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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.

So:

[tex]

{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] =

[/tex]

[tex]

- \left(2 \omega \alpha \underline r_I + \omega^2 \underline a_I\right),

[/tex]

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

[tex]

{d\over dt} \underline a_I =

- \omega^2 \left(-\omega^2 \cdot \underline r_I\right) =

\omega^4 \underline r_I

[/tex]

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

Cheers, Kuba

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# Homework Help: Jerk of the point of contact of a ball rolling on a flat surface

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