Time taken to slide down a circular path (with friction)

[eddiezhang]

Homework Statement

A point mass (m) is acted on by gravity, the normal force, and friction (μN). It starts at the point A(0,1). It then slides down a circular 'ramp', stopping when it touches the x-axis at B(1,0). It does not leave the surface of the ramp.

Find the general time of descent.

Relevant Equations

Centripetal force, friction etc.

This is for a math report that I'm supposed to write, which means I'm not supposed to use conservation of energy. This makes life much harder... so please bear with me. I am interested to see how you'd solve this purely kinematically though (if it can be solved that way).

Please tell me if this is posted to the wrong place. I asked a similar question in a math forum to no luck (they told me to post on a physics forum, so here I am).

Attached is a pdf with some diagrams and my working.

Thanks so much for your help,
Ed

 

[PeroK]

My first thought is that I'd be surprised if this has a closed form solution. Especially with friction involved.

Conservation of energy would give you the final speed, but not the time.

 

[PeroK]

Your differential equation is almost correct. The mass must cancel completely from the equations. And, by taking a dimensionless unit as radius, your equation has become dimensionally suspect. Better to take the radius as $R$.

 

[PeroK]

PS if you take $\mu =0$ you'll see that you may have got sines and cosines mixed up as well.

 

[erobz]

After you fix it up. You can make it first order-linear if you do a few variable substitutions.

First $\dot \theta \to \omega$, Then $\ddot \theta \to \dot \omega$

The you use anther sub like $u = \omega^2$ and its implications for the derivative $\frac{du}{d \theta}$ via the Chain Rule.

It doesn't get you time, but I think it gets something analytical... which is a step in the right direction? It's at least a worthy exploration in its own right.

EDIT: it might even get you time via $\omega ( \theta) = \frac{d \theta}{dt}$. Its surely going to be a fight(which is what you are looking for).

 

[PeroK]

PS with $\mu =0$ the problem reduces to a pendulum. Without the small angle approximation, the solution involves elliptic integrals.

 

[erobz]

PS with $\mu =0$ the problem reduces to a pendulum. Without the small angle approximation, the solution involves elliptic integrals.

So bad news for $t$? But still plenty of material for exploration in the report.

 

[eddiezhang]

PS if you take $\mu =0$ you'll see that you may have got sines and cosines mixed up as well.

oops that's pretty embarrassing. I take it the correct DE is gcos(φ) - φ'' = μ(φ')^2 +μgsin(φ)?

 

[PeroK]

oops that's pretty embarrassing. I take it the correct DE is gcos(φ) - φ'' = μ(φ')^2 +μgsin(φ)?

The differential equation must be (I'll use the more standard $\theta$ for the angle.
$$R\ddot \theta = g\cos \theta - \frac{F_f}{m} = g\cos \theta - \mu \frac N m$$Where $F_f$ is the frictional resistance and $N$ the magnitude of the normal force. And, we must have:
$$N = mg\sin \theta + mR\dot \theta^2$$That gives you:
$$\ddot \theta = \frac g R(\cos \theta - \mu \sin \theta) - \mu\dot \theta^2$$Which is what you have, given the missing $R$.

 

[eddiezhang]

After you fix it up. You can make it first order-linear if you do a few variable substitutions.

First $\dot \theta \to \omega$, Then $\ddot \theta \to \dot \omega$

The you use anther sub like $u = \omega^2$ and its implications for the derivative $\frac{du}{d \theta}$ via the Chain Rule.

It doesn't get you time, but I think it gets something analytical... which is a step in the right direction? It's at least a worthy exploration in its own right.

EDIT: it might even get you time via $\omega ( \theta) = \frac{d \theta}{dt}$. Its surely going to be a fight(which is what you are looking for).

I think I follow the chain rule part - would I be correct in saying $\frac{du}{d \theta} =2\ddot{\theta}$?

You may need to walk me through the rest of that approach... how exactly do you get it first-order linear? Similarly, if you start me off on the $\omega ( \theta) = \frac{d \theta}{dt}$ thing I'd be more than happy to try and bash out the rest of the algebra myself.

