Significance of the fact the the final expression for acceleration in rolling motion over a slope is independent of friction

In accelerated rolling motion on an incline , once we simplfy the equations we find that:

$$f_s = \frac{I_{cm}}{r^2} \left(\frac{g \sin \theta}{ 1 + \frac{I_{cm} }{mr^2} }\right) \tag{1}$$

What is strange to me is that here the expression seems completely independent of the normal force applied. Usually we have we speak of friction, we find that the frictional force is proportional to the weight which is applied on the surface and it takes on the form:

$$f_s = \mu N$$

But why is it different for rolling on an incline?

Refer

• I think it does depend on N. It's just that instead of having writing N in the equation, we have m, which makes sense because when m increases Icm increases (assuming r is constant) and a larger frictional force is needed for pure rolling. Mar 8 '21 at 13:27
• You can replace m using the relationship N = mgcos(theta). Mar 8 '21 at 13:29
• @LuoZeyuan It depends on the moment of inertia (which depends on mass) and the force exerted along the slope (which is the component of weight along the slope, which also depends on mass). It just so happens that mass is related to the normal force, but otherwise the normal force is not relevant here. When the question claims that the normal force is relevant, it shows a misunderstanding of static friction, as I explain in my answer. Mar 8 '21 at 13:35

What is strange to me is that here the expression seems completely independent of the normal force applied.

You are correct, the expression does seem to be completely independent of coefficient of friction or of the normal force ($$mgcos\theta$$ in this case). So let's go back to the derivation.

On having another look at the assumptions on the basis of which this formula has been derived, you will notice a seemingly obvious assumption but very important one, the motion of the object is pure rolling.

What this assumption means is simple, the object is not slipping which directly implies that, the translational acceleration of the object is equal to the rotational acceleration of the object. But what is not so simple is that this also means that the maximum limit of the frictional force is such that it encompasses the frictional force sufficient to provide a rotational acceleration equal to the translational acceleration.

This means that the value of limiting friction, $$\mu N$$, is always such that it encompasses the frictional force, $$f_s$$, necessary to make the motion take place.

This also means that if $$f_s=\frac {I_{cm}a}{r^2}$$ is such that its value is greater than $$\mu N$$,the motion will be a combo of rolling and sliding, and you should see a term of $$\mu$$ and $$N$$ in the expression of $$a_{cm}$$.

First you have to understand that friction can increase its value only upto a certain value known as limiting friction. Once you get that, what i mean by a combo of rolling and sliding is that even when the frictional forces are applied to its maximum value, it will not be able to provide with sufficient rotational acceleration to be equal to translational acceleration. What you get in this case is a motion where the object is accelerating in translational motion faster than it is able to roll down. This means that the object will have covered a horizontal distance of $$2\pi r$$ even before a complete revolution has been made.

This motion will continue on until some other external force is applied to change its motion. It should be noted that, in this type of motion, we will have to constantly provide a source of external energy to make up for the losses made by friction. This is different from pure rolling as in pure rolling, there are no losses due to friction. For the motion of an object rolling down the ramp, this source of external energy will be the gravitational potential energy of the object.

• " But what is not so simple is that this also means that the maximum limit of the frictional force is such that it encompasses the frictional force sufficient to provide a rotational acceleration equal to the translational acceleration." This part is not very clear to me, could you rephrase it? Mar 8 '21 at 14:41
• "he motion will be a combo of rolling and sliding" How can you be both rolling and sliding at the same time? You are either rolling (linear variable= angular) or sliding (linear variable> angular) if I am not mistaken Mar 8 '21 at 14:42
• So, ultimately what happens? if it will lower the rotational acceleration but then again because of gravity (or some other force) it maybe that the acceleration of body increases beyond which friction can handle, so this process you said will repeat? Can you give a reference for this? I think this idea came in some problems I did but I haven't read in book Mar 8 '21 at 18:31
• I deleted my comments as i saw i made a blunder when trying to explain my point. I will edit my answer to better explain my point Mar 9 '21 at 3:04
• The blunder I made in my comment is that friction will not decrease the value of translational acceleration but it will chip away the object's energy. the motion will be a combo of rolling and sliding until we tamper with the external forces applied in the object or the object comes to a complete stop due to the loss of all its kinetic energy which is not a problem in rolling down a ramp as in that motion, gravitational potential energy of the object keeps on changing into kinetic energy, thus providing a good source of kinetic energy. Mar 9 '21 at 3:32

I think that as long as $$f_s \le \mu N$$

there is no sliding and therefore no energy in the system is lost to heat. Under this condition friction can be ignored in the conservation dynamics of the description.

The first example in this link describes the situation in some detail: rolling motion

Usually we have we speak of friction, we find that the frictional force is proportional to the weight which is applied on the surface and it takes on the form:

$$f_s = \mu N$$

There are two big problems with this statement:

1. Friction is proportional to the normal force, which may not be equal to the weight of any object, especially when a slope is involved.
2. Only with kinetic friction is the actual force proportional to the normal force. With static friction, the maximum value of the force is proportional to the normal force. The actual force can take on any value up to this maximum, as required to prevent slipping.

The other answers seem to be focusing on the fact that this situation only involves static friction (independent of normal force), as opposed to other situations that involve kinetic friction (proportional to normal force). This is a good argument, but contradicted by the question (which uses the subscript ‘s’ for both situations), so let us emphasise something here. Where the question says:

$$f_s = \mu N$$

It should say either of the following:

\begin{align} f_s &\leq \mu N \\ f_k &= \mu N \end{align}

Actually, the coefficients are normally different, so we should distinguish them:

\begin{align} f_s &\leq \mu_s N \\ f_k &= \mu_k N \end{align}

• Ah that was a big blunder on my part, I've added the case which I'm talking about explicitly now Mar 8 '21 at 13:17
• Here there is no expression for normal for directly coming, though the angle which controls friction here is also controlling normal force but the angle is controlled by the incline ... soo Mar 8 '21 at 13:21

What is strange to me is that here the expression seems completely independent of the normal force applied. if you look at the free body diagram the normal force N and the constraint force $$~f_s~$$ are perpendicular.

the equations to obtain the constraint force are:

$$m\ddot s=-f_s+m\,g\,\sin(\theta)\\ I_{cm}\,\alpha=f_s\,r\\ \ddot s=r\,\alpha\tag 1$$

from here you obtain your result

$$f_s={\frac {I_{{{\it cm}}}\sin \left( \theta \right) mg}{m{r}^{2}+I_{{{ \it cm}}}}}$$

the normal force is $$~N=m\,g\,\cos(\theta)~$$ is not involved in equation (1).

that the friction force proportional to the weight is just a Ansatz in this case you don't have any more the rolling condition, but instead these equations:

$$m\ddot s=-f_s+m\,g\,\sin(\theta)\\ I_{cm}\,\alpha=f_s\,r\\ f_s=\mu\,N=m\,g\,\cos(\theta)$$

for

$$\mu=\frac{I_{cm}}{I_{cm}+m\,r^2}\,\tan(\theta)$$

you obtain the rolling condition $$~s=r\,\phi$$