What is the speed down a path on a rough sphere with gravity and friction?

[TimJ]

Hi.

My problem is:
On the surface of half of a rough sphere there is a known path $$\theta = \theta(\phi)$$.
I would like to known wath is the speed down the curve at any $$\theta$$ if there are the force of gravity (in the direction of -z) and the force of friction.

http://www.shrani.si/f/3G/Hc/41LLD0wT/path.jpg

I tried but with no success to use the formulas for a 2D curve that are on a picture below and applying them to spherical coordinates in 3D.

http://www.shrani.si/f/3v/yA/4EjmK6lQ/formule.jpg

Thank you for your answers.

 

[gabbagabbahey]

Why are you using the formulas for a 2D curve when your curve is clearly 3D?

 

[TimJ]

I came to the next equations. Can somebody tell me if I have done it right?

$$\mathbf{T}=\frac{dx}{ds}\mathbf{i}+\frac{dy}{ds}\mathbf{j}+\frac{dz}{ds}\mathbf{k}$$

$$\mathbf{N}=(-\frac{dz}{ds}-\frac{dy}{ds})\mathbf{i}+(-\frac{dz}{ds}+\frac{dx}{ds})\mathbf{j}+(\frac{dx}{ds}+\frac{dy}{ds})\mathbf{k}$$

$$F_g = mg \mathbf{k}$$

$$F_{friction}= - \mu (F_g \cdot \mathbf{N})\mathbf{T} =-\mu m g (\frac{dx}{ds} +\frac{dy}{ds})$$

$$\frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g (\frac{dx}{ds} +\frac{dy}{ds})$$

After integrating and converting to spherical coordinates i get:

$$v=\sqrt{2g((\cos{\theta_0}-\cos \theta)-\mu((\cos \phi_0 \sin \theta_0 - \cos \phi \ \sin \theta)+(\sin \phi_0 \sin \theta_0 - \sin \phi \sin \theta)) )}$$

 

[gabbagabbahey]

I came to the next equations. Can somebody tell me if I have done it right?

$$\mathbf{T}=\frac{dx}{ds}\mathbf{i}+\frac{dy}{ds}\mathbf{j}+\frac{dz}{ds}\mathbf{k}$$

That's better (assuming of course that $\textbf{T}$ represents the unit tangent to the particle's curved path), but since you are planning on using spherical coordinates, why not simply write this as

$$\textbf{T}=\frac{d\textbf{r}}{ds}$$

$$\mathbf{N}=(-\frac{dz}{ds}-\frac{dy}{ds})\mathbf{i}+(-\frac{dz}{ds}+\frac{dx}{ds})\mathbf{j}+(\frac{dx}{ds}+\frac{dy}{ds})\mathbf{k}$$

What does $\textbf{N}$ represent here? Is it the unit normal vector to the spherical surface $\textbf{N}_s$ or the unit normal vector of the particle's curved trajectory $\textbf{N}_c$? Either way, this expression does not look correct to me...how did you arrive at it?

$$F_g = mg \mathbf{k}$$

Surely you mean $\textbf{F}_g=-mg\textbf{k}$, right?

$$F_{friction}= - \mu (F_g \cdot \mathbf{N})\mathbf{T} =-\mu m g (\frac{dx}{ds} +\frac{dy}{ds})$$

Why are you equating a vector, $- \mu (F_g \cdot \mathbf{N})\mathbf{T}$ to a scalar, $-\mu m g (\frac{dx}{ds} +\frac{dy}{ds})$?

Unless I'm mistaken, the force of friction along a curved surface is proportional to the magnitude of the normal force $F_n$, and directed antiparallel to the unit tangent vector of the trajectory (so that it directly opposes the particle's motion):

$$\textbf{F}_f=-\mu F_n\textbf{T}$$

And the magnitude of normal force is given by $F_n=|\textbf{F}_g\cdot\textbf{N}_s|$, where $\textbf{N}_s$ is the unit normal vector to the surface

 

[TimJ]

What does $\textbf{N}$ represent here? Is it the unit normal vector to the spherical surface $\textbf{N}_s$ or the unit normal vector of the particle's curved trajectory $\textbf{N}_c$? Either way, this expression does not look correct to me...how did you arrive at it?

