# Minimizing Friction for a Ball in a Rough Bowl: Accelerations and Velocity

• Karol
In summary, the ball of mass m and radius r is released from horizontal position and the tangential and radial accelerations and velocities are calculated. The friction coefficient must be minimal at an angle θ so that the ball won't slip.
$$u=y^2~~\rightarrow~~\dot\theta=\sqrt{u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-3\mu e^{-2\mu x}]}$$

Yes.

TSny said:
Some typos here?
$$F_N=mR\dot\theta^2+mg\sin(\theta),~~T=F_N\mu r$$
$$T=I_c\alpha~~\rightarrow~~\alpha=\left( \frac{R\dot\theta^2+g\sin\theta}{kr^2} \right)\mu$$
TSny said:
You know ##\dot{\theta}(\theta)## and you know ##\alpha(\theta)## explicitly from your post #38 (after correcting the typo)
I have to insert ##u(x)=\dot\theta^2## from post #69 into the above and then use it in:
$$\omega(\theta) = \large \int \frac{\alpha(\theta)}{\dot{\theta}(\theta)} \small d\theta$$

Karol said:
$$\alpha=\left( \frac{R\dot\theta^2+g\sin\theta}{kr^2} \right)\mu$$
I believe this is off by a factor of ##r##.

I have to insert ##u(x)=\dot\theta^2## from post #69 into the above and then use it in:
$$\omega(\theta) = \large \int \frac{\alpha(\theta)}{\dot{\theta}(\theta)} \small d\theta$$
Yes. As you can see, it is a very difficult integral to evaluate.

TSny said:
I believe this is off by a factor of rrr.
$$T=I_c\alpha~~\rightarrow~~\alpha=\left( \frac{R\dot\theta^2+g\sin\theta}{kr} \right)\mu$$

Looks good.

VMP said:
After a bit of algebraic manipulation we find this result which we will use in a bit:
$$mR\dot{\theta}=\frac{10}{7}mg\cdot sin(\theta)$$
But i find:
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_{cm}\dot{\varphi}^2=\frac{1}{2}mR^2\dot\theta^2+\frac{1}{2}\frac{2}{5}mr^2\frac{R^2}{r^2}\dot\theta^2$$
$$mgR\sin\theta=\frac{7}{10}R\dot\theta^2~~\rightarrow~~g\sin\theta=\frac{7}{10}R\dot\theta^2$$

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Correction, just a minor mistake, it should be:
$$mR\dot{\theta}^2=\frac{10}{7}mg\cdot sin(\theta)$$
And i find, at the end
$$\mu=\frac{17}{5}\frac{1}{\tan(\theta)}$$

VMP
@VMP, shouldn't you measure the angle θ from the vertical x axis? maybe we should change x with y directions?

@Karol
It doesn't really matter how you label the axes as long as you know what you're doing.
Thank you for correcting the $mR\dot{\theta}^2$ term,
I'm not sure I made a mistake with the second term you mentioned and my conclusion regarding the functional relation was wrong. Only problem with the $\mu$ statement in post #61 is that it diverges for $\theta=0$. Either way, the result shows what was intended.

I gave the problem some thought and this is what I derived. Firstly, relabel the axes like this:

If you review the initial problem, ball dissipates energy until it reaches some critical angle $\theta_{0}$
And after $t\rightarrow \infty$ the ball will oscillate in such way that at critical angle both kinetic energy of $CoM$ and rotational kinetic energy are equal to $0$.

Now, let's start again.
I'm assuming that this critical angle is invariant to excess velocity and rotational energy because force of friction $\vec{f}$can be written as a function of $\theta$.
Suppose that you begin from the bottom of the cup with some initial velocity $v_{cm}(t=0)$,
some angular speed (about the center of mass) $\dot{\varphi}(t=0)$ and we want the ball to roll without slipping, then $R\dot{\theta}=v_{cm}=r\dot{\varphi}$ as I've wrote before in the older post.

Our goal is to find how $\mu$ is related to $\theta$, we can do this by writing out equations of motion and energy.

$$E=const.=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\dot{\varphi}^2-mgR(1-cos(\theta))\; (1)$$
$$m\ddot{\vec{R}}=-mR\dot{\theta}^2\cdot\hat{r}+mR\ddot{\theta}\cdot\hat{\theta}=(mgcos(\theta)-N)\cdot\hat{r}+(f-mgsin(\theta))\cdot\hat{\theta}\;(2)$$

From equation (1) you should be able to derive that $\ddot{\theta}=\frac{5g}{7R}sin(\theta)$.
From equation (2) $\ddot{\theta}=\frac{g}{R(4\mu^2+1)}(3\mu cos(\theta)+(\mu^2-1)sin(\theta))$.

