1. ## [SOLVED] Centripetal Force

A car navigates a turn in the road at 60 km/h without sliding. The road is inclined at an angle $\theta$. What does the coefficient of static friction need to be so that the car can take the turn at 90 km/h without skidding?

2. ## Motion in a circle

Hello topher0805
Originally Posted by topher0805
A car navigates a turn in the road at 60 km/h without sliding. The road is inclined at an angle $\theta$. What does the coefficient of static friction need to be so that the car can take the turn at 90 km/h without skidding?
When the car is travelling at 60 km/h (v = 50/3 m/sec), assuming there is no tendency to slide:

R(horiz) $N\sin\theta = \frac{mv^2}{r}= \frac{2500m}{9r}$, where $N$ is normal contact force with road and $r$ = radius of bend

R(vert) $N\cos\theta = mg$

Eliminate $N$:

$\Rightarrow \tan\theta = \frac{2500}{9gr}$ (1)

When speed = 90 km/h (v = 25 m/sec) and car is on point of sliding:

R(horiz) $\mu N\cos\theta + N\sin\theta = \frac{mv^2}{r} = \frac{625m}{r}$ (2)

R(vert) $N\cos\theta = mg + \mu N\sin\theta$

$\Rightarrow N (\cos\theta -\mu\sin\theta) = mg$ (3)

Eliminate $N$ from (2) and (3):

$\frac{\mu\cos\theta + \sin\theta}{\cos\theta -\mu\sin\theta} = \frac{625}{gr}$

$\Rightarrow \frac{\mu + \tan\theta}{1 - \mu\tan\theta} =\frac{625}{gr}$

$= \frac{9}{4}\tan\theta$, from (1)

Now solve for $\mu$ and you're done.