Question

If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking ?

Answer 1

In the diagram

R cos $\theta$ = mg ... (i)

R sin $\theta =\left(\frac{M{v}^2}{r}\right)$ ....(ii)

Dividing equation (i) with equation (ii), we get,

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

Here, v = 36 km/hr = 10 m/sec, r = 30 m

$\therefore tan\theta =\frac{v^2}{rg}=\frac{100}{30\times 10}=\left(\frac{1}{3}\right)\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{3}\right)$