It was guesing. I am also not sure about that normal vector. I tried the dot product $\textbf{N} \cdot \textbf{T}$ and it was 0 as it should be. I made a graph and it looked OK.
Aren't the the unit normal vector to the spherical surface $\textbf{N}_s$ and the unit normal vector of the particle's curved trajectory $\textbf{N}_c$ the same. The path is on the sphere so don't they have the same normal?

Surely you mean $\textbf{F}_g=-mg\textbf{k}$, right?
Why are you equating a vector, $- \mu (F_g \cdot \mathbf{N})\mathbf{T}$ to a scalar, $-\mu m g (\frac{dx}{ds} +\frac{dy}{ds})$?

Unless I'm mistaken, the force of friction along a curved surface is proportional to the magnitude of the normal force $F_n$, and directed antiparallel to the unit tangent vector of the trajectory (so that it directly opposes the particle's motion):

$$\textbf{F}_f=-\mu F_n\textbf{T}$$

And the magnitude of normal force is given by $F_n=|\textbf{F}_g\cdot\textbf{N}_s|$, where $\textbf{N}_s$ is the unit normal vector to the surface

It is true: $\textbf{F}_g=-mg\textbf{k}$.
If I use the normal that i have written in the previouse post and use your formulas:
$F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{k}|=mg (\frac{dx}{ds} +\frac{dy}{ds})$

$\textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{T}$
This is what I have written. I just forgot to put $\textbf{T}$. Or I am missing something?

That's better (assuming of course that $\textbf{T}$ represents the unit tangent to the particle's curved path), but since you are planning on using spherical coordinates, why not simply write this as

$$\textbf{T}=\frac{d\textbf{r}}{ds}$$

I also think that this would be probably the best way to do it but I just can't find the normal vector that would have the derivative d/ds. Can you help me with that?

There is one thing that I forgot to mention erlier. The path is $$\theta=\theta(\phi)$$ but I don't know what the exact relation is. This could complicate things a little. I will have to find the function $$\theta(\phi)$$ after I know the speed because I am trying to find the path $$\theta(\phi)$$ that minimizes the integral $$\int (ds)/(v)$$. I know how to this. The only problem that I have is that I can't find the speed at any $$\theta(\phi)$$.

 

[gabbagabbahey]

It was guesing.

Not typically a good way to tackle a math/physics problem.

Aren't the the unit normal vector to the spherical surface $\textbf{N}_s$ and the unit normal vector of the particle's curved trajectory $\textbf{N}_c$ the same. The path is on the sphere so don't they have the same normal?

No, if you look up "normal vector to a surface" and "normal vector to a curve in 3-space", you will see two very different definitions. It is only under very special circumstances that the two will be equal, and simply having the curve on the sphere is not sufficient for the two to be equal.

In fact, it is always a good idea to look up the definitions of terms when you are unsure how to calculate them.

I also think that this would be probably the best way to do it but I just can't find the normal vector that would have the derivative d/ds. Can you help me with that?

Look up the definition of "normal vector for a surface". After a quick calculation (which you should do!), you should easily find that the normal vector for a spherical surface is

$$\textbf{N}_s=\frac{\textbf{r}}{||\textbf{r}||}=\frac{\textbf{r}}{r}$$

There is one thing that I forgot to mention erlier. The path is $$\theta=\theta(\phi)$$ but I don't know what the exact relation is. This could complicate things a little. I will have to find the function $$\theta(\phi)$$ after I know the speed because I am trying to find the path $$\theta(\phi)$$ that minimizes the integral $$\int (ds)/(v)$$. I know how to this. The only problem that I have is that I can't find the speed at any $$\theta(\phi)$$.

