Question

A train has to negotiate a curve of $400$m. By how much should the outer sail be raised w.r.t. the inner rail for a speed of $48$ km/h? (given: Distance between rail $=1$m].

Answer 1

$v=48kmph=\frac{40}{3}m/s$

We know that

$\tan \theta =v^2/rg$ , where $r=$radius of curvature

From figure it is clear that $\tan \theta =\frac{h}{1}=\frac{v^2}{rg}$

for $r=400m,v=\frac{40}{3}$

$h=\frac{(40/3{)}^2}{400\times 10}$

$h=\frac{2}{45}m$