We all are familiar with the classic ball rolling down the incline exercise in rotational dynamics. Here is quite a tricky conceptual problem:

You have an incline of fixed height, but the angle of inclination may vary. Consider the total kinetic energy $K$ of the ball at the bottom of the incline. Describe the graph of $K$ vs. the angle of inclination $\theta$. We can assume for simplicity that the static and kinetic friction coefficient are the same.

Here are some conceptual observations. Now, for $\theta < \theta_s$ where $\theta_s$ is the minimum angle at which the ball slips, friction does not do any work on the ball (rolling friction), so $K=mgh$ (the graph is a straight line) But for $\theta > \theta_s$, the ball both slips and rolls, so some of the kinetic energy is lost to slipping. Thus we should the graph to decrease. As $\theta$ increases, the friction force decreases (it is proportional to $\cos(\theta)$, so we should expect the graph to increase after some point. For $\theta=90$, we are back to $K=mgh$. I also suspect that we have some quadratic-like behavior for $\theta_s<\theta<90$, but I don't know exactly how to quantify the behavior of the ball in this region as it is both slipping and rolling, which makes things somewhat complicated.

One might naively say that the energy lost due to slipping is $fd$ where $f$ is the friction force and $d$ is the distance along the incline which the ball travels. However I believe this is not the case, as the effective distance over which friction acts, call it $d_{eff}$ is less than $d$, and depends on the relationship between the angular velocity and the translational velocity of the ball.

Note really that this problem can be solved if one has a clear understanding of the mechanics for rolling and slipping scanrios, so it may be helpful to say a few things about this.

  • 3
    $\begingroup$ I don't see a real question here (there isn't even a question mark in the post), just a prompt to solve a problem. This is not conceptual, it's homework-like and thus off-topic. You admitting that you're just too lazy doesn't make me any more sympathetic, either. $\endgroup$
    – ACuriousMind
    Feb 24, 2015 at 14:15
  • $\begingroup$ I can't tell if you're serious. If you didn't notice I posted my approach which works. Its not a homework type problem its a problem which can be solved mathematical but I was asking for some physical intuition or conceptual descriptions. I shared my thoughts on the problem in the op. Its a fantastic problem and I had some good ideas on how to solve it, but was unsure about the details of rolling and slipping simultaneously. Floris thought it was a challenge as well. Its not off topic. $\endgroup$
    – math_lover
    Feb 24, 2015 at 16:42

2 Answers 2


The dynamics of a ball rolling down an incline is interesting. Let's start by figuring out the forces that come into play for the non-slipping case (mass m, radius R, angle of ramp $\theta$):

enter image description here

If we consider the motion of the ball as a rotation about point $P$, then the torque is given by

$$\Gamma = mgR\sin\theta$$

and the moment of inertia about $P$ is the moment of inertia about $C$ plus $mR^2$ (from the parallel axes theorem). Since $I=\frac25 mR^2$ for a sphere, that means that the moment of inertia about P is

$$I_P = \frac75 mR^2$$

The angular acceleration, $\dot{\omega}$ is

$$\dot{\omega} = \frac{\Gamma}{I_P} \\ = \frac{mgR\sin\theta}{\frac75 mR^2}\\ = \frac57 \frac{g\sin\theta}{R}$$

We can now compute the response force $f_f$ along the surface, since the torque that appears about the center $C$ should give the same acceleration:

$$f_f\ R=I_C\ \dot\omega = \left(\frac25 mR^2\right)\left( \frac57 \frac{g\sin\theta}{R}\right)\\ f_f = \frac27 m g \sin \theta$$

Checking for consistency, the linear acceleration of the center of mass is given by the net force, so

$$\begin{align} m a &= f_a - f_f \\ &= mg \sin \theta - \frac27 m g \sin \theta \\ &= \frac57 mg \sin\theta\\ a &= \frac57 g \sin \theta \end{align}$$

Of course without slipping, we know that $\dot\omega R = a$, and indeed this expression for $a$ agrees with the earlier one for $\dot\omega$.

Now we add sliding motion. Clearly, the sphere will slide when $f_f > \mu f_n$, which means

$$\frac27 mg \sin \theta > \mu m g \cos \theta\\ \mu < \frac27 \tan \theta$$

Note that this is much lower than the usual condition for sliding when there is no rolling.

If the force of friction is less than the $f_f$ needed to maintain rolling contact, we know it is constant at

$$f_f = \mu m g \cos \theta$$

We can now compute the acceleration of the ball down the slope:

$$\begin{align} a &= \frac{f_a - f_f}{m}\\ &= g \left(\sin \theta - \mu \cos \theta\right) \end{align}$$

The distance $d$ from top to bottom, given a constant height $h$, is

$$d = \frac{h}{\sin \theta}$$

so the time taken is

$$\begin{align} t &= \sqrt{\frac{2d}{a}}\\ &=\sqrt{\frac{2h}{g \sin\theta (\sin\theta - \mu\cos\theta)}} \end{align}$$

and at that point the velocity is $$\begin{align} v &= at\\ &=\sqrt{2ad}\\ &=\sqrt{\frac{2g \left(\sin \theta - \mu \cos \theta\right)h}{\sin\theta}} \end{align}$$

