Question

A rail track is banked for speed v by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. the radius of curvature of the track is r, then

• A. $\frac{h}{d}=\frac{v^2}{rg}$
• B. $tan\left({sin}^{-1}\frac{h}{d}\right)=\frac{v^2}{rg}$
• C. ${tan}^{-1}\frac{h}{d}=\frac{v^2}{rg}$
• D. $\frac{h}{d}=\frac{v^2}{dg}$

Answer 1

The correct option is B $tan\left({sin}^{-1}\frac{h}{d}\right)=\frac{v^2}{rg}$

Horizontal component of $N$ provides the necessary centripetal force. So applying Newton's laws in horizontal and vertical directions. we get

$N\sin \theta =\frac{m{v}^2}{r}$ ... $\left(I\right)$ and $N\cos \theta =mg$ ... $\left(II\right)$

from these two equations, we get

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

and $AB=d;BC=h$

$\sin \theta =\frac{h}{d}\Rightarrow \theta ={\sin }^{-1}\left(\frac{h}{d}\right)$ ... $\left(IV\right)$

By $\left(III\right)$ and $\left(IV\right)$

$\tan \left({\sin }^{-1}\frac{h}{d}\right)=\frac{v^2}{rg}$