Question

A motorcyclist wants to drive on the vertical surface of wooden ${}^{\prime }wel{l}^{\prime }$ of radius $5m$, 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

As the walls are vertical so friction alone will balance weight:

$\Rightarrow f=mg\to \left(1\right)$

Also normal force exerted by wall will provide the centripetal force:

$N=\frac{m{v}^2}{r}\to \left(2\right)$

Now $f=\mu N$ (where $\mu =$ coefficient of friction)

$\Rightarrow \mu =\frac{mg}{\frac{m{v}^2}{r}}$ (Using equation 1 and 2)

$\Rightarrow \mu =\frac{rg}{v^2}=\frac{5\times 10}{(5\sqrt{5}{)}^2}=0.4$