...solving for $\mu=\frac{3cos(\theta)-\sqrt{9cos^2(\theta)-\frac{624}{49}sin^2(\theta)}}{\frac{26}{7}sin(\theta)}$
This is the minimum value of $\mu$ for critical angle $\theta_{0}$.
Note the graph:

Interesting thing to note is that maximum value of $\theta_{0}\approx40^{\circ}$.

Cheers!

P.S. I didn't bother checking the math, so don't mind if the result is off by a numerical constant.

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@VMP, you wrote:
$$E=const.=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\dot{\varphi}^2-mgR(1-cos(\theta))\; (1)$$
How did you get to:
VMP said:
From equation (1) you should be able to derive that: ##\ddot{\theta}=\frac{5g}{7R}sin(\theta)##
$$\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\dot{\varphi}^2=mgR(1-cos(\theta))\; (1)$$
$$\frac{1}{2}mR^2\dot\theta^2+\frac{1}{2}\frac{7}{5}mr^2\frac{R^2}{r^2}\dot\theta^2=mgR(1-cos(\theta))~~\rightarrow~~\frac{7}{5}R\dot\theta^2=g(1-\cos(\theta))$$
You start from the bottom of the bowl and the ball deaccelerates. why did you take as positive ##+mR\ddot{\theta}\cdot\hat{\theta}## and also f in:
$$m\ddot{\vec{R}}=-mR\dot{\theta}^2\cdot\hat{r}+mR\ddot{\theta}\cdot\hat{\theta}=(mgcos(\theta)-N)\cdot\hat{r}+(f-mgsin(\theta))\cdot\hat{\theta}\;(2)$$
As i understand, if the acceleration is in the direction of decreasing θ it should be negative, and here the ball tries to fall back. isn't the positive direction of ##\hat{\theta}## further from the x axis, towards y, in your sketch counterclockwise?

Karol said:
@VMP
You start from the bottom of the bowl and the ball deaccelerates. why did you take as positive ##+mR\ddot{\theta}\cdot\hat{\theta}## and also f in:
Yea, I just realized i screwed up the differential equation solution, sorry. The energy equation should also contain the plus sign. I'll try to redeem myself when I get the time.

Cheers

@VMP said:
$$E=const.=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\dot{\varphi}^2-mgR(1-cos(\theta))\; (1)$$
From this i derive:
$$(\rm A)~\dot\theta^2=\frac{2g(1-\cos\theta)}{(k+1)R}~~\rightarrow~~(\rm B)~\ddot\theta=\frac{2g(1+\dot\theta\sin(\theta))}{2(k+1)R\dot\theta}$$
From:
$$m\ddot{\vec{R}}=-mR\dot{\theta}^2\cdot\hat{r}+mR\ddot{\theta}\cdot\hat{\theta}=(mgcos(\theta)-N)\cdot\hat{r}+(f-mgsin(\theta))\cdot\hat{\theta}$$
I derive:
$$\left\{ \begin{array}{l} N-mg\cos\theta=mR\dot\theta^2 \\ N\mu-mg\sin\theta=mR\ddot\theta \end{array} \right.$$
I insert B:
$$\mu g\cos\theta+\mu R\dot\theta^2-g\sin\theta=\frac{2g(1+\dot\theta\sin(\theta))}{2(k+1)\dot\theta}$$
It involves ##\dot\theta^3##

Hey, as I've said, I came back to redeem myself. I totally neglected the nature of $\mu$ and thus blundered spectacularly.
The solution to initial question "What should be the minimal friction coefficient at angle θ so that the ball won't slip?" I will mark with (*) for clarity. Now I will focus on understanding of the issue at hand.

Okay, so I will set up the whole problem as it was given in post #1, $\theta$ begging from horizontal, initial velocity and rotation being equal to $0$.

From initial system of equations: $$(1)\left\{\begin{matrix} mR\dot{\theta}^2=N-mg\, sin\theta\\ mR\ddot{\theta}=mg\,cos\theta-f\;\\ f=\mu\,N\end{matrix}\right.$$
One very important thing to notice is that $\mu$ is a function of $\theta$,$$\mu=\left\{\begin{matrix} \mu_s(\theta)\\ \mu_{k}=const.\end{matrix}\right.$$
In interval between $[0,\theta_{0})$, the ball is slipping and $\mu=\mu_{k}$, from which you can find (with a bit of work and wolframalpha ) the following diff. equation:$$\ddot{\theta}+\mu_{k}\,\dot{\theta}^2=\frac{g}{R}(cos\theta-\mu_{k}\,sin\theta)\;,\dot{\theta}(0)=0\;,\theta(0)=0$$
and it's "solution" $$\dot{\theta}^2=\frac{2g}{R(4\mu_{k}^2+1)}(-3\mu_{k}e^{-2\mu_{k}\theta}+sin\theta\,(1-2\mu_{k}^2)+3\mu_{k}\,cos\theta)\;(2)$$ From angle $\theta_{0}$ ball is rolling and thus energy is conserved:$$\frac{1}{2}mR^2\dot{\theta}^2+\frac{1}{5}mR^2\dot\theta^2+mgR(1-sin\theta)=mgR(1-sin\theta_{0})+\frac{1}{2}mR^2\omega_{0}^2+\frac{1}{5}mR^2\omega_{0}^2$$ $$\Rightarrow \dot{\theta}^2=\frac{10g}{7R}(sin\theta-sin\theta_{0})+\omega_{0}^2\;(3)$$
From equation (1) and (3) you can find $\mu_{s}$ for curiosity sake, notice how it depends on angle:$$\mu_{s}(\theta)=\frac{2\,cos\theta}{17\,sin\theta-10\,sin\theta_{0}+\frac{7R}{g}\omega_{0}^2}\;(*)$$
This really depressed me when I saw velocity dependence (which was against my assumption in previous post).
To find $\omega_{0}^2$ you simply need to plug in $\theta_{0}$ in equation (2).

