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.

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    $\begingroup$ I'm going to have to think about this some more. My initial thought is that the kinetic friction energy loss associated with this intermediate (some rolling some slipping) condition is somehow different than that associated with pure sliding. It would seem to be the same in the case of pure slipping, but I'm not sure about the intermediate condition. In order for the end result to be the same (all reaching at same time), it almost seems the friction loss would have to be greater when the moment of inertia is less. But on this point, I'm not sure. $\endgroup$
    – Bob D
    Nov 23, 2019 at 15:40
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    $\begingroup$ Based on my analysis of a free body diagram, if the objects do roll then the book answer can’t be correct. In other words, if the objects roll as well as slides, then the results of the race should be the same as if there is pure rolling of the objects- solid sphere wins. But it also seems to me you can't have rolling and slipping simultaneously, while one can follow the other (rolling following slipping). $\endgroup$
    – Bob D
    Nov 23, 2019 at 21:35
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    $\begingroup$ I'm thinking about a bowling ball thrown on a horizontal lane. It can start out purely sliding and then if it slows down enough due to friction loss, eventually as some point start rolling when static friction takes over. But it is also possible if thrown hard enough to slide all the way down the lane without rolling. $\endgroup$
    – Bob D
    Nov 23, 2019 at 21:35
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    $\begingroup$ In the case at hand, it seems to me the only way they can all reach the bottom at the same time is if they all start out sliding and continue to slide to the bottom without rolling. But I can’t prove how this can happen, so I am hesitant about posting this as an answer. Perhaps others seeing this comment can shed some light. $\endgroup$
    – Bob D
    Nov 23, 2019 at 21:35
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    $\begingroup$ “as the initial potential energy is totally converted to translational kinetic energy “. Not so. Some is lost as heat due to friction work, though it’s the same loss for all the objects of same mass. On simultaneous slipping and rolling I’ll see if I can find the analysis I saw some time ago $\endgroup$
    – Bob D
    Nov 24, 2019 at 4:56

1 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.


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.

  • $\begingroup$ +1: Thank you for your answer. I understood your answer completely. But, it would be great if you could explain why the final outcome is biased to that of one of the extreme cases i.e., all objects are sliding down a frictionless surface? On the lines of mathematics, I agree with your answer. But, why should I not expect the same result to hold good for the other extreme case (when pure rolling without slipping is happening)? 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? $\endgroup$
    – Vishnu
    Nov 24, 2019 at 11:55
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    $\begingroup$ @Intellex Hi, check my edit. $\endgroup$
    – Ajay Mohan
    Nov 24, 2019 at 12:57
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    $\begingroup$ beautiful answer i dont know why people downvoted it, your answer helped me and my entire class thankyou $\endgroup$ Jul 27, 2021 at 15:34

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