# 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.

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

The condition for pure rolling of a cylinder on an inclined plane is that the point of contact does not slip with respect to the incline. So friction will be static friction

So acceleration of point of contact is zero. So Mgsinθ - f = Ma and fR = I alpha (I = 0.5MR^2) and acceleration of point of contact = a -R*alpha = zero

Solving these we get the value of friction = f = Mgsinθ/3 and Normal by inclined plane must be Mgcosθ

So the condition f <= mu*N must also be satisfied . ( Here mu is coefficient of static friction) If this fails then friction will be kinetic friction and therefore friction will be (coefficient of kinetic friction)*N so the cylinder will accelerate downwards with constant acceleration = gsinθ- (coefficient of kinetic friction)*gcosθ.

Therefore the cylinder will rotate due to torque of friction with angular acceleration = 2*(mu)gcosθ/R. (mu = coefficient of kinetic friction). But this will not be in pure rolling as the point of contact is not at rest with respect to the inclined plane