I don't think it will complicate things too much.

 

[TimJ]

Now for the last time and then I will give up. What about this:

$$\textbf{T}=\frac{d\textbf{r}}{ds} = \frac{dx}{ds}\textbf{i}+ \frac{dy}{ds}\textbf{j}+ \frac{dz}{ds}\textbf{k}$$

$$\textbf{N}_s=\frac{\textbf{r}}{||\textbf{r}||}=\frac{\textbf{r}}{r} = \cos \phi \sin \theta\textbf{i}+\sin \phi \sin \theta\textbf{j}+ \cos \theta\textbf{k}$$

$\textbf{F}_g=-mg\textbf{k}$

$F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg \cos \theta \textbf{k}|=mg \cos \theta$

$\textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg \cos \theta\textbf{T}$

$$\frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g \cos \theta$$

$$\frac{1}{2}d(v^2)=g dz-\mu g \cos \theta ds$$

$$ds= R d\phi \sqrt{\sin^2 \theta+\theta'^2}$$

$$\frac{1}{2}(v^2)=g (\cos \theta_0 - \cos \theta) - \mu g R\cos \theta \sqrt{\sin^2 \theta+\theta'^2} (\phi - \phi_0)$$

 

[gabbagabbahey]

Now for the last time and then I will give up. What about this:

$$\textbf{T}=\frac{d\textbf{r}}{ds} = \frac{dx}{ds}\textbf{i}+ \frac{dy}{ds}\textbf{j}+ \frac{dz}{ds}\textbf{k}$$

$$\textbf{N}_s=\frac{\textbf{r}}{||\textbf{r}||}=\frac{\textbf{r}}{r} = \cos \phi \sin \theta\textbf{i}+\sin \phi \sin \theta\textbf{j}+ \cos \theta\textbf{k}$$

$\textbf{F}_g=-mg\textbf{k}$

$F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg \cos \theta \textbf{k}|=mg \cos \theta$

$\textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg \cos \theta\textbf{T}$

Right

$$\frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g \cos \theta$$

Not quite. Remember, Newton's 2nd law is a vector equation:

$$m\frac{d\textbf{v}}{dt}=\textbf{F}_g+\textbf{F}_f$$

So, you will get 3 differential equations from it---- one for each component of $\textbf{v}$

If you let $\textbf{v}\equiv v_x\textbf{i}+v_y\textbf{j}+v_z\textbf{k}$, then you will get

$$m\frac{dv_z}{dt}=\left(\textbf{F}_g\right)_z+\left(\textbf{F}_f\right)_z=-mg-m\mu g \cos \theta \frac{dz}{ds}$$

And two other equations for $v_x$ and $v_y$...the speed down the path will be given by $v=||\textbf{v}||=\sqrt{v_x^2+v_y^2+v_z^2}$

 

[TimJ]

Thank you very much for your help and patience.

Here is the finished product:

$$\textbf{T}=\frac{d\textbf{r}}{ds} = \frac{dx}{ds}\textbf{i}+ \frac{dy}{ds}\textbf{j}+ \frac{dz}{ds}\textbf{k}$$

$$\textbf{N}_s=\frac{\textbf{r}}{||\textbf{r}||}=\frac{\textbf{r}}{r} = \cos \phi \sin \theta\textbf{i}+\sin \phi \sin \theta\textbf{j}+ \cos \theta\textbf{k}$$

$\textbf{F}_g=-mg\textbf{k}$

$F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg \cos \theta \textbf{k}|=mg \cos \theta$

