| Thread Closed |
Constrained to move Horizontally |
Share Thread | Thread Tools |
| Aug8-10, 10:39 PM | #1 |
|
|
Constrained to move Horizontally
1. The problem statement, all variables and given/known data
![]() 2. Relevant equations v = [tex]\dot{r}[/tex] [tex]\hat{r}[/tex] + r[tex]\dot{\vartheta}[/tex] [tex]\hat{\vartheta}[/tex] 3. The attempt at a solution [tex]\dot{r}[/tex] = ? \vartheta = 80° v = 55mm/s So I guess I just use the formula above. v = [tex]\dot{r}[/tex] [tex]\hat{r}[/tex] + r[tex]\dot{\vartheta}[/tex] [tex]\hat{\vartheta}[/tex] 55² = [tex]\dot{r}[/tex]² + rΘ' And so you try and solve for [tex]\dot{r}[/tex] r'² = 55² - (r*Θ')² r' = sqrt(55² - (r*Θ')²) And then I get stuck. I am either missing something. Or not doing something right. I guess this isn't really r theta, it is more a conversion from r theta to x-y. Not 100% sure how to do that though. Cheers |
| Aug8-10, 11:08 PM | #2 |
|
Recognitions:
|
first you can write sevreal expresisons in tex as follows
[tex]\vec{v} = \dot{r} \hat{r} + r \dot{\vartheta}\hat{\vartheta}[/tex] so knowing theta and |v| you should be able to decompose v into components in the orthogonal directions [itex]\hat{r}, \hat{\vartheta}[/itex] |
| Aug11-10, 01:10 AM | #3 |
|
|
I manage to get the question.
Thanks :D |
| Thread Closed |
| Thread Tools | |
Similar Threads for: Constrained to move Horizontally
|
||||
| Thread | Forum | Replies | ||
| particle constrained to move on a sphere | Advanced Physics Homework | 1 | ||
| Constrained Maximization | Calculus & Beyond Homework | 3 | ||
| constrained extrema | Calculus & Beyond Homework | 3 | ||
| Constrained Least Squares | General Math | 23 | ||
| constrained PDE : 3D -> 1D | Introductory Physics Homework | 1 | ||