PX155 - E4 - centripetal acceleration

vectors do not have position, so they can be drawn anywhere
$\vec{v}$ is always perpendicular to $\vec{r}$
$\vec{r}$ moves in a circle with angular speed $ω, ∴$ so does $\vec{v}$
hence, if $v = ω r$ (speed at which $\vec{r}$ rotates), $a = ω v$ (speed at which $\vec{v}$ rotates)
$a = ω v = ω^{2} r$
$a = \frac{v}{r} v = \frac{v^{2}}{r}$
- $\vec{a}$ points in the $\hat{\vec{r}}$ direction towards $C$
eg: draw forces and accelerations on a car driving in the circle at constant speed. find the maximum speed the car can go around in a circle with no skidding
- centripetal acceleration is due to static friction between the tyres and the road
- PX155 - A2 - newton's second law:
  - horizontally: $F = m a = \frac{m v^{2}}{r}$
  - vertically: $N = m g$
- $F \leq μ_{s} N$
  so, $\frac{m v^{2}}{r} \leq μ_{s} m g$
$v \leq \sqrt{μ_{s} g r}$
eg: $μ_{s} = 0.8$ , $R = 50 m$ $⟹ v \leq 19.9 m s^{-} 1 (45 m p h)$