$\textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg \cos \theta\textbf{T}$

$$m\frac{d\textbf{v}}{dt}=\textbf{F}_g+\textbf{F}_f$$

z:

$$\frac{dv_z}{dt} =\frac{1}{2}\frac{d(v_z^2)}{ds}=-g\frac{dz}{ds} -\mu g \cos \theta \frac{dz}{ds}$$

$$v_z^2=2g(\cos \theta_0 -\cos \theta) (1+ \mu \cos \theta)$$

y:

$$\frac{1}{2}\frac{d(v_y^2)}{ds}= -\mu g \cos \theta \frac{dy}{ds}$$

$$v_y^2=2g\mu \cos \theta (\sin \phi_0 \sin \theta_0 - \sin \phi \sin \theta)$$

x:

$$\frac{1}{2}\frac{d(v_x^2)}{ds}= -\mu g \cos \theta \frac{dx}{ds}$$

$$v_x^2=2g\mu \cos \theta (\cos \phi_0 \sin \theta_0 - \cos \phi \sin \theta)$$

$v=||\textbf{v}||=\sqrt{v_x^2+v_y^2+v_z^2}$

$v= (\;2g\mu\cos\theta((\cos\phi_0\sin\theta_0-\cos\phi\sin\theta)+$
$+ (\sin\phi_0 \sin\theta_0 - \sin \phi \sin \theta))+$
$+(\cos \theta_0 -\cos \theta) (1+ \mu \cos \theta)\;)^{1/2}$

 

[gabbagabbahey]

$$\frac{dv_z}{dt} =\frac{1}{2}\frac{d(v_z^2)}{ds}=-g\frac{dz}{ds} -\mu g \cos \theta \frac{dz}{ds}$$

I hate to be the bearer of bad news, but...

$$\frac{d v_z}{d t}=\frac{ds}{dt}\frac{dv_z}{ds}=v\frac{dv_z}{ds}=\sqrt{v_x^2+v_y^2+v_z^2}\frac{dv_z}{ds}\neq\frac{1}{2}\frac{d v_z^2}{ds}$$

And similarly for your other two equations, which leaves you with 3 coupled differential equations which will be much harder to solve.

 

[TimJ]

Thank you for warning me.

$$\frac{d v_z}{d t}=\frac{ds}{dt}\frac{dv_z}{ds}=v\frac{dv_z}{ds}=\sqrt{v_x^2+v_y^2+v_z^2}\frac{dv_z}{ds}\neq\frac{1} {2}\frac{d v_z^2}{ds}$$

And similarly for your other two equations, which leaves you with 3 coupled differential equations which will be much harder to solve.

This really complicates things a lot.

 

[gabbagabbahey]

I think you can simplify things at least a little bit by working in spherical coordinates...Since $r$ is a constant all along the curve, you have:

$$\textbf{v}=\frac{d\textbf{r}}{dt}=\frac{d}{dt}(r\mathbf{\hat{r}})=r\frac{d}{dt}\left(\sin\theta\cos\phi\textbf{i}+\sin\theta\sin\phi\textbf{j}+\cos\theta\textbf{k}\right)=r\dot{\theta}\mathbf{\hat{\theta}}+r\sin\theta\dot{\phi}\mathbf{\hat{\phi}}=r\theta'\dot{\phi}\mathbf{\hat{\theta}}+r\sin\theta\dot{\phi}\mathbf{\hat{\phi}}$$

While,

$$\textbf{T}=\frac{d\textbf{r}}{ds}=r\theta'\frac{d\phi}{ds}\mathbf{\hat{\theta}}+r\sin\theta\frac{d\phi}{ds}\mathbf{\hat{\phi}}$$

I'll leave it to you to express $\frac{d\textbf{v}}{dt}$ and $\textbf{k}$ in spherical coordinates and plug it into $m\frac{d\textbf{v}}{dt}=-mg\textbf{k}-\mu mg\cos\theta\textbf{T}$...

 

[TimJ]

I'll leave it to you to express $\frac{d\textbf{v}}{dt}$ and $\textbf{k}$ in spherical coordinates and plug it into $m\frac{d\textbf{v}}{dt}=-mg\textbf{k}-\mu mg\cos\theta\textbf{T}$...

