# If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 80 m radius curve banked at 15.0. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 25.0 km/h?

This problem aims to find the velocity of a car running on a curved surface. Also, we are to find the coefficient of friction between the car’s tires and the road. The concept required to solve this problem is related to introductory dynamic physics, which includes velocity, acceleration, coefficient of friction, and centripetal force.

We can define the centripetal force as the force that keeps an object stay in a curvilinear motion which is headed towards the center of the rotational axis. The formula for centripetal force is shown as mass $(m)$ times the square of tangential velocity $(v^2)$ over the radius $(r)$, given as:

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

However, the coefficient of friction is just the ratio of the frictional force $(F_f)$ and the normal force $(F_n)$. It is usually represented by mu $(\mu)$, shown as:

$\mu = \dfrac{F_f}{F_n}$

To start with, if the car bears a curved bank below the ideal speed, some amount of friction is required to hold it from skating inwards of the curve. We are also given some data,

The radius of the curved bank $r = 80m$ and,

The angle of the curved bank $\theta = 15^{\circ}$.

Using the trigonometric formula for $\tan\theta$, we can find the ideal speed $v_i$:

$\tan(\theta) = \dfrac{v_i^2}{r\times g}$

Rearranging for $v_i$:

$v_i^2 = \tan(\theta)\times rg$

$v_i = \sqrt{\tan(\theta)\times rg}$

$v_i = \sqrt{\tan(15)\times 80.0\times 9.8}$

$v_i = 14.49\space m/s$

To determine the coefficient of friction, we will use the formula of frictional force given by:

$F_f = \mu\times F_n$

$F_f = \mu\times mg$

The centripetal force acting on the car with velocity $(v_1)$ can be found by:

$F_1 = m\times a_1 = \dfrac{mv_1^2}{r}$

Substituting the values:

$F_1 = \dfrac{m\times (14.49)^2}{80}$

$F_1 = 2.62m\space N$

Similarly, the centripetal force acting on the car with velocity $(v_2)$ can be found by:

$F_2 = m\times a_2 = \dfrac{mv_2^2}{r}$

Substituting the values:

$F_2 = \dfrac{m\times (6.94)^2}{80}$

$F_2 = 0.6m\space N$

Now the frictional force acting due to the centripetal force can be given as:

$F_f = |F_1 – F_2|$

Substituting the values into the above equation:

$\mu\times m\times g = |2.62m – 0.6m|$

$\mu\times m\times 9.8 = 2.02m$

$\mu= \dfrac{2.02m}{9.8m}$

$\mu = 0.206$

## Numerical Result

Part a: The ideal speed to cover the curved banked is $v_i = 14.49\space m/s$.

Part b: The coefficient of friction needed for the driver is $\mu = 0.206$.

## Example

Imagine that the radius $(r)$ of a curve is $60 m$ and that the advised speed $(v)$ is $40 km/h$. Find the angle $(\theta)$ of the curve to be banked.

Suppose a car of mass $(m)$ covers the curve. The car’s weight, $(mg)$,  and the surface normal $(N)$ can be related as:

$N\sin\theta = mg$

Here $g = \dfrac{v^2}{r}$,

$N\sin\theta = m\dfrac{v^2}{r}$

Which gives:

$\tan\theta = \dfrac{v^2}{rg}$

$\theta = \tan^{-1}(\dfrac{v^2}{rg})$

$\theta = \tan^{-1}(\dfrac{(40\times 1000/3600)^2}{60\times 9.8})$

$\theta = 11.8^{\circ}$