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I came upon this while wondering whether the friction of a rolling cylinder on an inclined plane depends on the value of friction coefficient.

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now, $$f\leqq\upsilon N$$

Again after calculating I found that the $$f=\frac{mg\sin\theta}{3}$$ So,$$\frac{mg\sin\theta}{3}\leqq \upsilon mg\cos\theta$$ $$\Rightarrow \tan\theta\leqq 3\upsilon$$ But, I can set the angle and mu to be such that this relation does not hold!

What will happen then? Please explain.

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  • $\begingroup$ I cannot think of an appropriate title.Any help will be appreciated. $\endgroup$ – soumyadeep Feb 8 '14 at 16:08
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    $\begingroup$ If the condition is violated (because $\theta$ is too big) the contact point of the cylinder with the plane has not vanishing velocity and so... $\endgroup$ – Valter Moretti Feb 8 '14 at 17:57
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This problem is an example of rolling without slipping. A very good explanation of this concept is given here. In this case, it implies that rolling without slipping occurs if $\tan \theta \leq 3 \mu$. The expression validates one's intuition too. Its easy to observe that a cylinder tends to roll without slipping when kept on a wedge with lesser slope.

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  • $\begingroup$ But i have never assumed while finding f that it is rolling without slipping!i cannot understand what you are trying to tell. $\endgroup$ – soumyadeep Feb 15 '14 at 14:37
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    $\begingroup$ You did when using the equivalence between angular speed and linear speed. $\endgroup$ – Davidmh Mar 11 '14 at 6:04

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