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

[ehild]

The solution I get now is $\dot\theta^2=\frac g{(R-r)(1+4\mu_k^2)}[6\mu_k(\cos(\theta)-e^{-2\mu_k\theta})+\sin(\theta)]$

And I got $$\frac{2g}{(R-r)(1+4μ^2)} \left( -3μ e^{-2μθ}+3μ \cos(θ)+(1-2μ^2) \sin(θ) \right)$$...

I had an other idea what the problem maker might mean.
Assume pure rolling starts at theta. That means a=rα. It is sure that mv²≤2mg(R-r)sin(θ) at that angle. From the force-acceleration equations, we find f, the force of friction.
$f=\frac{k}{1+k}mg\cos(θ)$
As N=mv²/(R-r)+mgsin(θ), N ≤ 3mg sin(θ), an upper limit for N. As sliding just stops at θ , f=μN, so
$f=\frac{k}{1+k}mg\cos(θ) ≤ μ(3mg\sin(θ))$, yielding a lower limit for μ.

 

[haruspex]

And I got $$\frac{2g}{(R-r)(1+4μ^2)} \left( -3μ e^{-2μθ}+3μ \cos(θ)+(1-2μ^2) \sin(θ) \right)$$...

I had an other idea what the problem maker might mean.
Assume pure rolling starts at theta. That means a=rα. It is sure that mv²≤2mg(R-r)sin(θ) at that angle. From the force-acceleration equations, we find f, the force of friction.
$f=\frac{k}{1+k}mg\cos(θ)$
As N=mv²/(R-r)+mgsin(θ), N ≤ 3mg sin(θ), an upper limit for N. As sliding just stops at θ , f=μN, so
$f=\frac{k}{1+k}mg\cos(θ) ≤ μ(3mg\sin(θ))$, yielding a lower limit for μ.

Neat, but there are two reasons I do not believe this was the intent of the question.
It does not ask for a lower bound, it asks for the minimum value.
More significantly, we would still need to use the DE solution to answer the first part, where it asks for the radial acceleration.

Either this was intended as quite a difficult question or the setter blundered.

 

[Karol]

The normal force N:
$$N=m\frac{v^2}{R-r}+mg\sin\theta=m\left( \frac{v^2}{R-r}+g\sin\theta \right)$$
Torque-angular acceleration:
$$T=I_c\alpha:~~Nr=kmr^2\alpha~~\rightarrow~~\alpha=\frac{1}{rk}\left[ \frac{v^2}{R-r}+g\mu\sin\theta \right]$$
Tangential acceleration (W_t is gravity's tangential component):
$$F_t=ma_t:~~W_t-f=ma_t:~~W_t-N\mu=ma_t:~~mg\cos\theta-m\mu\left( \frac{v^2}{R-r}+g\sin\theta \right)$$
$$\rightarrow~~a_t=(\cos\theta+\mu\sin\theta)g-\frac{v^2}{R-r}\mu$$
If i add the non slip relation $a_t=\alpha r$:
$$\left( 1-\frac{1}{k} \right)\mu g\sin\theta+g\cos\theta=\left( 1+\frac{1}{k} \right)\frac{v^2}{R-r}$$
I feel there should be another connection between v and θ but i don't find it. i hoped to find the point of pure rolling by using $a_t=\alpha r$ but i remained with v and θ again.

 

[TSny]

If i add the non slip relation $a_t=\alpha r$

This is not the condition for onset of rolling without slipping. Think about the classic bowling ball problem where the bowling ball is initially skidding on a horizontal surface. In this case $a_t$ and $\alpha$ are constants during the slipping phase and $a_t$ never equals $\alpha r$ while the ball is slipping. Only after slipping stops do you suddenly get $a_t=\alpha r$ satisfied (trivially because $a_t$ and $\alpha$ are then both zero for the bowling ball).

Torque-angular acceleration:
$$T=I_c\alpha:~~Nr=kmr^2\alpha~~\rightarrow~~\alpha=\frac{1}{rk}\left[ \frac{v^2}{R-r}+g\mu\sin\theta \right]$$

Some typos here? Is N the force that produces the torque? Also, you have a factor of $\mu$ in the last term but not in the first term of the [ ] brackets.

$$\rightarrow~~a_t=(\cos\theta+\mu\sin\theta)g-\frac{v^2}{R-r}\mu$$

How did the $\sin\theta$ term and the $v^2$ term end up with opposite signs?

