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Speed down a path on a sphere

  1. Aug 21, 2009 #1
    Hi.

    My problem is:
    On the surface of half of a rough sphere there is a known path [tex]\theta = \theta(\phi) [/tex].
    I would like to known wath is the speed down the curve at any [tex]\theta [/tex] 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 [Broken]

    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 [Broken]

    Thank you for your answers.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 22, 2009 #2

    gabbagabbahey

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    Why are you using the formulas for a 2D curve when your curve is clearly 3D?
     
  4. Aug 23, 2009 #3
    I came to the next equations. Can somebody tell me if I have done it right?

    [tex]\mathbf{T}=\frac{dx}{ds}\mathbf{i}+\frac{dy}{ds}\mathbf{j}+\frac{dz}{ds}\mathbf{k} [/tex]

    [tex]\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} [/tex]

    [tex]F_g = mg \mathbf{k} [/tex]

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

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

    After integrating and converting to spherical coordinates i get:

    [tex] 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)) )} [/tex]
     
    Last edited: Aug 23, 2009
  5. Aug 23, 2009 #4

    gabbagabbahey

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    That's better (assuming of course that [itex]\textbf{T}[/itex] 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

    [tex]\textbf{T}=\frac{d\textbf{r}}{ds}[/tex]

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

    Surely you mean [itex]\textbf{F}_g=-mg\textbf{k}[/itex], right?


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

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

    [tex]\textbf{F}_f=-\mu F_n\textbf{T}[/tex]

    And the magnitude of normal force is given by [itex]F_n=|\textbf{F}_g\cdot\textbf{N}_s|[/itex], where [itex]\textbf{N}_s[/itex] is the unit normal vector to the surface
     
    Last edited: Aug 23, 2009
  6. Aug 24, 2009 #5
    It was guesing. I am also not sure about that normal vector. I tried the dot product [itex]\textbf{N} \cdot \textbf{T}[/itex] 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 [itex]\textbf{N}_s[/itex] and the unit normal vector of the particle's curved trajectory [itex]\textbf{N}_c[/itex] the same. The path is on the sphere so don't they have the same normal?

    It is true: [itex]\textbf{F}_g=-mg\textbf{k} [/itex].
    If I use the normal that i have written in the previouse post and use your formulas:
    [itex]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}) [/itex]

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

    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 [tex]\theta=\theta(\phi)[/tex] but I don't know what the exact relation is. This could complicate things a little. I will have to find the function [tex]\theta(\phi)[/tex] after I know the speed because I am trying to find the path [tex]\theta(\phi)[/tex] that minimizes the integral [tex]\int (ds)/(v)[/tex]. I know how to this. The only problem that I have is that I can't find the speed at any [tex]\theta(\phi)[/tex].
     
    Last edited: Aug 24, 2009
  7. Aug 24, 2009 #6

    gabbagabbahey

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

    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.


    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

    [tex]\textbf{N}_s=\frac{\textbf{r}}{||\textbf{r}||}=\frac{\textbf{r}}{r}[/tex]

    I don't think it will complicate things too much.
     
  8. Aug 25, 2009 #7
    Now for the last time and then I will give up. What about this:

    [tex]
    \textbf{T}=\frac{d\textbf{r}}{ds} = \frac{dx}{ds}\textbf{i}+ \frac{dy}{ds}\textbf{j}+ \frac{dz}{ds}\textbf{k}
    [/tex]

    [tex]
    \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}
    [/tex]

    [itex]
    \textbf{F}_g=-mg\textbf{k}
    [/itex]

    [itex]
    F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg \cos \theta \textbf{k}|=mg \cos \theta
    [/itex]

    [itex]
    \textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg \cos \theta\textbf{T}
    [/itex]

    [tex]
    \frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g \cos \theta
    [/tex]

    [tex]
    \frac{1}{2}d(v^2)=g dz-\mu g \cos \theta ds
    [/tex]

    [tex]
    ds= R d\phi \sqrt{\sin^2 \theta+\theta'^2}
    [/tex]

    [tex]
    \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)
    [/tex]
     
  9. Aug 25, 2009 #8

    gabbagabbahey

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    Right:approve:

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

    [tex]m\frac{d\textbf{v}}{dt}=\textbf{F}_g+\textbf{F}_f[/tex]

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

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

    [tex]
    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}
    [/tex]

    And two other equations for [itex]v_x[/itex] and [itex]v_y[/itex]....the speed down the path will be given by [itex]v=||\textbf{v}||=\sqrt{v_x^2+v_y^2+v_z^2}[/itex]
     
    Last edited: Aug 25, 2009
  10. Aug 26, 2009 #9
    Thank you very much for your help and patience.