Thanks so much

 

[erobz]

I think I follow the chain rule part - would I be correct in saying $\frac{du}{d \theta} =2\ddot{\theta}$?

You may need to walk me through the rest of that approach... how exactly do you get it first-order linear? Similarly, if you start me off on the $\omega ( \theta) = \frac{d \theta}{dt}$ thing I'd be more than happy to try and bash out the rest of the algebra myself.

Thanks so much

Use chain rule on $\frac{d \omega}{dt}$ to switch to $\theta$ as the independent variable (keep in terms of $\omega$ for the dependent)Once you see the trick you won’t forget it.

 

[eddiezhang]

Use chain rule on $\frac{d \omega}{dt}$ to switch to $\theta$ as the independent variable. Once you see the trick you won’t forget it.

Attempt:

$\frac{d \omega}{dt} = \frac{d \omega}{d\theta} \times \frac{d \theta}{dt} = \omega \frac{d \omega}{d\theta}$

Am I barking up the wrong tree?

I'll probably kick myself hard, but I'm just not seeing it yet...

 

[haruspex]

Attempt:

$\frac{d \omega}{dt} = \frac{d \omega}{d\theta} \times \frac{d \theta}{dt} = \omega \frac{d \omega}{d\theta}$

That’s it.

 

[eddiezhang]

That’s it.

Oh yay!

OK so looking at the DE again:

$\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2$

$\dot{\theta}$ substitutes out directly, but how do I deal with $\ddot{\theta}$? Plus don't the trig terms still make it non-linear?

Edit: I don't have much experience with ODEs - you can probably tell.

 

[haruspex]

, but how do I deal with $\ddot{\theta}$?

Using your formula in post #12.

 

[erobz]

Oh yay!

OK so looking at the DE again:

$\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2$

$\dot{\theta}$ substitutes out directly, but how do I deal with $\ddot{\theta}$? Plus don't the trig terms still make it non-linear?

Edit: I don't have much experience with ODEs - you can probably tell.

It linear ODE by my understanding of classification. The ODE will no longer have nonlinear terms in the dependent variable.

$$ y’ + p(x)y = f(x) $$

Trig functions in $x$ as don’t change the classification to non-linear. They can make the integration harder.

 

[eddiezhang]

Using your formula in post #12.

True.

OK... I'm not sure how exactly:

$\ddot{\theta} = \frac{d}{dt} (\omega \frac{d \omega}{d\theta})$

Bear with me because I haven't done much calculus... multiplying infinitesimals still gives me the heebie-jeebies. Is this right?

$\ddot{\theta} = \frac{d\omega}{dt} \times \frac{d \omega}{d\theta} = \omega (\frac{d \omega}{d\theta})^2$

 

[erobz]

OK... I'm not sure how exactly:

$\ddot{\theta} = \frac{d}{dt} (\omega \frac{d \omega}{d\theta})$

Bear with me because I haven't done much calculus... multiplying infinitesimals still gives me the heebie-jeebies. Is this right?

$\ddot{\theta} = \frac{d\omega}{dt} \times \frac{d \omega}{d\theta} = \omega (\frac{d \omega}{d\theta})^2$

if thats how you want to think about it - not saying it’s mathematically sound- does that eliminate time?

 

[eddiezhang]

if thats how you want to think about it - not saying it’s mathematically sound- does that eliminate time?

It does... I see. OK I'll see if I can solve the ODE with the subs. and (maybe?) get the time of descent (too optimistic?)

 

[haruspex]

You have

$\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2$

and of course
$\dot\theta=\omega, \ddot{\theta} =\dot\omega$,
and

$\frac{d \omega}{dt} = \frac{d \omega}{d\theta} \times \frac{d \theta}{dt} = \omega \frac{d \omega}{d\theta}$

Substitute as necessary to get an equation relating $\theta, \omega, \frac{d\omega}{d\theta}$.

 

[erobz]

It does... I see. OK I'll see if I can solve the ODE with the subs. and (maybe?) get the time of descent (too optimistic?)