 

[Karol]

This is not the condition for onset of rolling without slipping

I know but i hoped to find θ at which sliding stops. i hoped to find a value for θ.

 

[pgardn]

Holy cow I never took that into account. The last part of the question now makes a lot more sense. It's a rolling and slipping problem. I should have known from the last question.

Thanks.

Very sorry Karol.

I will just observe.

Well I stepped out and for good reason. Way out of my league..

If I may be so humble to ask...

What course or class is this? I am trying to understand what introductary physics means.

 

[TSny]

$v = \omega r$ must be satisfied at the angle $\theta_0$ where slipping stops. If $v$ and $\omega$ were known as functions of $\theta$ during the slipping phase, then $v = \omega r$ would give the equation for $\theta_0$.

$v$ as a function of $\theta$ is easily found from haruspex's solution for $\dot{\theta}^2$ as written in ehild's post #36. However, finding $\omega$ as a function of $\theta$ seems to lead to an integral that is too messy to do.

$\alpha = \frac{d\omega}{ dt }= \frac{d\omega }{d\theta} \dot{\theta}$

So, $\omega(\theta) = \large \int \frac{\alpha(\theta)}{\dot{\theta}(\theta)} \small d\theta$

You know $\dot{\theta}(\theta)$ and you know $\alpha(\theta)$ explicitly from your post #38 (after correcting the typo). But, as you can see, this integral is too hard to evaluate.

 

[Karol]

@ehild, why friction is $f=\frac{k}{1+k}mg\cos(θ)$? indeed i think it should be f=Nμ
And what is the explanation to $yy'+ky^2=A\cos(x)-B\sin(x)$? how is it derived?

 

[ehild]

@ehild, why friction is $f=\frac{k}{1+k}mg\cos(θ)$? indeed i think it should be f=Nμ

The friction is static during rolling, and it is $f_s=\frac{k}{1+k}mg\cos(θ)$. During sliding, kinetic friction acts, and it is f_k=μN. f_s≤f_k
Edit: assuming the same value of both static and kinetic coefficients of friction.

And what is the explanation to $yy'+ky^2=A\cos(x)-B\sin(x)$? how is it derived?

That was the derivation of Haruspex. Ask him.
I think the writer of the problem did not think it over. I suggest you to drop the problem. It is too difficult as it is written. Are you sure you copied the problem text correctly?

 

[Karol]

But also if it's static friction, isn't it still f=Nμ? the velocity term is $m\frac{v^2}{R-r}$ and it's valid, no?

 

[ehild]

μ

$v = \omega r$ must be satisfied at the angle $\theta_0$ where slipping stops. If $v$ and $\omega$ were known as functions of $\theta$ during the slipping phase, then $v = \omega r$ would give the equation for $\theta_0$.

$v$ as a function of $\theta$ is easily found from haruspex's solution for $\dot{\theta}^2$ as written in ehild's post #36. However, finding $\omega$ as a function of $\theta$ seems to lead to an integral that is too messy to do.

The problem asked the questions:

What are the tangential and radial accelerations at angle θ and what is the velocity.
What should be the minimal friction coefficient at angle θ so that the ball won't slip

During rolling, also a=rα is true. From this, we can derive the force of static friction, f_s, but it is not the the same as f_k. Although f_k ≥ f_s (assuming μ_s=μ_k) so f_s is a lower limit for f_k, we can not say that the minimal coefficient of kinetic friction ensuring pure rolling at θ₀ is equal to f_s/N.

It was not specified which coefficient of friction was the question, so the problem maker might assume (wrong) that the ball rolled from the beginning, as @TSny suggested in Post #30.

 

[haruspex]

Although fk ≥ fs

Backwards.

 

[ehild]

But also if it's static friction, isn't it still f=Nμ? the velocity term is $m\frac{v^2}{R-r}$ and it's valid, no?

No, the static and kinetic frictions are different.
Edit: f(static)≤μ_sN, and it does not need to be the same as f_k even in case μ_s=μ_k.

 

[ehild]

Backwards.

I meant f_s≤μ_sN. And assumed equal static and kinetic coefficients of friction.

 

[haruspex]

what is the explanation to $yy'+ky^2=A\cos(x)-B\sin(x)$? how is it derived?