    Here is the finished product:

    [tex]
    \textbf{T}=\frac{d\textbf{r}}{ds} = \frac{dx}{ds}\textbf{i}+ \frac{dy}{ds}\textbf{j}+ \frac{dz}{ds}\textbf{k}
    [/tex]

    [tex]
    \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}
    [/tex]

    [itex]
    \textbf{F}_g=-mg\textbf{k}
    [/itex]

    [itex]
    F_n=|\textbf{F}_g\cdot\textbf{N}_s|=|-mg \cos \theta \textbf{k}|=mg \cos \theta
    [/itex]

    [itex]
    \textbf{F}_f=-\mu F_n\textbf{T}=-\mu mg \cos \theta\textbf{T}
    [/itex]

    [tex]
    m\frac{d\textbf{v}}{dt}=\textbf{F}_g+\textbf{F}_f
    [/tex]

    z:

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

    [tex]
    v_z^2=2g(\cos \theta_0 -\cos \theta) (1+ \mu \cos \theta)
    [/tex]

    y:

    [tex]
    \frac{1}{2}\frac{d(v_y^2)}{ds}= -\mu g \cos \theta \frac{dy}{ds}
    [/tex]

    [tex]
    v_y^2=2g\mu \cos \theta (\sin \phi_0 \sin \theta_0 - \sin \phi \sin \theta)
    [/tex]

    x:

    [tex]
    \frac{1}{2}\frac{d(v_x^2)}{ds}= -\mu g \cos \theta \frac{dx}{ds}
    [/tex]

    [tex]
    v_x^2=2g\mu \cos \theta (\cos \phi_0 \sin \theta_0 - \cos \phi \sin \theta)
    [/tex]

    [itex]
    v=||\textbf{v}||=\sqrt{v_x^2+v_y^2+v_z^2}
    [/itex]

    [itex]
    v= (\;2g\mu\cos\theta((\cos\phi_0\sin\theta_0-\cos\phi\sin\theta)+
    [/itex]
    [itex]
    + (\sin\phi_0 \sin\theta_0 - \sin \phi \sin \theta))+
    [/itex]
    [itex]
    +(\cos \theta_0 -\cos \theta) (1+ \mu \cos \theta)\;)^{1/2}
    [/itex]
     
  11. Aug 26, 2009 #10

    gabbagabbahey

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

    [tex]\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}[/tex]

    And similarly for your other two equations, which leaves you with 3 coupled differential equations which will be much harder to solve.
     
  12. Aug 27, 2009 #11
    Thank you for warning me.

    This realy complicates things a lot. :frown:
     
    Last edited: Aug 27, 2009
  13. Aug 27, 2009 #12

    gabbagabbahey

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

    [tex]\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}}[/tex]

    While,

    [tex]\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}}[/tex]

    I'll leave it to you to express [itex]\frac{d\textbf{v}}{dt}[/itex] and [itex]\textbf{k}[/itex] in spherical coordinates and plug it into [itex]m\frac{d\textbf{v}}{dt}=-mg\textbf{k}-\mu mg\cos\theta\textbf{T}[/itex]...
     
    Last edited: Aug 27, 2009
  14. Aug 28, 2009 #13

    If I am right the vecor [itex]\textbf{k}[/itex] in spherical coordinates is

    [tex]\textbf{k}=\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}}[/tex]

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

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

    and the force of friction is

    [tex]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}})[/tex]

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

    So [tex]\frac{d \textbf{v}}{dt}= ||\textbf{v}|| \frac{d \textbf{v}}{ds}[/tex]

    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 [tex]\frac{d\phi}{dt}[/tex] in the equation for velocity because I don't want that the velocity is dependent from time in the end.
    What do you think?
     
  15. Aug 28, 2009 #14

    gabbagabbahey

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    Yes.

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

    Where did the [itex]\sin\theta[/itex] in your first term on the RHS come from?

