously same position but not necessarily same derivatives of position. Here, the velocities of I_1 and I_2 match instantaneously but e.g. acceleration doesn't - on body 2 it has centripetal acceleration, o body one it has zero acceleration.

    On body 1 (ground), the instant center moves with a fixed velocity on a straight line.

    On body 2 (the ball), the instant center moves along the cycloidal path and thus has resultant acceleration and jerk.

    The word "instant" in the instant center means that it's the center of relative rotation of the bodies at that point in time only. Instant centers can wander a lot in a typical mechanism.

    I'm actually looking at the instant center on body 2 (the rolling ball). I was trying to understand what's going on. Thanks for all your help. You've saved me an hour of head scratching over a stupidity. Thanks a lot!!!

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