Banking of A Road
Category : JEE Main & Advanced
For getting a centripetal force, cyclist bend towards the centre of circular path but it is not possible in case of four wheelers.
Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre.
In the figure (A) shown reaction R is resolved into two components, the component R cos \[\theta \] balances weight of vehicle
\[\therefore \] \[R\,\,\cos \,\,\theta =mg\] ...(i)
and the horizontal component R sin \[\theta \] provides necessary centripetal force as it is directed towards centre of desired circle
Thus \[R\,\,\sin \,\,\theta =\frac{m{{v}^{2}}}{r}\] ...(ii)
Dividing (ii) by (i), we have
\[\tan \,\theta =\frac{{{v}^{2}}}{r\,g}\] ...(iii)
or \[\tan \theta =\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}\] ...(iv) \[[As\,\,\upsilon =r\omega ]\]
If l = width of the road, h = height of the outer edge from the ground level then from the figure (B)
\[\tan \theta =\frac{h}{x}=\frac{h}{l}\] ...(v) [since \[\theta \] is very small]
From equation (iii), (iv) and (v)
\[\tan \theta =\frac{{{v}^{2}}}{rg}\]\[=\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}=\frac{h}{l}\]
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