# Question 5cce4

Jun 27, 2015

#### Explanation:

As you know, an object that's travelling along a circular motion is experiencing an outward force called the centrifugal force.

In order for a car to be able to turn around a curve of radius $r$, the force of friction exerted between the tires and the road must act as a centripetal force.

That is, the force of friction acts to prevent the car from sliding off the road, i.e. it counteracts the centrifugal force.

Mathematically, this is witten like this

${F}_{\text{centripetal" = F_"friction}}$

$\cancel{m} {v}^{2} / r = \mu \cdot \cancel{m} g \iff {v}^{2} / r = \mu g$

This means that the velocity of the car will be

v = sqrt(r * mu * g) = sqrt(30"m" * 9.8"m"/"s"^2 * 0.4) = "10.84 m/s"

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the radius of the curve and for the coefficient of friction

v_"car" = color(green)("11 m/s")

Jun 27, 2015

The maximum speed the car can have without skidding is 10.85 m/s or 39 km/hr

#### Explanation:

As the road is level, the only forces are the "reaction to the centripetal force" (directed outwards) and the frictional force (directed inwards).
We take the limiting case when the forces balance as only if the reaction is larger the car skids.

Reaction to centripetal Force $F = \frac{m {v}^{2}}{R}$
Force due to friction F= µR=µmg
Therefore µg=v^2/R
(independent of mass of vehicle)
or v=sqrt(µgR)#
substituting
$v = \sqrt{0.4 \times 9.81 \times 30}$
= 10.85 $m {s}^{-} 1$