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 tyre and ground? (Ans : 5° 51', 0.1020)

Answer 1

given, speed of motorcycle , v = 5m/s
radius of circle , r = 25m
acceleration due to gravity, g = 9.8 m/s²

Let angle of inclination is $\theta$.
coefficient of friction is $\mu$.

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

= $tan^{-1}\left(\frac{5^2}{25\times9.8}\right)$

= $tan^{-1}(0.1020)$ = 5°51'

now, we know, necessary centripetal force is provided by the friction between the road and the tyres.
so, F = mv²/r = $\mu mg$

$\mu$ = v²/2g = (5)²/2 × 9.8 = 0.102

Answer 2

Answer:

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