JEE Main & Advanced Physics Two Dimensional Motion Banking of A Road

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)              

\[\tan \theta =\frac{{{v}^{2}}}{rg}\]\[=\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}=\frac{h}{l}\]

Note : 

• If friction is also present between the tyres and road then \[\frac{{{v}^{2}}}{rg}=\frac{\mu +\tan \theta }{1-\mu \tan \theta }\]              
• Maximum safe speed on a banked frictional road \[v=\sqrt{\frac{rg(\mu +\tan \theta )}{1-\mu \tan \theta }}\]