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?