# Contradiction regarding friction of a rolling cylinder in an inclined plane

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

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.

• I cannot think of an appropriate title.Any help will be appreciated. – soumyadeep Feb 8 '14 at 16:08
• 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... – Valter Moretti Feb 8 '14 at 17:57

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.