Just to be clear, I’ll just give you this to be sure.

$$\ddot \theta = \frac{d\omega}{dt} = \frac{d\omega}{d \theta} \frac{d\theta}{dt} = \frac{d\omega}{d \theta}\omega $$

 

[erobz]

If you can get time, not saying that’s the case it would go through $\omega$ as a function of $\theta$. So finding that is your immediate goal.

 

[erobz]

Well, let’s just make sure you get the right ODE first.

 

[eddiezhang]

Just to be clear, I’ll just give you this to be sure.

$$\ddot \theta = \frac{d\omega}{dt} = \frac{d\omega}{d \theta} \frac{d\theta}{dt} = \frac{d\omega}{d \theta}\omega $$

Oh wait I don't know why I worded it that way but it's literally just the chain rule. Oops.

Thanks for the help so far... I might come back here if the algebra gets too rough.

 

[eddiezhang]

Well, let’s just make sure you get the right ODE first.

So $\dot{\theta} = \omega$ and $\ddot{\theta} = \omega \frac{d\omega}{d\theta}$ (I was clearly having a brain fart before with this part)

Substituting into $\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2$

Produces $\omega \frac{d\omega}{d\theta} = \frac{g}{R} \left(\cos\theta - \mu \sin\theta \right) - \mu \omega^2$

Which I should try solve to find $\omega(\theta)$

 

[erobz]

So $\dot{\theta} = \omega$ and $\ddot{\theta} = \omega \frac{d\omega}{d\theta}$ (I was clearly having a brain fart before with this part)

Substituting into $\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2$

Produces $\omega \frac{d\omega}{d\theta} = \frac{g}{R} \left(\cos\theta - \mu \sin\theta \right) - \mu \omega^2$

Which I should try solve to find $\omega(\theta)$

You still have another couple subs to make. What you have is still currently nonlinear.

Let $u = \omega^2$

The take the derivative of $u$ wrt $\theta$.

This is the step that converts it to linear ODE.

 

[eddiezhang]

You still have another couple subs to make. What you have is still currently nonlinear.

Let $u = \omega^2$

The take the derivative of $u$ wrt $\theta$.

This is the step that converts it to linear ODE.

Ah right...

$\frac{du}{d\theta} = \frac{d}{d\theta}\omega^2 = 2\omega \frac{d\omega}{d\theta} = 2\ddot{\theta}$

So the ODE turns into:
$\frac{1}{2} \times \frac{du}{d\theta} = \frac{g}{R} \left(\cos\theta - \mu \sin\theta \right) - \mu u^2$

And if I'm not mistaken, the strategy is to find u(θ) $\rightarrow$ find ω(θ) $\rightarrow$ see what happens from there wrt $\frac{d\theta}{dt}$?

 

[erobz]

Ah right...

$\frac{du}{d\theta} = \frac{d}{d\theta}\omega^2 = 2\omega \frac{d\omega}{d\theta} = 2\ddot{\theta}$

So the ODE turns into:
$\frac{1}{2} \times \frac{du}{d\theta} = \frac{g}{R} \left(\cos\theta - \mu \sin\theta \right) - \mu u^2$

You made a sub for $u$ incorrectly in there.

And if I'm not mistaken, the strategy is to find u(θ) $\rightarrow$ find ω(θ) $\rightarrow$ see what happens from there wrt $\frac{d\theta}{dt}$?

yea, first comes $u(\theta)$

 

[haruspex]

$\frac{1}{2} \times \frac{du}{d\theta} =$

Having eliminated t, I would simplify the notation by writing $u'$.

the strategy is to find u(θ) $\rightarrow$ find ω(θ) $\rightarrow$ see what happens from there wrt $\frac{d\theta}{dt}$?

Yes, but generally the equation of velocity as a function of displacement is effectively the energy equation. Don’t be too hopeful about finding displacement as a function of time, or v.v., from there.

 

[eddiezhang]

You made a sub for $u$ incorrectly in there.