Friction "coefficient" really is the culprit of the problem. An idea came to my mind to take the derivate of (2) and (3) equate them and evaluate them at $\theta_{0}$. However, finding $\mu_{k}$ explicitly is not possible. Maybe you could use Taylor expansion, but playing around with https://www.desmos.com/calculator ensured me that such $\mu_{k}$ really does exist.

Cheers!

VMP said:
From angle ##\theta_{0}## ball is rolling and thus energy is conserved:
$$\frac{1}{2}mR^2\dot{\theta}^2+\frac{1}{5}mR^2\dot\theta^2+mgR(1-sin\theta)=mgR(1-sin\theta_{0})+\frac{1}{2}mR^2\omega_{0}^2+\frac{1}{5}mR^2\omega_{0}^2$$
The second and third R's on the right side of the equation should be r's=the ball's radius, not R=the radius from the center of the bowl to the ball's center

Karol said:
The second and third R's on the right side of the equation should be r's=the ball's radius, not R=the radius from the center of the bowl to the ball's center
No, first term is kinetic energy of CoM, second term is rotational kinetic energy. Recall, if the ball is rolling without slipping, then $v_{cm}=R\dot{\theta}=r\dot{\varphi}$.

Right. but i get a different expression for ##\mu_s## but it's tedious. i understand you differentiated (3) to get ##\ddot\theta=\frac{1}{2}\cos\theta## and inserted it into
$$\ddot{\theta}+\mu_{k}\,\dot{\theta}^2=\frac{g}{R}(cos\theta-\mu_{k}\sin\theta)$$
You renamed ##\mu_k## to ##\mu_s##, right?

Karol said:
Right. but i get a different expression for ##\mu_s## but it's tedious. i understand you differentiated (3) to get ##\ddot\theta=\frac{1}{2}\cos\theta## and inserted it into
$$\ddot{\theta}+\mu_{k}\,\dot{\theta}^2=\frac{g}{R}(cos\theta-\mu_{k}\sin\theta)$$
You renamed ##\mu_k## to ##\mu_s##, right?

Time derivative of equation (3) in post #84 is $2\dot{\theta}\ddot{\theta}=\frac{10g}{7R}cos\theta\dot{\theta}$.
You'll have to be more specific with your last $\mu$ related question.

VMP said:
You'll have to be more specific with your last ##\mu## related question.
I mean in set of equations (1) ##\mu_k## appears, but in eq' (*) ##\mu_s##. i understand we can interchange between them when we deal with non slip.

Karol said:
I mean in set of equations (1) ##\mu_k## appears, but in eq' (*) ##\mu_s##. i understand we can interchange between them when we deal with non slip.
In system of equations marked by (1) $\mu$ is one function which is defined differently on different intervals.

For clarity sake: $$\mu(\theta)=\left\{\begin{matrix} (\,\mu_{k}=const.\,),[0,\theta_{0})\bigcup (\angle,\pi]\;(*)\\ \mu_{s}(\theta),[\theta_{0},\angle]\end{matrix}\right.$$ Also, note that $\mu_{s}^{(max)}\geq \mu_{k}$

(*)$\angle$ is just some arbitrary angle.

Cheers!

after ##\theta_0## the ball reaches the other side of the bowl, to ##\pi-\theta_0##:
$$\mu_{s}(\pi-\theta_0)=\frac{2\,cos(\pi-\theta_0)}{17\,sin(\pi-\theta_0)-10\,sin\theta_{0}+\frac{7R}{g}\omega_{0}^2}=\frac{-2\cos\theta_0}{7\sin\theta_0+\frac{7R}{g}\omega_0^2}<0$$
Also how do you know that for this particular problem, not in general: ##\mu_{s}^{(max)}\geq \mu_{k}##
We don't know ##\mu_k## and not ##\mu_s## simce we don't know ##\theta_0## and ##\omega_0##

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