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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.

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closed as off-topic by Pranav Hosangadi, Kyle Kanos, Martin, ACuriousMind, BMS Feb 25 '15 at 19:09

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    $\begingroup$ 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 '15 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. $\endgroup$ – Joshua Benabou Feb 24 '15 at 16:42
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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.

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  • $\begingroup$ Beautiful. Very cool. $\endgroup$ – Joshua Benabou Feb 19 '15 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 '15 at 23:08
  • $\begingroup$ This was actually a multiple choice problem on the 2015 F=MA exam. I found it very interesting and tricky. 5 different graphs were given as answer choices. I narrowed down the answer to two graphs: one is the first graph you got with the drop discontinuity (before you corrected the mistake), and the other is the graph you have now. The main difference between these graphs is that the kinetic energy lost to friction when the ball begins to slip is: $\endgroup$ – Joshua Benabou Feb 20 '15 at 22:09
  • $\begingroup$ in the first graph, a lot of energy is lost to friction very quickly, and then decreases as the ramp becomes steeper. in the second graph the energy lost to friction increases smoothly with the ramp angle increasing, and then decreases smoothly to 0. So, how might we determine conceptually/intuitively which of the two answers is correct? Of course, I did not have time to go through all the math on the test! $\endgroup$ – Joshua Benabou Feb 20 '15 at 22:12
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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 '15 at 22:15
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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.

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  • $\begingroup$ I know it's a typo. $\endgroup$ – Joshua Benabou Feb 18 '15 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 $t$. $\endgroup$ – Joshua Benabou Feb 20 '15 at 18:51
  • $\begingroup$ I have fixed it. The curves make much more sense now. Thanks for spotting the problem. $\endgroup$ – Floris Feb 20 '15 at 20:50

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