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

[Karol]

Homework Statement

[/B]
The ball of mass m and radius r is released from horizontal position, the edge of the bowl.
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

Homework Equations

Radial acceleration: $a_r=\frac{v^2}{r}$

The Attempt at a Solution

Conservation of energy: $mgh=\frac{1}{2}mv^2~~\rightarrow~~mg(R-r)\sin\theta=\frac{1}{2}mv^2$
$$v^2=2g(R-r)\sin\theta$$
$$a_t=\frac{mg\cos\theta}{m}=g\cos\theta$$
$$a_r=\frac{v^2}{r}=\frac{2g(R-r)\sin\theta}{R}$$
In order not to slip the tangential forces of friction and the component of gravity must cancel:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$

 

[haruspex]

What about rotational KE?

 

[Karol]

$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~mg(R-r)\sin\theta=\frac{1}{2}m(k+1)r^2\omega$$
$$\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$
$$a_t=\frac{F_t}{m}=\frac{mg\cos\theta}{m}=g\cos\theta$$
$$a_r=\frac{v^2}{R-r}$$
The condition for minimum μ remains the same:
$$mg\mu=mg\cos\theta~~\rightarrow~~\mu=\cos\theta$$

 

[haruspex]

$$a_t=\frac{F_t}{m}=\frac{mg\cos\theta}{m}=g\cos\theta$$

What about friction?

 

[Karol]

At going up and also down the friction acts against gravity's component:
$$F_t-f=mg~~\rightarrow~~mg(\cos\theta-\mu\sin\theta)=ma_t$$

 

[pgardn]

I was also trying this.

OP: what is R?

And I think the other poster is saying you have to set gravitational potential energy = translational ke and rotational ke. For both radial and translational acceleration. We know translational acceleration will be smaller than if the ball was purely sliding.

 

[Karol]

$$M=I_A\alpha~~\rightarrow~~mrg\cos\theta=mr^2(k+1)\alpha$$
$$\alpha=\frac{g\cos\theta}{(k+1)r},~~v=\omega r$$
R is the bowl's radius.
$$a_t=\alpha r=\left( \frac{\cos\theta}{k+1} \right)g,~~a_r=\frac{v^2}{R}$$

 

[pgardn]

$$M=I_A\alpha~~\rightarrow~~mrg\cos\theta=mr^2(k+1)\alpha$$
$$\alpha=\frac{g\cos\theta}{(k+1)r},~~v=\omega r$$
R is the bowl's radius.
$$a_t=\alpha r=\left( \frac{\cos\theta}{k+1} \right)g,~~a_r=\frac{v^2}{R}$$

Can't you just set mgh = both types of kinetic energy and solve for v?

This will then give you radial acceleration. v^2/R

Then use kinematics to figure out tangential acceleration

 

[Karol]

$$EK_{(rot)}=\frac{1}{2}I_c\omega^2=\frac{1}{2}I_c\frac{v^2}{r^2}=\frac{1}{2}kmv^2$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2=\frac{1}{2}mv^2(k+1)$$
$$mg(R-r)\sin\theta=\frac{1}{2}mv^2(k+1)~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_r=\frac{v^2}{R-r}=\left( \frac{2\sin\theta}{k+1} \right)g$$
How do i use kinematics to find tangential acceleration? Isn't tangential acceleration only a function of the forces (gravity and friction) acting on the ball?
Kinematics deals with velocities, accelerations and distances, not forces. the tangential acceleration acts on the center, right?
The forces depend on the angle, so i can't use constant acceleration formulas.

 

[pgardn]

$$EK_{(rot)}=\frac{1}{2}I_c\omega^2=\frac{1}{2}I_c\frac{v^2}{r^2}=\frac{1}{2}kmv^2$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2=\frac{1}{2}mv^2(k+1)$$
$$mg(R-r)\sin\theta=\frac{1}{2}mv^2(k+1)~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_r=\frac{v^2}{R-r}=\left( \frac{2\sin\theta}{k+1} \right)g$$
How do i use kinematics to find tangential acceleration? Isn't tangential acceleration only a function of the forces (gravity and friction) acting on the ball?
Kinematics deals with velocities, accelerations and distances, not forces. the tangential acceleration acts on the center, right?
The forces depend on the angle, so i can't use constant acceleration formulas.

So to start with your diagram does not have R as a given, only h. Hopefully you were given R as it would make finding centripetal acceleration a lot easier. I also hope the question meant to say a ball of mass m WITH a radius of r? I am having some R, r confusion.