For simplicity, I write R for your R-r.
$F_N=mR\dot\theta^2+mg\sin(\theta)$
$F_f=\mu F_N$
$mR\ddot \theta=mg\sin(\theta)-F_f$
$R\ddot \theta=g\cos(\theta)-\mu R\dot\theta^2-\mu g\sin(\theta)$
Writing $x=\theta$, $y=\dot \theta$, $\ddot\theta=\frac d{dt}y=\frac {dx}{dt}\frac d{dx}y=y\frac {dy}{dx}=yy'$ (this is a standard trick for eliminating time in acceleration equations):
$yy'+\mu y^2=\frac gR(\cos(x)-\mu\sin(x))$
Note that my formulation using arbitrary constants A, B, k was too general. B=Ak.

 

[Karol]

Can you explain why f_s isn't Nμ_s? what's the explanation for $f_s=\frac{k}{1+k}mg\cos(θ)$, how did you derive it?
Friction, static or kinetic, to my opinion, is the normal force times the coefficient: Nμ, and if μ_s=μ_k then f_s=f_k always.

I don't understand this line: $mR\ddot \theta=mg\sin(\theta)-F_f$
The left side isn't the torque, so it isn't the formula $T=I_c\alpha$
And on the right side we subtract 2 rectangular forces? mgsin(θ) is a radial force and F_f is tangential

 

[ehild]

Can you explain why f_s isn't Nμ_s? what's the explanation for $f_s=\frac{k}{1+k}mg\cos(θ)$, how did you derive it?
Friction, static or kinetic, to my opinion, is the normal force times the coefficient: Nμ, and if μ_s=μ_k then f_s=f_k always.

No. The static force of friction is the force that opposes other forces so as the body in question does not slide. The force of static friction can not exceed μ_sN, but can be anything between zero and μ_sN, in contrast with the kinetic friction, which is definitely F_k=μ_kN.

 

[haruspex]

The left side isn't the torque

Quite so - it is the linear acceleration parallel to the slope. Remember, theta here is for the position of the ball within the bowl, and my R is your R-r.

 

[TSny]

I don't understand this line: $mR\ddot \theta=mg\sin(\theta)-F_f$

There's a "misprint" on the right side where sin should be cos.
But haruspex's next line $R\ddot \theta=g\cos(\theta)-\mu R\dot\theta^2-\mu g\sin(\theta)$ is correct.

 

[haruspex]

There's a "misprint" on the right side where sin should be cos.
But haruspex's next line $R\ddot \theta=g\cos(\theta)-\mu R\dot\theta^2-\mu g\sin(\theta)$ is correct.

Thanks - transcription error.

 

[Hamal_Arietis]

And at $\theta$ we have:
$$-mgsin\theta+N=m\frac{v^2}{R-r}$$
$$-F_r+mgcos\theta=m\gamma.(R-r)$$
And $\mu_{min}$ :
$$F_r≤\mu N$$

 

[Karol]

How to solve $yy'+ky^2=A\cos(x)-B\sin(x)$? where to find sources on the internet or elsewhere

 

[TSny]

How to solve $yy'+ky^2=A\cos(x)-B\sin(x)$? where to find sources on the internet or elsewhere

Let $u = y^2$.

 

[Karol]

I wonder if the problem wants you to assume (unrealistically) that the ball rolls without slipping from the beginning. Then, for an arbitrary value of θ, find the minimum coefficient of static friction that will prevent slipping at that θ. You can then see how $\mu^{\rm min}_s$ diverges as θ → 0.

According to which equation do you find that $\mu^{\rm min}_s$ diverges, still you don't have the equation for μ
Where do i find explanation about the Latex operator \rm that you used in $\mu^{\rm min}_s$? i understand it makes text look nice but i didn't find material about it

 

[TSny]

According to which equation do you find that $\mu^{\rm min}_s$ diverges

If you make the false assumption that the ball rolls without slipping from the start, then you can use conservation of energy to find $v$ as a function of $\theta$, as in your post # 9.

Using this expression for $v(\theta)$ in your first equation in post #38, you can find the normal force as a function of $\theta$, $N(\theta)$.

From $\sum F_t = ma_t$ , $\sum \tau_c = I_c \alpha$, and $a_t = r\alpha$, you can find the friction force as a function of $\theta$, $f(\theta)$.

For an arbitrary $\theta$ you can use $f(\theta)$ and $N(\theta)$ to find the minimum value of the coefficient of friction for that value of $\theta$, $\mu_s^{\rm min}(\theta)$. You get an expression $\mu_s^{\rm min}(\theta)$ that diverges as $\theta \rightarrow 0$.