If I am right the vecor $\textbf{k}$ in spherical coordinates is

$$\textbf{k}=\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}}$$

so the force of gravity in the direction of the tangent is:

$$F_g=-mg\textbf{k} \cdot \textbf{T}=mgr \theta'\sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}$$

and the force of friction is

$$F_f=(-\mu mg \cos \theta ) ( r \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+r \sin \theta \frac{d \phi}{ds} \mathbf{\hat{\phi}})$$

Now for the velocity $$\textbf{v}=r\theta'\dot{\phi}\mathbf{\hat{\theta}}+r\sin\theta\dot{\phi}\mathbf{\hat{\phi}}$$ we could use the same technic as before. So $$\textbf{v}=(v_{\theta},v_{\phi})$$ and $$||\textbf{v}||=\sqrt{v_{\theta}^2+v_{\phi}^2}$$

So $$\frac{d \textbf{v}}{dt}= ||\textbf{v}|| \frac{d \textbf{v}}{ds}$$

As I discovered before this is not the best way but I couldn't find another one. I have some concerns also about the differential $$\frac{d\phi}{dt}$$ in the equation for velocity because I don't want that the velocity is dependent from time in the end.
What do you think?

 

[gabbagabbahey]

If I am right the vecor $\textbf{k}$ in spherical coordinates is

$$\textbf{k}=\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}}$$

Yes.

so the force of gravity in the direction of the tangent is:

$$F_g=-mg\textbf{k} \cdot \textbf{T}=mgr \theta'\sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}$$

Sure, I guess, but why bother computing the tangential component of gravity at all?

and the force of friction is

$$F_f=(-\mu mg \cos \theta ) ( r \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+r \sin \theta \frac{d \phi}{ds} \mathbf{\hat{\phi}})$$

Where did the $\sin\theta$ in your first term on the RHS come from?

Now for the velocity $$\textbf{v}=r\theta'\dot{\phi}\mathbf{\hat{\theta}}+r\sin\theta\dot{\phi}\mathbf{\hat{\phi}}$$ we could use the same technic as before. So $$\textbf{v}=(v_{\theta},v_{\phi})$$ and $$||\textbf{v}||=\sqrt{v_{\theta}^2+v_{\phi}^2}$$

So $$\frac{d \textbf{v}}{dt}= ||\textbf{v}|| \frac{d \textbf{v}}{ds}$$

As I discovered before this is not the best way but I couldn't find another one. I have some concerns also about the differential $$\frac{d\phi}{dt}$$ in the equation for velocity because I don't want that the velocity is dependent from time in the end.
What do you think?

No need to introduce $v_\theta$ and $v_\phi$ at all here, just use the fact that $$\dot{\phi}=v\frac{d\phi}{ds}$$ and actually carry out the derivative $\frac{d\textbf{v}}{ds}$ to express everything in terms of $\frac{d\phi}{ds}$

 

[TimJ]

Sure, I guess, but why bother computing the tangential component of gravity at all?

Because the tangental component looks in the direction of the movement and because so doing I get $\frac{d \phi}{ds}$

Where did the $\sin\theta$ in your first term on the RHS come from?

It' s a mistake that I made when I was copying from the paper.

actually carry out the derivative $\frac{d\textbf{v}}{ds}$ to express everything in terms of $\frac{d\phi}{ds}$

I am not sure if I understand correctly. Did you ment doing this:

$$m \frac{d \textbf{v}}{dt}=m v \frac{d \textbf{v}}{ds} =m v^2 r \left(\frac{d\phi}{ds} \left(\hat{\mathbf{\phi}} \cos \theta \frac{d\theta}{ds}+\hat{\mathbf{\theta}}\frac{d \theta'}{ds} \right)+\left (\hat{\mathbf{\phi}} \sin\theta+\hat{\mathbf{\theta}} \theta' \right) \frac{d^2\phi}{ds^2} \right)$$

