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In this article...

https://billiards.colostate.edu/physics/Domenech_AJP_87%20article.pdf

...which analyzes rolling friction on a rolling ball, the author claims that the normal force on a ball rolling down an incline is given by:

enter image description here

Where $R_b$ is the regular radius of the ball and $R_e$ is the effective radius of the ball, which is less than the regular radius since the ball squishes under its own weight.

But I don't understand. Why does the fact that the ball "squishes" as it rolls change the magnitude of the normal force?

enter image description here

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Look at their Fig. 3. The normal forces are not directed orthogonally to the inclined plane. You might say that their vector sum is orthogonal to the inclined plane, but I guess the components tangential to the plane still affect the rolling friction in their first formula.

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  • $\begingroup$ hmm that makes sense. They should've explained themselves better. Why would it end up being that ratio though, Rb/Re? $\endgroup$ Dec 29 '18 at 20:42
  • $\begingroup$ I also thought it maybe had something to do with the surfaces (of the ball or of the plane) behaving a bit like a spring? Since its deforming? $\endgroup$ Dec 29 '18 at 20:43
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    $\begingroup$ @JoshuaRonis : If there are two normal forces with absolute values $N_1$, as in figure 3, and both of them are directed at an angle $\phi$ from the normal to the inclined plane, then $mg=2 N_1 \cos \phi$ and $\cos\phi=R_e/R_b$. $\endgroup$
    – akhmeteli
    Dec 29 '18 at 22:44
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    $\begingroup$ @JoshuaRonis : Well, I probably made a mistake in my previous comment, I should have written $m g \cos\theta=2 N_1 \cos\phi$. $\endgroup$
    – akhmeteli
    Dec 29 '18 at 23:53
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    $\begingroup$ @JoshuaRonis : $N=2N_1$. $\endgroup$
    – akhmeteli
    Dec 30 '18 at 0:47

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