# What is the minimal friction force required in order for a ball on an incline to roll without slipping?

I read an article now and it confused me completely. Since I thought I have read somewhere else a completely different solution.

You put a ball on an incline at rest. What is the minimal friction force so that the ball rolls without slipping?

I thought there is a simple and easy solution. All you need is that the sigma forces in X axis would be equal at least to zero. Which is the minimum, or else the friction could be stronger.

But I have read this article https://www.google.co.il/url?sa=t&source=web&rct=j&url=http://www.feynmanlectures.info/solutions/roll_without_slipping_sol_1.pdf&ved=0ahUKEwig-tinj4DbAhWCBZoKHextDoEQFghtMBQ&usg=AOvVaw1T-CXiuqp0CtO1W2NNFzbY and he has long calculations for it.

I don't really understand why????

I am trying to think about it and my solution seems perfectly correct. There isn't any force that rotates the ball apart from friction. So the ball will certainly rotate as long as it doesn't slip. And it doesn't slip when the friction is equal or stronger than the X component of the MG force of gravity.

Isn't it sufficient to say that MGsin(a)= F friction?

Isn't it enough? Why are there so much complexity in the article? And why is it correct?

• The calculation in the article is very detailed. Which steps do you disagree with and why? – sammy gerbil May 12 '18 at 13:07
• I don't even need and specifications. I just want someone to explain why isn't MY solution correct? If you ask me, I basically don't understand anything from what is written. Why do you need to calculate anything further beyond equating the friction force to X component of mg? – bilanush May 12 '18 at 13:10
• Well, I thought for a second that I understood you. But I regretted it. Picture a hoop on an incline, why wouldn't it rotate? The friction being equal to mg would only cause it to no slip. But as far as the rotational movement is concerned the mg doesn't act at all the only force acting is friction which WOULD rotate the hoop. I even remember that this is what I studied with similar example. – bilanush May 12 '18 at 13:22
• If the friction force equals $mg\sin\theta$ up the slope, and the component of gravity down the slope is $mg\sin\theta$, then the resultant force on the ball is zero. The ball does not accelerate. If placed at rest it will remain at rest. ... If the resultant force on any object is zero, its COM does not accelerate ($F=ma$). The object might rotate, but the COM does not accelerate. – sammy gerbil May 12 '18 at 13:22
• Why? The rotational force acting on the ball is only friction. Mg sin a doesn't cancel it. Them being equal only means it wouldn't slip. – bilanush May 12 '18 at 13:24

## 3 Answers

If the no slip condition is satisfied then the linear acceleration of the centre of mass of the ball is equal to the radius of the ball times the angular acceleration of the ball.

This means that there must be a net force down the slope on the ball and at the same time a net torque acting on the ball.

If $mg \sin \theta = F$ there is no net force down the slope and yet there is a torque on the ball about its centre of mass equal to $Fr$ where $r$ is the radius of the ball.
So there is no linear acceleration down the slope and yet there is an angular acceleration.

If there is no frictional force then there is a net force down the slope and the ball will accelerate down the slope but as there is no torque on the ball about its centre of mass there will not be an angular acceleration.

This shows that the required frictional force has a value between zero and $mg \sin \theta$.

The derivation has the net force down the slope, with the frictional force less than $mg \sin \theta$, causing a linear acceleration down the slope and the torque produced by the frictional force causing an angular acceleration each of the correct magnitude so that the no slip condition is satisfied.

• So what would happen if mgsinθ=F ? Would the ball rotate in its place? Since there is a torque angularly of friction only ? Moment of inertia multiplied by angular acceleration. – bilanush May 12 '18 at 15:21
• There would be no linear acceleration of the ball but there would be an angular acceleration of the ball. – Farcher May 12 '18 at 15:24
• So I am asking what would happen? Would the ball rotate? – bilanush May 12 '18 at 15:26
• The centre of mass of the ball would either be at rest or move down the slope at constant velocity and the ball would rotate with the angular speed increasing but this is not a realistic situation. – Farcher May 12 '18 at 15:32
• Sorry . I am asking a simple question. Would it rotate? What do you mean at rest? Would it rotate on its place? You didn't answer. – bilanush May 12 '18 at 15:35

Isn't it sufficient to say that MGsin(a)= F friction?

No, that is impossible.

• If $mg \sin(a)= f_s$, with $f_s$ being static friction, then all forces balance out along the incline. Then there can be no linear acceleration (due to Newton's 2nd law) and thus the ball wouldn't start moving at all. But we know that it will start moving - so clearly this is not the case. The forces do not balance out.

• Also, the static friction would never be larger than the weight component - because then there would be a net force up along the incline, and the ball would start rolling upwards. This also doesn't actually happen, so that also can't be true.

The true situation must be that the static friction is smaller than the weight component.

Why are there so much complexity in the article? And why is it correct?