I differentiated with $ds$.

$$F_g +F_f=mgr \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+ \left (-\mu mg \cos \theta \right ) \left ( r \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+r \sin \theta \frac{d \phi}{ds} \mathbf{\hat{\phi}} \right)$$

Now both equation together:
$$m \frac{d \textbf{v}}{dt}=F_g +F_f$$
$$m v^2 r \left(\frac{d\phi}{ds} \left(\hat{\mathbf{\phi}} \cos \theta \frac{d\theta}{ds}+\hat{\mathbf{\theta}}\frac{d \theta'}{ds} \right)+ \left (\hat{\mathbf{\phi}} \sin\theta+\hat{\mathbf{\theta}} \theta' \right ) \frac{d^2\phi}{ds^2} \right)=mgr \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+ \left (-\mu mg \cos \theta \right ) \left ( r \theta' \sin\theta \frac{d \phi}{ds}\mathbf{\hat{\theta}}+r \sin \theta \frac{d \phi}{ds} \mathbf{\hat{\phi}} \right)$$

We multiply both sides with $ds^2$ and we get:

$$v^2 \left( \left(d\theta' +\theta' \right) \hat{\mathbf{\theta}} + \left( \cos \theta d\theta +\sin \theta \right ) \hat{\mathbf{\phi}} \right ) d \phi = g \left( \left ( \theta' \sin \theta - \mu \cos \theta \theta' \right ) \mathbf{\hat{\theta}} + \left (- \mu \cos \theta \sin \theta \right ) \mathbf{\hat{\phi}}\right ) d\phi ds$$

If I am correct so far we multiply with $d \phi$ and we replace $ds$ with

$$ds= r d\phi \sqrt{\sin^2 \theta+\theta'^2}$$

and we get

$$v^2 \left( \left(d\theta' +\theta' \right) \hat{\mathbf{\theta}} + \left( \cos \theta d\theta +\sin \theta \right ) \hat{\mathbf{\phi}} \right )= r g \left( \left ( \theta' \sin \theta - \mu \cos \theta \theta' \right ) \mathbf{\hat{\theta}} + \left (- \mu \cos \theta \sin \theta \right ) \mathbf{\hat{\phi}}\right ) \sqrt{\sin^2 \theta+\theta'^2} d\phi$$

Now we can integrate the differentials that remain:

$$v^2 \left( \left(\left(\theta' - \theta'_0 \right) +\theta' \right) \hat{\mathbf{\theta}} + \left( \left( \sin \theta - \sin \theta_0 \right) +\sin \theta \right ) \hat{\mathbf{\phi}} \right )= r g \left( \left ( \theta' \sin \theta - \mu \cos \theta \theta' \right ) \mathbf{\hat{\theta}} + \left (- \mu \cos \theta \sin \theta \right ) \mathbf{\hat{\phi}}\right ) \sqrt{\sin^2 \theta+\theta'^2} \left ( \phi - \phi_0 \right)$$

This is to where I' ve come. Now I want to know what do you think about this. Is it correct? If I am now I don't know what to do with the final equation. The part that confuses me most are the unit vectors $$\mathbf{\hat{\theta}}$$ and $$\mathbf{\hat{\phi}}$$.

 

[gabbagabbahey]

I am not sure if I understand correctly. Did you ment doing this:

$$m \frac{d \textbf{v}}{dt}=m v \frac{d \textbf{v}}{ds} =m v^2 r \left(\frac{d\phi}{ds} \left(\hat{\mathbf{\phi}} \cos \theta \frac{d\theta}{ds}+\hat{\mathbf{\theta}}\frac{d \theta'}{ds} \right)+\left (\hat{\mathbf{\phi}} \sin\theta+\hat{\mathbf{\theta}} \theta' \right) \frac{d^2\phi}{ds^2} \right)$$

