Question

A motorcyclist wants to drive on the vertical surface of wooden 'well' of radius $5$m, with a minimum speed of $5\sqrt{5}$m/s. Find the minimum value of coefficient of friction between the tyres and the wall of the well. (Take $g=10m/s^2$)

Answer 1

Frictional force $=f=mg$

Centripetal Force = Contact force $=N=\frac{mv²}{r}$

$\mu =\frac{f}{N}$

$=\frac{mg}{\left(\frac{mv²}{r}\right)}$

$=\frac{rg}{v²}$

$=\frac{(5\times 10)}{\left(5\sqrt{5}\right)²}$

$=\frac{50}{125}=0.4$

Minimum coefficient of friction between the tyres and wall of the well must be $\mathrm{0.4.}$