Recent content by TimJ

Thank you very much for your help in this project.

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

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...

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...

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...

Because the tangental component looks in the direction of the movement and because so doing I get \frac{d \phi}{ds} 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: m \frac{d \textbf{v}}{dt}=m v \frac{d...

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...

Thank you for warning me. This really complicates things a lot. :frown:

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...

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...

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...

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}...

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...

Hi. Does anybody know how to solve next integral: \int \frac{\sqrt{(y_1-y)\cos{y}}}{\sqrt{c_1-(y_1-y)\cos{y}}} dy where y_1 and c_1 are constants. I am rewriting it, because it seems that latex is not working: ∫ (sqrt[(y_1-y) cos[y]]) / (sqrt[c_1 -(y_1-y) cos[y]]) dy where y_1...

Thank you for your help.