Sort of, but the spherical unit vectors themselves will also depend on position/arclength, so for example,

$$\frac{d}{ds}\left(r\theta'\dot{\phi}\mathbf{\hat{\theta}}\right)=\frac{d}{ds}\left(r\theta'v\frac{d\phi}{ds}\mathbf{\hat{\theta}}\right)=r\frac{d\theta'}{ds}v\frac{d\phi}{ds}\mathbf{\hat{\theta}}+r\theta'\frac{dv}{ds}\frac{d\phi}{ds}\mathbf{\hat{\theta}}+r\theta'v\frac{d^2\phi}{ds^2}\mathbf{\hat{\theta}}+r\theta'v\frac{d\phi}{ds}\frac{d\mathbf{\hat{\theta}}}{ds}$$

$$=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}+rv\theta'\left(\frac{d\phi}{ds}\right)\left[\frac{d}{ds}\left(\cos\theta\cos\phi\mathbf{\hat{x}}+\cos\theta\sin\phi\mathbf{\hat{y}}-\sin\theta\mathbf{\hat{z}}\right)\right]$$

$$=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}$$
$$+rv\theta'\left(\frac{d\phi}{ds}\right)\left[-\theta'\frac{d\phi}{ds}\sin\theta\cos\phi\mathbf{\hat{x}}-\frac{d\phi}{ds}\cos\theta\sin\phi\mathbf{\hat{x}}-\theta'\frac{d\phi}{ds}\sin\theta\sin\phi\mathbf{\hat{y}}+\frac{d\phi}{ds}\cos\theta\cos\phi\mathbf{\hat{y}}-\theta'\frac{d\phi}{ds}\cos\theta\mathbf{\hat{z}}\right]$$

$$=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}+rv\theta'\left(\frac{d\phi}{ds}\right)^2\left[-\theta'\mathbf{\hat{r}}+\cos\theta\mathbf{\hat{\phi}}\right]$$

The part that confuses me most are the unit vectors $$\mathbf{\hat{\theta}}$$ and $$\mathbf{\hat{\phi}}$$.

The idea is that when you compare the $r$, $\theta$ and $\phi$ components of $v\frac{d\textbf{v}}{ds}$ with the $\theta$ and $\phi$ components of $\textbf{F}_g+\textbf{F}_f$ you will get 3 simultaneous differential equations where everything is expressed in spherical coordinates.

The 3 DEs you got by comparing Cartesian components had a mix of Cartesian coordinates ($x$,$y$,$z$,$v_x$ etc.) and spherical coordinates ($\theta$, $\phi$ etc) all of which will have some dependence on $s$, making it very unclear as to how to solve them!

At least when you compare spherical components like this, you get everything in terms of spherical coordinates.

However, by looking at the DEs you get doing this, you can see it is still very difficult to solve them!:sad:

Luckily , this:

Because the tangental component looks in the direction of the movement and because so doing I get $\frac{d \phi}{ds}$

gives me an idea to simplify things considerably...

Recall that the acceleration along a curve can be written $\textbf{a}=\frac{d\textbf{v}}{dt}=\frac{dv}{dt}\textbf{T}+v^2\kappa\textbf{N}_c$ where $\kappa$ is the curvature and $\textbf{N}_c$ is the unit normal to the curve (not the normal to the surface).

Looking at the acceleration in this form, it should be clear that any changes in the speed of the particle will be due entirely to the tangential component of the applied force!

That means,

$$m\frac{dv}{dt}=\textbf{F}_g\cdot\textbf{T}+\textbf{F}_f\cdot\textbf{T}=mgr\sin\theta\theta' \frac{d \phi}{ds}-\mu mg\cos\theta$$

Which is much simpler than solving 3 coupled DEs!