Ooops. Just $\mu u$ at the end, right?

 

[erobz]

Ooops. Just $\mu u$ at the end, right?

Now put it in standard form and solve with integration factor method.

 

[kuruman]

Yes, but generally the equation of velocity as a function of displacement is effectively the energy equation. Don’t be too hopeful about finding displacement as a function of time, or v.v., from there.

I agree and energy considerations are not allowed as an approach to the solution.

What's v.v. ?

 

[haruspex]

What's v.v. ?

Vice versa.

energy considerations are not allowed as an approach to the solution.

Well, that's hard to define. There are often simple algebraic ways to obtain an equation which is effectively the energy conservation equation without using it as a principle. Likewise momentum, angular momentum.

 

[erobz]

It’s been a blind manipulation of symbols to try and solve the ODE that was derived using forces. If anyone has a direct route to solve non linear ODE then it should be shared too.

Furthermore what do the rules say about AI in the report? PF and whatever math forum he’s visiting are effectively that as far as the OP is concerned. We are just a bit more reserved in our response.

 

[kuruman]

Vice versa.

Thanks. I like the symmetry of the abbreviation which truly conveys the symmetry of the idea it's trying to express.

I thought I might gain some insight on how this has already been approached and searched for the swing time of a pendulum starting from the horizontal which is this problem in the no-friction limit. The first thing that popped up was, as is usually the case with searches nowadays, the AI Overview which I show below.

Clearly, here "AI" stands for Artificial Ignorance since an amplitude of 90° is outside the realm of the "small angle" approximation in which T = 2π * sqrt(L/g) can be used.

After digging a bit deeper, look what I found in the PF archives. Follow the link given in post #5 to the Wikipedia article, understand what is involved in getting the answer to the frictionless case, toss in the friction terms and solve. Easier said than done.

This is for a math report . . .

Was this problem given to you by a mathematician?

 

[eddiezhang]

It’s been a blind manipulation of symbols to try and solve the ODE that was derived using forces. If anyone has a direct route to solve non linear ODE then it should be shared too.

Furthermore what do the rules say about AI in the report? PF and whatever math forum he’s visiting are effectively that as far as the OP is concerned. We are just a bit more reserved in our response.

AI is technically allowed, but the point of the report is to show a) understanding of the math used and b) solving things from the ground up (which is why I was discouraged from using energy concepts).

 

[eddiezhang]

Thanks. I like the symmetry of the abbreviation which truly conveys the symmetry of the idea it's trying to express.

I thought I might gain some insight on how this has already been approached and searched for the swing time of a pendulum starting from the horizontal which is this problem in the no-friction limit. The first thing that popped up was, as is usually the case with searches nowadays, the AI Overview which I show below.

View attachment 355723
Clearly, here "AI" stands for Artificial Ignorance since an amplitude of 90° is outside the realm of the "small angle" approximation in which T = 2π * sqrt(L/g) can be used.

After digging a bit deeper, look what I found in the PF archives. Follow the link given in post #5 to the Wikipedia article, understand what is involved in getting the answer to the frictionless case, toss in the friction terms and solve. Easier said than done.

Thanks, I will investigate this.

Was this problem given to you by a mathematician?

I have to write this for a math class. This was the scenario I chose to investigate, but if it doesn't resolve itself nicely or is too much effort, I can always just change topics.

 

[PeroK]

AI is technically allowed, but the point of the report is to show a) understanding of the math used and b) solving things from the ground up (which is why I was discouraged from using energy concepts).

Energy is fundamental. Once you go beyond Newtonian mechanics, there are no forces and you need energy-based methods: Lagrangian, Hamiltomian etc.

 

[kuruman]

I have to write this for a math class. This was the scenario I chose to investigate, but if it doesn't resolve itself nicely or is too much effort, I can always just change topics.

I see. I would be curious to know whether the person in charge of your class has actually solved this problem analytically before presenting to you as a choice and, if so, how.