And the kinetic energy is

$$\begin{align}E &= \frac12 m v^2 \\ &= m g h \frac{\left(\sin \theta - \mu \cos \theta\right)}{\sin\theta}\\ &= mgh(1-\mu\cot\theta) \end{align}$$

The rolling kinetic energy is given by the rotational velocity of the ball. With a constant torque $\Gamma$ and time $t$, the energy is

$$\begin{align} E &= \frac12 I\omega^2\\ &= \frac12 I \left(\frac{\Gamma t}{I}\right)^2\\ &= \frac{\Gamma^2 t^2}{2I}\\ &= \frac{f_f^2 R^2}{\frac45 m R^2} \frac{2h}{g \sin\theta (\sin\theta - \mu\cos\theta)}\\ &= \frac{\mu^2 m^2 g^2 \cos^2\theta R^2}{\frac45 m R^2} \frac{2h}{g \sin\theta (\sin\theta - \mu\cos\theta)}\\ &= \frac{5 \mu^2 m g h\cos^2\theta}{2 \sin\theta (\sin\theta - \mu\cos\theta)} \end{align}$$

Plotting these for a couple of values of $\mu$, you get the following (note - this is updated - there was a factor 2 missing in my expression for $t$):

enter image description here

When the sphere starts slipping, you lose energy. As the ramp angle increases, the degree of slip becomes greater and so more energy is lost in heat. As the ramp becomes steeper still, the energy dissipated will become less, until there is none when the ramp is vertical.

  • $\begingroup$ Beautiful. Very cool. $\endgroup$
    – math_lover
    Feb 19, 2015 at 23:06
  • $\begingroup$ I thought it was an awesome question - surprised I was the only one who thought so. This cleared up a few issues for me - which is why it took me a bit of time to answer. $\endgroup$
    – Floris
    Feb 19, 2015 at 23:08
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    $\begingroup$ I think that when the ball is barely slipping, the relative motion between ball and slope is small, and so the amount of work done (the relative displacement between the two surfaces) is only very small. That's the intuitive explanation. I admit that when I posted the wrong answer, I was very annoyed there seemed to be a factor 2 missing - at the moment it starts to slip I was expecting a gradual increase in losses, not a sudden jump. But the math seemed to say otherwise. I think this is now correct - both intuitively and mathematically. Thanks for the challenge! $\endgroup$
    – Floris
    Feb 20, 2015 at 22:15
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    $\begingroup$ @Jon I am pretty sure “less” is the right word. The ball is accelerating, so there needs to be a certain amount of torque to increase the angular momentum in lockstep with the linear momentum. If there is not enough force of friction to supply the torque, the ball’s rotation will not keep up with the linear motion and it slips. And then there is constant force of friction. $\endgroup$
    – Floris
    May 14, 2021 at 21:11
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    $\begingroup$ @Jon the lateral force between the ball and the surface will indeed have a different form for the non-slipping ball; as you point out, there is no loss of energy in that case. I hesitate to give that force the name ‘friction’ because i associate that word with a lossy process. There is such a thing as rolling friction, but that’s not what is in play here. Instead it’s a static-friction-like force that ensures there is no slip at the contact point, and that the ball’s angular acceleration matches the linear acceleration. This should not change the graph. $\endgroup$
    – Floris
    May 14, 2021 at 21:50

Here is the approach:

Rolling and slipping simultaneously is not much different from rolling without slipping. All the equations are the same except you can't use $a=R\alpha$. Here is the method for solving the problem:

  1. Friction: $f=\mu mg\cos(\theta)$

  2. Torque to find angular acceleration: $\tau=I\alpha=fR$

  3. Linear acceleration: $ma=mg\sin(\theta)-f$

  4. Final linear speed: $v_{cm}^2=2ad$ where $d=H/\sin(\theta)$.

  5. Final angular speed: $w=\alpha t$ where $t$ is the time it takes to roll to the bottom. Note that we can't use $w^2=2\Delta\theta\alpha$ because we don't know $\Delta\theta$, because the ball is rolling AND slipping.

  6. Find the time $t$: $d=0.5at^2$.

  7. Total kinetic energy: simply use $K_{rot}=0.5Iw^2$ and $K_{tr}=0.5mv^2$.

The final formula is nasty but plugging into Wolfram Alpha we can visualize the graph.

  • $\begingroup$ I know it's a typo. $\endgroup$
    – math_lover
    Feb 18, 2015 at 19:28
  • $\begingroup$ @Floris: there is a mistake in your calculation. You are missing a 2 in the numerator of the radicand for the time $t$. $\endgroup$
    – math_lover
    Feb 20, 2015 at 18:51
  • $\begingroup$ I have fixed it. The curves make much more sense now. Thanks for spotting the problem. $\endgroup$
    – Floris
    Feb 20, 2015 at 20:50
  • $\begingroup$ The solution by @Floris implicitly assumes a rigid body, and for a rigid body there are no heating effects. The force of friction opposes motion and changes the kinetic energy (translation and rotational) of the body. See physics.stackexchange.com/questions/607942/… $\endgroup$
    – John Darby
    Aug 22, 2021 at 4:11

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