 

[TimJ]

$$m\frac{dv}{dt}=\textbf{F}_g\cdot\textbf{T}+\textbf{F}_f\cdot\textbf{T}=mgr\sin\theta\theta' \frac{d \phi}{ds}-\mu mg\cos\theta$$

Which is much simpler than solving 3 coupled DEs!

This is really a lot easier to solve.

I am just not sure about the part of the friction force $$-\mu mg\cos\theta$$. On the picture below is an example for two dimensions that I found in one book.

http://www.shrani.si/f/J/7b/4WSx9Z5N/formule3.jpg

If I do it like on the picture I get:

$$\textbf{k}=\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}}$$

$$F_g=mg \textbf{k} = mg (\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}})$$

$$\textbf{N}_s=\mathbf{\hat{\theta}}$$

$$F_n=-\mu(F_g \textbf{N}_s)\textbf{T} = -\mu m g (-\sin\theta ) \textbf{T}$$

$$F_n\textbf{T} = \mu m g \sin\theta$$

What do you think about this?

 

[gabbagabbahey]

Draw a picture and label the unit vector $$\mathbf{\hat{\theta}}$$ at any point on the surface...does it look normal to the surface?

 

[TimJ]

Sorry. I just don't know why I sad that the normal is $$\mathbf{\hat{\theta}}$$ when in real it is $$\mathbf{\hat{r}}$$. Lately I am just seeing thetas everywhere. So everything in your formula is correct.

The result is:

$$m\frac{dv}{dt}=m \frac{1}{2}\frac{dv^2}{ds}=mgr\sin\theta\theta' \frac{d \phi}{ds}-\mu mg\cos\theta$$

$$m\frac{1}{2}dv^2=mgr\sin\theta\theta' d \phi-\mu mg\cos\theta ds = mgr\sin\theta d \theta-\mu mg\cos\theta R \sqrt{\sin^2 \theta+\theta'^2} d \phi$$
where $$\theta' d \phi=\frac{d \theta}{d \phi}d \phi=d \theta$$

$$\frac{1}{2}(v^2)=g (\cos \theta_0 - \cos \theta) - \mu g R\cos \theta \sqrt{\sin^2 \theta+\theta'^2} (\phi - \phi_0)$$

This is the same result as in post #7 where we were using cartesian coordinates. We got the same result by using two different ways, so it must be the right result.

 

[TimJ]

I was talking to a friend and we discovered that this problem is very easy to solve with energy:

$$\frac{mv^2}{2}=mg\Delta z+\int F_t ds$$

$$\frac{mv^2}{2}=mg\Delta z-\int \mu mg\cos\theta R \sqrt{\sin^2 \theta+\theta'^2} d \phi$$

$$\frac{1}{2}(v^2)=g (\cos \theta_0 - \cos \theta) - \mu g R\cos \theta \sqrt{\sin^2 \theta+\theta'^2} (\phi - \phi_0)$$

 

[gabbagabbahey]

Yes, that is much simpler. However, since $\theta$depends on $\phi$, I don't think you can say that

$$\int \mu g\cos\theta R \sqrt{\sin^2 \theta+\theta'^2} d \phi=\mu g R\cos \theta \sqrt{\sin^2 \theta+\theta'^2} (\phi - \phi_0)$$

 

[TimJ]

Yes, that is much simpler. However, since $\theta$depends on $\phi$, I don't think you can say that

$$\int \mu g\cos\theta R \sqrt{\sin^2 \theta+\theta'^2} d \phi=\mu g R\cos \theta \sqrt{\sin^2 \theta+\theta'^2} (\phi - \phi_0)$$

I also think that this is not the right way. But the problem is how to integrate that without knowing the relation between $\theta$ and $\phi$

 

[gabbagabbahey]

I don't think you can evaluate the integral explicitly without knowing $\theta(\phi)$. You'll just have to leave it written as $\int \mu g\cos\theta R \sqrt{\sin^2 \theta+\theta'^2} d \phi$

 

[TimJ]

Thank you very much for your help in this project.