    No need to introduce [itex]v_\theta[/itex] and [itex]v_\phi[/itex] at all here, just use the fact that [tex]\dot{\phi}=v\frac{d\phi}{ds}[/tex] and actually carry out the derivative [itex]\frac{d\textbf{v}}{ds}[/itex] to express everything in terms of [itex]\frac{d\phi}{ds}[/itex]
     
  16. Aug 29, 2009 #15
    Because the tangental component looks in the direction of the movement and because so doing I get [itex]\frac{d \phi}{ds}[/itex]

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

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

    [tex]
    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)
    [/tex]

    I differentiated with [itex]ds[/itex].

    [tex]
    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)
    [/tex]

    Now both equation together:
    [tex]
    m \frac{d \textbf{v}}{dt}=F_g +F_f
    [/tex]
    [tex]
    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)
    [/tex]

    We multiply both sides with [itex]ds^2[/itex] and we get:

    [tex]
    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
    [/tex]

    If I am correct so far we multiply with [itex]d \phi[/itex] and we replace [itex]ds[/itex] with

    [tex]
    ds= r d\phi \sqrt{\sin^2 \theta+\theta'^2}
    [/tex]

    and we get

    [tex]
    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
    [/tex]

    Now we can integrate the differentials that remain:

    [tex]
    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)
    [/tex]

    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 [tex] \mathbf{\hat{\theta}}[/tex] and [tex] \mathbf{\hat{\phi}}[/tex].
     
  17. Aug 30, 2009 #16

    gabbagabbahey

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

    [tex]\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}[/tex]

    [tex]=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][/tex]

    [tex]=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}}[/tex]
    [tex]+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][/tex]

    [tex]=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][/tex]

    The idea is that when you compare the [itex]r[/itex], [itex]\theta[/itex] and [itex]\phi[/itex] components of [itex]v\frac{d\textbf{v}}{ds}[/itex] with the [itex]\theta[/itex] and [itex]\phi[/itex] components of [itex]\textbf{F}_g+\textbf{F}_f[/itex] 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 ([itex]x[/itex],[itex]y[/itex],[itex]z[/itex],[itex]v_x[/itex] etc.) and spherical coordinates ([itex]\theta[/itex], [itex]\phi[/itex] etc) all of which will have some dependence on [itex]s[/itex], 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 :smile:, this:
    gives me an idea to simplify things considerably....

    Recall that the acceleration along a curve can be written [itex]\textbf{a}=\frac{d\textbf{v}}{dt}=\frac{dv}{dt}\textbf{T}+v^2\kappa\textbf{N}_c[/itex] where [itex]\kappa[/itex] is the curvature and [itex]\textbf{N}_c[/itex] 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,

    [tex]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[/tex]

    Which is much simpler than solving 3 coupled DEs!
     
  18. Aug 31, 2009 #17
    This is realy a lot easier to solve.

    I am just not sure about the part of the friction force [tex]-\mu mg\cos\theta[/tex]. 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 [Broken]

    If I do it like on the picture I get:

    [tex]
    \textbf{k}=\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}}
    [/tex]

    [tex]
    F_g=mg \textbf{k} = mg (\cos\theta \mathbf{\hat{r}}-\sin\theta \mathbf{\hat{\theta}})
    [/tex]

    [tex]
    \textbf{N}_s=\mathbf{\hat{\theta}}
    [/tex]

    [tex]
    F_n=-\mu(F_g \textbf{N}_s)\textbf{T} = -\mu m g (-\sin\theta ) \textbf{T}
    [/tex]

    [tex]
    F_n\textbf{T} = \mu m g \sin\theta
    [/tex]

    What do you think about this?
     
    Last edited by a moderator: May 4, 2017
  19. Aug 31, 2009 #18

    gabbagabbahey

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    Draw a picture and label the unit vector [tex]\mathbf{\hat{\theta}}[/tex] at any point on the surface.....does it look normal to the surface?
     
  20. Sep 1, 2009 #19
    Sorry. I just don't know why I sad that the normal is [tex]\mathbf{\hat{\theta}}[/tex] when in real it is [tex]\mathbf{\hat{r}}[/tex]. Lately I am just seeing thetas everywhere. So everything in your formula is correct.

    The result is:

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

    [tex]
    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
    [/tex]
    where [tex]\theta' d \phi=\frac{d \theta}{d \phi}d \phi=d \theta[/tex]

    [tex]
    \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)
    [/tex]

    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.
     
    Last edited: Sep 1, 2009
  21. Sep 1, 2009 #20
    I was talking to a friend and we discovered that this problem is very easy to solve with energy:

    [tex]\frac{mv^2}{2}=mg\Delta z+\int F_t ds[/tex]

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

    [tex]\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)[/tex]
     
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