# Difference between revisions of "Circular motion/Notes"

From Maths

(Created page with "==Acceleration== * {{MM|a(t)\eq p(t)\cdot\left(\frac{r' '(t)}{r(t)}-(\theta'(t))^2\right)+\big(\theta' '(t)\cdot r(t)+2\theta'(t)\cdot r'(t)\big)\cdot\left[\begin{array}{r}-\s...") |
m |
||

Line 12: | Line 12: | ||

#** But notice: | #** But notice: | ||

#*** {{M|a(t)\eq \theta' '(t)\left(\left[\begin{array}{r}-r(t)\cdot\sin(\theta(t)) \\ r(t)\cdot\cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)}} | #*** {{M|a(t)\eq \theta' '(t)\left(\left[\begin{array}{r}-r(t)\cdot\sin(\theta(t)) \\ r(t)\cdot\cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)}} | ||

− | #***: {{M|\eq \theta' '(t)\left(\left[\begin{array}{r} - | + | #***: {{M|\eq \theta' '(t)\left(\left[\begin{array}{r} -p_y(t)\\p_x(t)\end{array}\right]-\theta' '(t)\cdot p(t)\right)}} |

There must be a geometric interpretation for this! As the vector here is {{M|p(t)}} reflected in the line {{M|x\eq 0}}! | There must be a geometric interpretation for this! As the vector here is {{M|p(t)}} reflected in the line {{M|x\eq 0}}! | ||

+ | * {{XXX|This is wrong actually}} as the {{M|x}} component comes from cos in {{M|p(t)}}, not sin. Which is not a reflection in {{M|x\eq 0}}, a matrix might be able to represent this linear combination better | ||

+ | ** I've corrected the formula [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 18:56, 13 September 2018 (UTC) |

## Latest revision as of 18:56, 13 September 2018

## Acceleration

- [math]a(t)\eq p(t)\cdot\left(\frac{r' '(t)}{r(t)}-(\theta'(t))^2\right)+\big(\theta' '(t)\cdot r(t)+2\theta'(t)\cdot r'(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/math], or:
- substituting in [ilmath]p(t)[/ilmath] by it's definition:
- [math]a(t)\eq \left(\frac{r' '(t)}{r(t)}-(\theta'(t))^2\right)\cdot\left[\begin{array}{r}r(t)\cdot\cos(\theta(t)) \\ r(t)\cdot\sin(\theta(t))\end{array}\right]+\big(\theta' '(t)\cdot r(t)+2\theta'(t)\cdot r'(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/math]
- However in many special cases it is useful to consider the first form with [ilmath]p(t)[/ilmath] in it.

- substituting in [ilmath]p(t)[/ilmath] by it's definition:

### Special cases

- unchanging radius, [ilmath]r(t):\eq r_0\in\mathbb{R}_{>0} [/ilmath]
- obviously, now [ilmath]r'(t)\eq 0[/ilmath] and [ilmath]r' '(t)\eq 0[/ilmath], thus:
- [ilmath]a(t)\eq -(\theta' '(t))^2\cdot p(t)+\big(\theta' '(t)\cdot r(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/ilmath]
- [ilmath]\eq \theta' '(t)\left(r(t)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

- But notice:
- [ilmath]a(t)\eq \theta' '(t)\left(\left[\begin{array}{r}-r(t)\cdot\sin(\theta(t)) \\ r(t)\cdot\cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]
- [ilmath]\eq \theta' '(t)\left(\left[\begin{array}{r} -p_y(t)\\p_x(t)\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

- [ilmath]a(t)\eq \theta' '(t)\left(\left[\begin{array}{r}-r(t)\cdot\sin(\theta(t)) \\ r(t)\cdot\cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

- [ilmath]a(t)\eq -(\theta' '(t))^2\cdot p(t)+\big(\theta' '(t)\cdot r(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/ilmath]

- obviously, now [ilmath]r'(t)\eq 0[/ilmath] and [ilmath]r' '(t)\eq 0[/ilmath], thus:

There must be a geometric interpretation for this! As the vector here is [ilmath]p(t)[/ilmath] reflected in the line [ilmath]x\eq 0[/ilmath]!

- TODO: This is wrong actuallyas the [ilmath]x[/ilmath] component comes from cos in [ilmath]p(t)[/ilmath], not sin. Which is not a reflection in [ilmath]x\eq 0[/ilmath], a matrix might be able to represent this linear combination better