Originally Posted by

**craig** I'm not too good on circular motion questions so please bare with me ;), here's my attempt.

Sort a few things out first:

$\displaystyle 80 km h^{-1} = \frac{200}{9}ms^{-1}$

In case we need it later, angular speed, $\displaystyle \omega = \frac{v}{r} = \frac{200}{9 \times 200} = \frac{1}{9}$

At $\displaystyle 80 km h^{-1}$ the car starts to slip inwards, so this means that the friction is acting **up the slope**, and that the friction is also equal to $\displaystyle \mu N$ where $\displaystyle N$ is the normal reaction of the slope on the car?

Resolving vertically I get $\displaystyle N = mg\cos{\theta}$

Resolving horizontally I get $\displaystyle F = mg\sin{\theta}$

We can rewrite the second equation as $\displaystyle \mu N = mg\sin{\theta}$, and substituting in our first equation we get $\displaystyle \mu mg\cos{\theta} = mg\sin{\theta}$, therefore $\displaystyle \mu = \tan{\theta} = \frac{\sqrt{3}}{3}$?

I have a feeling that Newton's $\displaystyle F = ma$ needs to be used but I'm not sure where?

Not sure how to approach the second question either, this has me a bit stumped.

Thanks in advance for the help