What is the angle of banking of a railway track of radius of curvature $250 \mathrm{~m}$, if the maximum…
What is the angle of banking of a railway track of radius of curvature $250 \mathrm{~m}$, if the maximum velocity of the train is $90 \mathrm{~km} / \mathrm{h}$ ? $\left(g=10 \mathrm{~ms}^{-2}\right)$
$\theta=\tan ^{-1}\left(\frac{1}{2}\right)$
$\theta=\tan ^{-1}\left(\frac{1}{3}\right)$
$\theta=\tan ^{-1}\left(\frac{1}{4}\right)$
$\theta=\tan ^{-1}\left(\frac{1}{5}\right)$
Solution
Given, radius of curvature, $r=250 \mathrm{~m}$
Maximum velocity of train, $v=90 \mathrm{kmh}^{-1}$
$
=\frac{90 \times 5}{18} \mathrm{~m} / \mathrm{s}=25 \mathrm{~m} / \mathrm{s}
$
Let $\theta$ be the angle of banking of railway track.
Then, by using an expression of maximum safe speed,
$
v_2=r g \tan \theta \Rightarrow \tan \theta=\frac{v^2}{r g}
$
By substituting the values, we get
$
\begin{aligned}
& \tan \theta=\frac{(25)^2}{250 \times 10}=\frac{625}{2500}=\frac{1}{4} \\
\therefore \quad \theta & =\tan ^{-1}\left(\frac{1}{4}\right)
\end{aligned}
$