I was thinking that this might be what they wanted you to do. It shows how $\mu_s^{\rm min}(\theta)$ would need to increase as $\theta$ decreases. And it shows that $\mu_s^{\rm min}(\theta)$ would need to go to infinity to prevent slipping at the beginning. Since that's impossible, it just shows (as expected) that you can't have rolling without slipping from the start.

Where do i find explanation about the Latex operator \rm that you used in $\mu^{\rm min}_s$? i understand it makes text look nice but i didn't find material about it

I can't remember where I first saw \rm for roman text. What little Latex I know I have learned from posting here at PF. Here is a link that lists the command \rm
https://en.wikibooks.org/wiki/LaTeX/Command_Glossary

 

[VMP]

According to which equation do you find that $\mu^{\rm min}_s$ diverges, still you don't have the equation for μ ...

If you assume that the ball is doing pure rolling (which it is not) then we can set up the coordinate system like this:

Since the ball is rolling without skidding, no energy is lost hence:
$$0=\frac{1}{2}mv^2+\frac{1}{2}I_{cm}\dot{\varphi}^2-mgh,$$ it can easily be shown that $$h=sin(\theta)\cdot R,\; I_{cm}=\frac{2}{5}mr^2,\;v_{cm}=R\cdot \dot{\theta}=r\cdot\dot{\varphi}$$ where $\dot{\varphi}$ is angular velocity about center of mass (cm).
After a bit of algebraic manipulation we find this result which we will use in a bit. $$mR\dot{\theta}=\frac{10}{7}mg\cdot sin(\theta)$$
Okay, since we know that position vector $\vec{r}$ has a constant magnitude $\left | \vec{r} \;\right |=R=const.$ then, $$m\ddot{\vec{r}}=-mR\dot{\theta}^2\cdot\hat{r}+mR\ddot{\theta}\cdot\hat{\theta}.$$$$N-mg\cdot sin(\theta)=mR\dot{\theta}^2 \;(1),\;mg\cdot cos(\theta)-\mu N=mR\ddot{\theta}\;(2).$$ After you solve for N in equation (1), place it into the second equation, take the time derivative of equation (1) and replace it with right hand side of equation (2) you will find the following result: $$\frac{2}{17tan(\theta)}=\mu$$
Here is your equation for $\mu$ as you can see it, it is not a coefficient i.e. a "constant function" and it diverges for $\theta=0.$

Regarding the differential equations other fine members mentioned, here is a link for a quick guide:http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

Hope this helped, cheers!

 

[Karol]

Let $u=y^2$

$$yy'+\mu y^2=\frac gR(\cos(x)-\mu\sin(x))$$
$$u=y^2,~~y=\sqrt{u},~~u'=2yy'~~\rightarrow~~y'=\frac{u'}{2y}=\frac{u'}{2\sqrt{u}}$$
$$yy'+\mu y^2=\sqrt{u}=\frac{u'}{2\sqrt{u}}+ku$$
$$\rightarrow~~yy'+\mu y^2=\frac{1}{2}u'+ku$$
$$\frac{1}{2}u'+ku=\frac gR(\cos(x)-\mu\sin(x))$$
I have to solve this differential equation

 

[TSny]

$$\frac{1}{2}u'+ku=\frac gR(\cos(x)-\mu\sin(x))$$
I have to solve this differential equation

Yes, if R is actually R_bowl - r.

 

[haruspex]

I have to solve this differential equation

Tip: just assume there are solutions which are linear combinations of cos(x) and sin(x). Plug that in and determine the coefficients.

Edit: I forgot to point out that this will only give you a particular solution. To this must be added the general solution of the homogeneous equation (setting RHS to zero). Thanks to TSny for adding that.