 

[haruspex]

At the least, we can use dimensional analysis to discover the form of the relationship (assuming the start and finish angles are fixed): $t=\sqrt{\frac rg}f(\mu)$. Likewise the velocity at the bottom, but I feel the most interesting result is the angle at which it will come to rest (having started at the horizontal), since that will be a function of $\mu$ only.
So in principle we could use simulation with arbitrary values of r and g (both=1, say) to build up a table of results.
Doing that, I found the critical value for just making it to the bottom is very close to $\mu=0.6$.

 

[kuruman]

At the least, we can use dimensional analysis to discover the form of the relationship (assuming the start and finish angles are fixed): $t=\sqrt{\frac rg}f(\mu)$. Likewise the velocity at the bottom, but I feel the most interesting result is the angle at which it will come to rest (having started at the horizontal), since that will be a function of $\mu$ only.
So in principle we could use simulation with arbitrary values of r and g (both=1, say) to build up a table of results.
Doing that, I found the critical value for just making it to the bottom is very close to $\mu=0.6$.

I used the "bottom" expression $$\omega(\theta)= \left \{\left[ \frac{v_0^2}{R^2}- \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi} + \frac{2g}{R} \frac{(2\mu^2\cos\theta -\mu\sin\theta+\cos\theta)e^{2\mu \theta}}{4\mu^2+1} \right\}^{1/2}e^{-\mu~\theta}\tag{1}$$ derived here. Equation (1) gives the angular speed of the sliding block as a function of $\theta$ starting from the horizontal position $\left(\theta =-\frac{\pi}{2}\right)$ with initial speed $v_0.$ Setting $v_0=0$ and $\theta=0$, the angular speed at the bottom of the track for this case is $$\omega(0)= \left \{\left[- \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi}+ \frac{2g}{R} \frac{(2\mu^2 +1)}{4\mu^2+1} \right\}^{1/2}.\tag{2}$$ To find at what value of $\mu$ this is equal to zero, we set the right hand side of equation (2) equal to zero and see if we can solve the resulting equation for $\mu.$ After the obvious simplifications, we get $$\begin{align}
& 2\mu^2 +1=\mu e^{-\mu \pi} \nonumber \\
& \ln(2\mu^2 +1)=\ln(\mu) -\mu \pi. \nonumber \\
\nonumber \end{align}$$The left-hand side in the last equation is always positive whilst the right-hand side is always negative. It appears that the block cannot come to rest at the bottom even if it starts from rest at the top.

Shown below is $\omega(\theta)$ for several values of $\mu$ with $g=10~\text{m/s}^2$ and $R=1~\text{m}.$

If there is an error in the derivation of equation (1), I have not been able to find it.

 

[haruspex]

It appears that the block cannot come to rest at the bottom even if it starts from rest at the top.

Which is clearly infeasible for sufficiently large $\mu$. See my reply on that other thread.

 

[kuruman]

The corrected expression for the "bottom" track is
$$\omega(\theta)= \left \{\left[ \frac{v_0^2}{R^2}- \frac{6g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi} - \frac{2g}{R} \frac{[(2\mu^2-1)\cos\theta +3\mu\sin\theta]e^{2\mu \theta}}{4\mu^2+1} \right\}^{1/2}e^{-\mu~\theta}$$

Doing that, I found the critical value for just making it to the bottom is very close to $\mu=0.6.$

Shown below is a revised plot of $\omega(\theta)$ using the corrected expression with $\mu=0.62.$ The plot is, of course, unrealistic because the mass should reach zero velocity asymptotically.

Edited to fix typos.

 

[haruspex]

The plot is, of course, unrealistic because the mass should reach zero velocity asymptotically.

Asymptotically wrt time, yes, but that is consistent with your plot. Approaching $\theta=0$ it looks linear wrt $\theta$, $\omega\approx -c\theta$, so $\theta \approx Ae^{-ct}$.
However, I suspect that linearity is an artefact of the digital plot. For the case where it stops right at the bottom, the acceleration should be approximately constant at the end, leading to the angular speed varying as the square root of the (absolute) angle. My spreadsheet model produces a curve just like that in post #43 except that both ends appear vertical.