What is your answer for v at the h vertical level? All you need is conservation of energy, no forces or angles required IMO. This v is really just speed in variable form. You can convert omega to v. (omega x r = v) The parenthetical is a key part of your mess up I believe.

And I think if you do know v at h, tangential acceleration is easy at h below the starting vertical level. ( I am assuming the ball starts from rest) You just use the instantaneous speed you found using conservation of energy kinematically. Again, no forces or angles required. Now the last part is going to require forces.

 

[haruspex]

All you need is conservation of energy, no forces or angles required I

The question asks for accelerations, though. I suppose you could use a = v dv/dx, but the direct approach using forces and moments looks easier to me.

 

[ehild]

What is your answer for v at the h vertical level? All you need is conservation of energy, no forces or angles required IMO. This v is really just speed in variable form. You can convert omega to v. (omega x r = v)

The ball starts from the edge of the bowl, with zero speed. What is the normal force and friction at the uppermost position of the ball? Can the ball roll from the beginning? What makes the ball roll ?
If it is not pure rolling, the ball skids at the beginning part of its track, and you can not apply conservation of energy.

 

[Karol]

From post #9:
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{1}{2}mv^2+\frac{1}{2}kmv^2~~\rightarrow~~v^2=\frac{2(R-r)\sin\theta}{k+1}$$
$$a_t=\dot{v}=\frac{(R-r)\cos\theta}{(k+1)\sqrt{\frac{2(R-r)}{k+1}\sin\theta}}$$

but the direct approach using forces and moments looks easier to me.

Do you mean what i wrote in post #7:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int \alpha$$
For that i need $\theta(t)=?$

Can the ball roll from the beginning? What makes the ball roll ?

The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll.

 

[ehild]

The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll.

Then it slides at some part of its track. Can you derive the speed from conservation of energy?

 

[haruspex]

For that i need θ(t)=?θ(t)=?\theta(t)=?

If work is conserved (see below) you can get the velocities in terms of theta from that, then use the velocities and moments etc. to find acceleration, again as a function of theta. You are not asked to find anything as a function of time, and indeed that might be much harder.

The normal force at the beginning is zero and the ball cannot roll. the friction makes it roll

If there is no normal force, how can there be friction? Yet you need angular acceleration to match the linear acceleration to roll.

 

[Karol]

Then it slides at some part of its track. Can you derive the speed from conservation of energy?

But immediately after the beginning there is normal force and it starts rolling, so what i did is correct:
$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$

 

[haruspex]

But immediately after the beginning there is normal force and it starts rolling, so what i did is correct:
$$mgh=\frac{1}{2}I_A\omega^2~~\rightarrow~~\omega=\frac{2g(R-r)\sin\theta}{(k+1)r^2},~~v=\omega r$$

There is linear acceleration right from the start, so for rolling contact there should be rotational acceleration from the start. But since there is no normal force then there is no frictional torque to provide it. In order to make the later transition to rolling contact, it will then have to go through a sliding phase, in which kinetic friction brings it up to rotational speed.

 

[Karol]

Then it slides at some part of its track. Can you derive the speed from conservation of energy?

Sliding makes things worse since i have to add friction losses which i don't know:
$$mgh=\frac{1}{2}I_A\omega^2+E_f$$

 

[pgardn]

The ball starts from the edge of the bowl, with zero speed. What is the normal force and friction at the uppermost position of the ball? Can the ball roll from the beginning? What makes the ball roll ?
If it is not pure rolling, the ball skids at the beginning part of its track, and you can not apply conservation of energy.

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.

 

[pgardn]

There is linear acceleration right from the start, so for rolling contact there should be rotational acceleration from the start. But since there is no normal force then there is no frictional torque to provide it. In order to make the later transition to rolling contact, it will then have to go through a sliding phase, in which kinetic friction brings it up to rotational speed.

Thanks for this addition also.

Sorry Karol.

 

[Karol]

To find where pure rolling starts:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int_0^{\theta} \alpha=\left( -\frac{g\mu}{kr^2} \right)\cos\theta$$
$$F_t-f=ma_t:~~mg\cos\theta-mg\mu\sin\theta=ma_t$$
$$\rightarrow~~v=\int\frac{F_t}{m}=g\int_0^{\theta}(\cos\theta-\mu\sin\theta)=-g(\sin\theta+\mu\cos\theta)$$
The dimensions aren't correct, and what does the minus sign mean in ω and v?

 

[haruspex]

Then it slides at some part of its track. Can you derive the speed from conservation of energy?

@ehild, can you see a way to solve the first part? Eliminating time, I get an equation of the form $yy'+ky^2=A\cos(x)-B\sin(x)$. The equation with time has $\theta, \dot\theta, \ddot\theta$ in it, so there seems to be no way to find the accelerations without solving it.