 

[Karol]

Tip: just assume there are solutions which are linear combinations of cos(x) and sin(x). Plug that in and determine the coefficients.

$$yy'+\mu y^2=A\cos(x)-B\sin(x),~~yy'+\mu y^2=\frac gR(\cos(x)-\mu\sin(x))$$
$$\rightarrow~~\frac{1}{2}u'+\mu u=\frac gR(\cos(x)-\mu\sin(x))$$
I assume $u=C\cdot \sin(x)+D\cdot \cos(x)~~\rightarrow~~u'=C\cdot \cos(x)-D\cdot\sin(x)$
$$\frac{1}{2}(C\cdot \cos (x)-D\cdot\sin (x))+\mu C\cdot\sin(x)+\mu \cdot\cos(x)=A\cos(x)-B\sin(X)$$
$$\left( \frac{C}{2}+\mu D \right)\cos(x)+\left( \mu C-\frac{D}{2} \right)\sin(x)=A\cos(x)-B\sin(X)$$
$$\rightarrow~\frac{C}{2}+\mu D = A,~~\mu C-\frac{D}{2}=-B$$
$$C=2A-2\mu D,~~D=\frac{4\mu A+2B}{4\mu^2+1}$$
$$u=y^2~~\rightarrow~~y=\sqrt{C\cdot\sin(x)+D\cdot\cos(x)}=...$$
But $A=\frac{g}{R},~~B=\frac{\mu g}{R}$:
$$C=\frac{2(1-2\mu^2)g}{(4\mu^2+1)R},~~D=\frac{6\mu g}{(4\mu^2+1)R}$$
Where did the term $e^{-2\mu_k\theta}$ in the solution:
$$\dot\theta^2=\frac g{(R-r)(1+4\mu_k^2)}[6\mu_k(\cos(\theta)-e^{-2\mu_k\theta})+\sin(\theta)]$$
Come from?

 

[TSny]

Your solution u_p(x) = C sinx + D cosx is a "particular" solution of the differential equation, but it is not the most general solution. (Note that this solution does not satisfy the initial condition that the ball starts from rest.)

Suppose you can find a solution U(x) to the differential equation (1/2)U'(x) + μU(x) = 0 where the right hand side has been set to zero.

Show that the function u(x) = u_p(x) + kU(x) satisfies the original equation (1/2)u' + μu = A cosx - B sinx for any value of the constant k.

 

[Karol]

Suppose you can find a solution U(x) to the differential equation (1/2)U'(x) + μU(x) = 0

$$U(x)=\frac{1}{2}e^{-2\mu x},~~u(x)=u_p(x)+kU(x)=\frac{1}{2}u'+\mu u+C\cdot\sin(x)+D\cdot\cos(x)$$
Initial angular velocity $\dot\theta=0$:
$$u(0)=D+k=0~~\rightarrow~~k=-D$$
From post #65:
$$C=\frac{2(1-2\mu^2)g}{(4\mu^2+1)R},~~D=\frac{6\mu g}{(4\mu^2+1)R}$$
$$\rightarrow~u(x)=u_p(x)+kU(x)=C\cdot\sin(x)+D\cdot\cos(x)-D\frac{1}{2}e^{-2\mu x}$$
$$u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-\frac{3}{2}\mu e^{-2\mu x}]$$

 

[TSny]

$$U(x)=\frac{1}{2}e^{-2\mu x}$$

OK, but you don't need to include the factor of 1/2 here.

$$u(x)=u_p(x)+kU(x)=\frac{1}{2}u'+\mu u+C\cdot\sin(x)+D\cdot\cos(x)$$

Couple of typos: The second "=" should not be there. The next to last "+" should be "=".

Initial angular velocity $\dot\theta=0$:
$$u(0)=D+k=0~~\rightarrow~~k=-D$$

If you include the 1/2 factor in U(x), then you would not get k = -D.

$$u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-\frac{3}{2}\mu e^{-2\mu x}]$$

The factor of 3/2 in the last term is not correct. This is due to your including the 1/2 factor in U(x). Note your expression does not satisfy u(0) = 0.

But, otherwise your work looks very good.

 

[Karol]

$$u(x)=e^{-2\mu x},~~u'(x)=-2\mu e^{-2\mu x},~~\frac{1}{2}u'+\mu u=0$$
$$u(x)=u_p(x)+kU(x)=C\cdot\sin(x)+D\cdot\cos(x)+ke^{-2\mu x}$$
$$u(0)=D+k=0~~\rightarrow~~k=-D$$
$$C=\frac{2(1-2\mu^2)g}{(4\mu^2+1)R},~~D=\frac{6\mu g}{(4\mu^2+1)R}$$
$$\rightarrow~u(x)=u_p(x)+kU(x)=C\cdot\sin(x)+D\cdot\cos(x)-De^{-2\mu x}$$
$$u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-3\mu e^{-2\mu x}]$$

 

[TSny]

$$u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-3\mu e^{-2\mu x}]$$

That looks right.