# Question #c6aa7

May 16, 2014

First we'll start by looking at the forces acting on the car. There is the centripetal force causing the car to turn in a circle, the weight of the car and the normal force. In this case, the centripetal force is due to the friction between the car and the road. If the centripetal force is greater than the force of friction, the car won't be able to make the turn.

Mathematically this can be expressed using Newton's 2nd Law
$\sum F = m a$
Centripetal acceleration is equal to ${v}^{2} / r$, the frictional force is equal to $\mu {F}_{N}$, since the road is flat, normal force is equal to the weight of the car $m g$

Combining all of these together we get:
$\sum F = m a$
${F}_{f}$ $= m {v}^{2} / r$
$\mu {F}_{N}$ $= m {v}^{2} / r$
$\mu m g = m {v}^{2} / r$ <-- mass can be canceled out
$\mu g = {v}^{2} / r$
$v = {\left(\mu g r\right)}^{\frac{1}{2}}$

Looking up the value for $\mu$, assuming rubber tires on asphalt, we get a value of 0.85. We're using static friction because the car isn't slipping.

Plugging in what we know:
$v = {\left(0.85 \cdot 9.81 \frac{m}{s} ^ 2 \cdot 25 m\right)}^{\frac{1}{2}}$
$v = 14.4 \frac{m}{s}$