 

[haruspex]

To find where pure rolling starts:
$$M=I_A\alpha~~\rightarrow~~\alpha=\frac{g\cos\theta}{(k+1)r},~~\omega=\int_0^{\theta} \alpha=\left( -\frac{g\mu}{kr^2} \right)\cos\theta$$
$$F_t-f=ma_t:~~mg\cos\theta-mg\mu\sin\theta=ma_t$$
$$\rightarrow~~v=\int\frac{F_t}{m}=g\int_0^{\theta}(\cos\theta-\mu\sin\theta)=-g(\sin\theta+\mu\cos\theta)$$
The dimensions aren't correct, and what does the minus sign mean in ω and v?

I don't understand the bit after $\int_0^{\theta}\alpha$. How are you integrating that with respect to time without knowing alpha as a function of time?

 

[Karol]

How are you integrating that with respect to time without knowing alpha as a function of time?

I don't know if i integrate with respect to time or θ. i hoped that if α=α(θ) then i need to integrate with respect to θ.
@haruspex, how did you get $yy'+ky^2=A\cos(x)-B\sin(x)$ and what is x, is it the angle θ?

 

[ehild]

@ehild, can you see a way to solve the first part? Eliminating time, I get an equation of the form $yy'+ky^2=A\cos(x)-B\sin(x)$. The equation with time has $\theta, \dot\theta, \ddot\theta$ in it, so there seems to be no way to find the accelerations without solving it.

During the sliding part of the motion, the accelerations a and α can be expressed with the forces:
ma=mgcos(θ) - μN
kmr²α=rμN.
Knowing N(θ), we know the accelerations.
Solving the equation $yy'+ky^2=A\cos(x)-B\sin(x)$ for y=(dθ/dt) in terms of x=theta, we find N=mv² /(R-r) + mg sin(θ).

 

[haruspex]

i hoped that if α=α(θ) then i need to integrate with respect to θ.

$\alpha=\dot\omega$, the derivative wrt time, so to get ω from α it would have to be an integral wrt time.

what is x, is it the angle θ

Yes, and y is ω. y' is dy/dx. But the k I wrote is not your k, just some constant.

 

[haruspex]

y=(dθ/dt)²

I don't think that's quite what you meant, but yes, I see how to solve it now, thanks. (Doh!)

we find N= ...

No, that's where I came from to get my differential equation. To find N we need to find v, so need to find omega...
The solution I get now is $\dot\theta^2=\frac g{(R-r)(1+4\mu_k^2)}[6\mu_k(\cos(\theta)-e^{-2\mu_k\theta})+\sin(\theta)]$
Wow. Can it really be that complicated? Can't find a mistake in my working.

 

[TSny]

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

I get a very similar solution except for the coefficient of sinθ. Where you have sinθ, I get (1-2μ²)sinθ. But I could be mistaken.

 

[haruspex]

I get a very similar solution except for the coefficient of sinθ. Where you have sinθ, I get (1-2μ²)sinθ. But I could be mistaken.

You are right, I dropped a factor.

 

[TSny]

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.

Probably not going to get many to concur with this view.

 

[haruspex]

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.

Probably not going to get many to concur with this view.

The problem with that is that to figure out if it will commence rolling at some theta we need to calculate the acquired rotation to that point, i.e. while sliding. So we are back having to use that rebarbative equation.
Assuming it doesn't slip until then doesn't work. If it hasn't slipped earlier, it is not going to start slipping at theta.

 

[TSny]

The problem with that is that to figure out if it will commence rolling at some theta we need to calculate the acquired rotation to that point, i.e. while sliding. So we are back having to use that rebarbative equation.
Assuming it doesn't slip until then doesn't work. If it hasn't slipped earlier, it is not going to start slipping at theta.

Yes, I see your point.

 

[ehild]

I don't think that's quite what you meant, but yes, I see how to solve it now, thanks. (Doh!)

Yes, I corrected it. I was a bit ahead, as we can make a first order equation for (dθ/dt)².

No, that's where I came from to get my differential equation. To find N we need to find v, so need to find omega...

You mean to find (dθ/dt). But you found it if you solved the de for y.

 

[ehild]

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.

Probably not going to get many to concur with this view.

I thought the same :) No problem makers are correct. And the text of the problem is not quite clear.

 

[haruspex]

You mean to find (dθ/dt). But you found it if you solved the de for y.

OK, I was confused by your writing "solving the equation ... we find ...". You meant, solving the equation, we can plug the velocity into ... and hence find N.

 

[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.