Question

A motor cyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with vertical. What is the value of coefficient of friction between type and ground ?

Answer 1

Given, speed of motorcycle , v = 5 m/s

radius of circle, r = 25 m

acceleration due to gravity , g = 9.8 $m/s^2$

Let angle of inclination is $\theta$.

coefficient of friction is $\mu$.

$\theta =ta{n}^{-1}\left(\frac{v^2}{rg}\right)$

= $ta{n}^{-1}\left(\frac{5^2}{25\times 9.8}\right)$

= $ta{n}^{-1}\left(0.1020\right)=5^0{51}^{\prime }$

Now, we know,

necessary centripetal force is provided by the friction

between the road and the tyres,

so, F = $\frac{m{v}^2}{r}$ = $\mu mg$

$\mu =\frac{v^2}{2g}$

= $\frac{5^2}{2\times 9.8}$

$=0.102$