If I have two metal rods arranged like a ramp with coefficient of friction $\mu$, at an angle $\theta$ to the ground, and a solid metal ball or radius $r$ and mass $m$ rolling down between the two tracks in such a way that it touches both of them, how does the separation between the tracks $d$ affect the acceleration $a$ of the ball, assuming a uniform gravitational field of strength $g$?

I know that I can model a box sliding down a ramp using equations with energy but how do I calculate for a rolling ball, and take into account the separation between the tracks?


The separation of the rods does affect the acceleration even for constant separation, because (assuming a rolling motion) the relative amount of energy that goes into linear and angular motion is affected by the spacing.

If the rolling is assumed to be without slipping, we can solve the problem by conservation of energy: $$ mg \Delta h = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 \tag{*} \,,$$ with the relationship between $v$ and $\omega$ set by the spacing of the rods through the radius of contact during rolling.1 You seek to determine the effective radius $r_{eff}$ of the rolling. For rods of diameter much less than the radius of the ball and spacing $d$ you can use $r_{eff}^2 = r^2 - (d/2)^2$. If you allow that the diameter of the rods is non-trivial, the geometry is a little more complicated.

In either case use $v = r_{eff}\omega$ in (*) and insert the appropriate moment of inertia ($\frac{2}{5}mr^2$ for a solid sphere), and you are home and dry.

A further complication is that if the separation of the rods vary in space the center of gravity of the ball can drop (for increasing separation) or rise (for decreasing separation) relative the plane of the rods.

There is a toy that employs this effect to let you make a bearing roll "uphill" by gently and carefully controlling the separation of the ends.

1 Recall that for rolling down a ramp the relationship is $v = r\omega$.


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