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:

\[ \mu = \dfrac{F_f}{F_n}\]

## Expert Answer

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,** $(mg)$,Â 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:**

\[\tan\theta = \dfrac{v^2}{rg}\]

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

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

\[\theta = 11.8^{\circ}\]