Rolling race where objects roll with slipping

Question

One of the interesting demonstrations of Moment of Inertia includes the "Rolling Race" where objects of same mass and radii but having different Moments of Inertia, are allowed to roll down an incline without slipping, and seeing which one crosses the finish line first. We know that, the object with least moment of inertia wins the race.

The following question is from the book "Concepts of Physics" by Dr.H.C.Verma, from the chapter "Rotational Mechanics", which considers the case where objects roll with slipping:

^{Page 194, Objective I, Question 24}

A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by

(a) the solid sphere

(b) the hollow sphere

(c) the disc

(d) all will take same time

I approached the problem as described below:

If it were rolling without slipping in the above question, then the winner would have been the solid sphere (as its Moment of Inertia is $\frac 2 5 mr^2$ whereas for hollow sphere and disc it's $\frac 2 3 mr^2$ and $\frac 1 2 mr^2$ respectively). If it were entirely slipping i.e., if no friction were present then we don't need to worry about rolling or moment of inertia, and all objects reach the bottom at the same time (no one wins).

The above question is an intermediate case where objects roll as well as slip due to insufficient friction. So, I concluded that the solid sphere wins the race (but the margin of winning will be less when compared to rolling without slipping), but the answer is given as - all will reach the bottom at the same time. The outcome is similar to the case of "entirely slipping and no rolling".

The following are my doubts regarding this:

• Why is this approach leading to the incorrect answer even though it seems reasonable? Is this an incorrect method?

• Why must the outcome be similar to the case where no friction is present? Why should it be biased to one of the extreme cases?

In many sources which I read so far, only the case where the object rolls without slipping is being discussed. If possible, kindly provide useful links for further reading regarding rolling race with slipping, as I could not find one. Kindly clarify my above doubts.

Thank you in advance.

Please Note: Even though this question is based on an exercise problem, I don't think this is off-topic. I am asking about the concept of rolling with slipping which is not covered in most of the sources. So, I believe this question will be helpful for a broader audience. Further, I have showed my own effort in solving this problem. For your kind information, I asked this question after reading this meta page - How do I ask homework questions on Physics Stack Exchange?.

If you still feel this question must be closed, kindly state the reason in the comments, so that I could understand your homework policies and avoid such circumstances in future.

Answer 1

$$\underline{\text{The Case of Pure Rolling}}$$ Let's understand this case first. What is the frictional force acting on an object (let's consider a disc of radius $R$ and mass $M$) that's pure-rolling down an incline (inclination angle $\theta$)?

Since the point of contact is instantaneously at rest (no slipping), the frictional force acting on the object is static in nature ($f_s \leq \mu_s N$). It is given by $(1)$. $$f_s = \frac{mg\sin\theta}{1+\frac{mR^2}{I_{CM}}} \tag{1}$$ $$ \text{Requirement for pure rolling : }\mu_s \geq \frac{f_s}{N}=\frac{\tan \theta}{1+\frac{mR^2}{I_{CM}}}=\mu_c \tag{2}$$

The friction on the surface tries to ensure no slipping when the disc is released from rest at the top of the incline. And it is successful only if $(2)$ is satisfied. But what if $(2)$ is not satisfied?

$$\underline{\text{The Case of Rolling with Slipping}}$$ If $(2)$ is not satisfied, the point of contact will start to slip and therefore, the frictional force acting on the disc will not be static but will instead be kinetic. $$f_k = \mu_k N \tag{3}$$ $$mg\sin \theta -f_k = ma \Rightarrow a = g\sin \theta - \mu_kg \cos \theta\tag{4}$$

Now, to get to your problem, the question states that the friction coefficient ($\mu_s \approx \mu_k$) between the objects and the incline is the same for all the three objects and that it is not sufficient for pure rolling (meaning $(2)$ is not satisfied for each of the three objects). Therefore, it is $(4)$ that gives the acceleration of the three objects and since $\mu_k$ is the same for all three, they roll down with the same acceleration and reach the bottom at the same time.

$$\textbf{EDIT}$$

Why would the same result not hold for pure rolling?

For the case of pure rolling, the static frictional force acting on the object depends on the $I_{CM}$ of the object [$(1)$] : the frictional force will not be the same for the three objects. Which, in turn, implies that the objects will experience different accelerations [$(5)$] (and hence, they don't reach the bottom at the same time). This is the reason why the object with the least moment of inertia wins the race : because the frictional force acting on that object is the least. $$a=g\sin \theta \left( \frac{mR^2}{I_{CM} + mR^2} \right) \tag{5}$$

And why not a case where the solid sphere wins the race but the margin of winning is less than the case of pure rolling - occurs?

It is important to keep in mind that there are three $\mu_s$'s (call them $\mu_1, \mu_2, \mu_3$) involved here : one for each object. Let's call the critical value of $\mu$ below which the object cannot pure-roll down the incline, $\mu_c$ [See $(2)$] (Again, there is one for each object : $\mu_{c1}, \mu_{c2}, \mu_{c3}$).

For the case of pure rolling of all three, the following must be satisfied. $$\mu_1 > \mu_{c1} \;|\; \mu_2 > \mu_{c2} \;|\; \mu_3 > \mu_{c3} \tag{6}$$

Let's say we're modifying the surface so as to bring down the value of $\mu_1$, $\mu_2$ and $\mu_3$ (by smoothening the surface perhaps?). As long as the final $\mu_1$, $\mu_2$ and $\mu_3$ (which now each have lower values than what they had previously) satisfy $(6)$ by remaining above the critical values, there is absolutely no difference in the motion of the objects : the time difference between the arrival of the objects at the bottom of the incline doesn't shrink as I lower the values of $\mu$ as long as $(6)$ is satisfied. This is because the frictional force acting on the objects in pure rolling, do not depend on $\mu_s$ [See $(1)$].

Your intuition (as many people's intuition would be as well) about a continuous decrease in the difference of arrival times with the continuous decrease of $\mu$ lead you astray. If the frictional force behaved in some other way (obeyed some other empirical result, I mean), then maybe your intuition would correspond to reality.

But, yes, to hit the nail one last time : There is a discrete behavior shift of the frictional force in the problem. If $\mu < \mu_c$, the frictional force is given by $(3)$ and if $\mu > \mu_c$, the frictional force is given by $(1)$.

why is the final outcome biased to that of one of the extreme cases i.e., all objects are sliding down a frictionless surface?

Since each object experiences the same frictional force in the rolling with slipping case, their accelerations are the same and they reach the bottom at the same time. It is not exactly the same as the case of a frictionless surface : in which case the objects don't rotate and also reach the bottom more quickly as compared to the case of "rolling+slipping" since there's no opposing force.