To solve it we must ask ourselves what is required for rolling without slipping. From the above we just know that the static friction must be smaller than the weight component - that is not specific enough. So, what else can we use?

Another indicator we can use is that the contact point is stationary! The part of the periphery that touches the ground does not move. (This is how we know that there must be static friction rather than kinetic friction).

• While the ball rolls down with a centre-of-mass acceleration $a_{cm}$, the rotation must correspond to a "counter-acceleration" of the periphery, so to speak; a "counter-acceleration" of the periphery so that the acceleration of the contact point cancels to zero. This happens to be achieved by a periphery acceleration of $a_{cm}$ relative to the centre. Then the top point accelerates with $a_{cm}$ forwards and the bottom point with $a_{cm}$ backwards relative to the centre. Relative to the ground, the top point then accelerates with $a_{cm}+a_{cm}=2 a_{cm}$ and the bottom point - the contact point - with $a_{cm}-a_{cm}=0$.

Perfect. Now we have a clear indicator that only is the case when we have rolling without slipping. Since the periphery follows the relation $a=\alpha r$, then we can use: $$a_{cm}=\alpha r$$ as out indicator. When we plug that into some formulas somewhere, then those formulas only hold true for rolling without slipping.

Now we only need some formulas. Since we are looking for $\mu_s$, then I would set up some Newton's laws, either linearly or rotation-wise, and step by step see if I can isolate $\mu_s$ and plug in the above expression and remove all unknowns.

• Comments are not for extended discussion; this conversation has been moved to chat. – rob May 14 '18 at 6:22

What is the minimal friction force required in order for a ball on an incline to roll without slipping?

As you have suggested in a comment, one way of looking at this is to model the ball as a regular polygon, and to gradually increase the number of sides.

For any polygon the object will slide if $\tan\theta \gt \mu$. This is a standard result found by comparing the component of force down the incline $mg\sin\theta$ with the maximum friction force $F= \mu N$ where $N=mg\cos\theta$ is the normal reaction.

However, as the slope angle $\theta$ is increased the disk could topple before it reaches the condition for sliding if its COG falls outside of its base. Toppling may lead to subsequent tumbling or rolling down the incline. There might also be some sliding depending on dynamical factors, which are difficult to analyse but less significant as the number of sides $n$ gets larger. See Motion of a Hexagonal Pencil on an Inclined Plane.

A square disk will topple if $\theta \gt \frac{\pi}{4}$ provided there is sufficient friction to avoid sliding. So as $\theta$ is increased if $\mu \gt \tan\frac{\pi}{4}=1$ the disk will topple first and subsequently tumble down the incline. Otherwise is will slide before the toppling condition is reached.

A hexagonal disk will topple if $\theta \gt \frac{\pi}{6}$. So as $\theta$ is increased if $\mu \gt \tan\frac{\pi}{6} \approx 0.577$ the disk will topple first and roll. Otherwise it will slide before the toppling condition is reached.

In general, an $n$-sided disk will topple (roll) first if $\mu \gt \tan\frac{\pi}{n}$. For $n=36$ the minimum coefficient of friction is $0.0875$. In the ideal case of a perfect circle ($n\to \infty$) the minimum friction can be vanishingly small.

• So what happens when friction force is indeed equal to the force downwards? – bilanush May 12 '18 at 20:23
• With the n-sided disk, if $\theta$ is gradually increased from 0 then friction will equal force down slope, both increasing together, with nothing happening until either the object slides or it topples. If $\mu \lt \tan\frac{\pi}{n}$ then the friction limit is reached before the toppling condition is met, and the disk will slide. If $\mu \gt \tan\frac{\pi}{n}$ then the disk will topple (roll) before reaching the friction limit. For a perfect circle $\tan\frac{\pi}{n} \to 0$ so even a very small amount of friction is enough to make the disk topple before the friction limit is reached. – sammy gerbil May 12 '18 at 20:52
• I don't understand what you mean by topple. I don't care about other disks. Only a perfect one. Please explain why wouldn't it rotate? – bilanush May 12 '18 at 21:07
• Toppling is the same as rolling or rotating about an edge. ... There are 3 things the n-disk can do as $\theta$ is increased : 1. remain at rest (when friction = force down slope), 2. slide (when friction $\le$ force down slope), 3. topple/roll (when also friction $\le$ force down slope). For an $n$-disk if $n$ is finite the disk is at rest (condition 1) initially. But if $n$ is infinite (perfect circle) then the disk topples (rolls) immediately even for very small angles $\theta$ and small values of $\mu$. It is never at rest, and friction is always less than force down slope. – sammy gerbil May 12 '18 at 21:43
• What does that mean? Isn't the friction is a fixed number each material gets? Why can't I equate the mg to the friction? – bilanush May 12 '18 at 22:03

## protected by Qmechanic♦May 12 '18 at 19:33

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?