Question

A car of mass 1,325 kg makes a circular turn of radius 16.18 m along a level roadway. The coefficient of friction is 0.826 between the tires and the road.How fast (in m/s) can the car go without skidding off the turn?(Use the approximation that g ≈ 10 m/s^2)Answer: ________ m/s (round to the nearest hundredth)

Answer 1

Given:

The mass of the car is,

$m=1325\text{ kg}$

The radius of the circular path is,

$r=16.18\text{ m}$

The coefficient of friction between the tires and the road is,

$\mu=0.826$

The acceleration due to gravity is,

$g=10\text{ m/s}^2$

To find:

How fast (in m/s) can the car go without skidding off the turn

Explanation:

The frictional force balances the centripetal force.

The frictional force is,

$f=\mu mg$

The centripetal force is,

$F=\frac{mv^2}{r}$

Here, v is the speed of the car without skidding.

We can write,

$\begin{gathered} \frac{mv^2}{r}=\mu mg \\ v=\sqrt{\mu gr} \end{gathered}$

Substituting the values we get,

$\begin{gathered} v=\sqrt{0.826\times10\times16.18} \\ v=11.6\text{ m/s} \end{gathered}$

Hence, the required speed is 11.6 m/s.