Question

If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that scooter going at 18 km/hr does not skid?

Answer 1

The minimum friction coefficient so that scooter going at 18 km/hr does not skid is 0.3.

Explanation:

If the road horizontal (no banking)

We have

$\begin{equation}\frac{m v^{2}}{r}=f_{z}$

Where $f_s$ is the force of friction

N=mg

N is the normal reaction  

If the μ is the friction coefficient  

We have  

Friction force, $f_s$ = μN

So ,

$\begin{equation}\frac{m v^{2}}{r}=\mu N\end$

( N = mg )

$\begin{equation}\frac{m v^{2}}{r}=\mu m g\end$

$\begin{equation}\frac{v^{2}}{r}=\mu g \ldots \ldots \ldots e q^{n}(1)\end$

Given values in previous question  

It’s rider moving at 36 km/hr

Speed ( v ) = 36 km /hr  

Converting km / hr to meter per second  

$\begin{equation}\mathrm{Km} / \mathrm{hr}=\frac{36000 \mathrm{m}}{3600 \mathrm{s}}$

Speed (v) = 10 m/s  

Radius, r = 30 m    

Let g = 10 m/s

Put the values in $eq" ( 1 )$

$\begin{equation}\begin{aligned}&\frac{10^{2}}{30}=10 \mu\\&\frac{100}{30}=10 \mu\\&\frac{100}{300}=\mu\end{aligned}$

$\mu=\frac{1}{3} = 0.3$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

The minimum friction cofficent so that the scooter going with 18 km/hr

does not skid is 0.3