Question

A railway track goes around a curve having a radius of curvature of 1km the distance between the rails is 1 m find the elevation of the outer rail above the inner rail so that there is no side pressure against the rail when a train goes round the curve At 36km/hr (ans =1.02)​

Answer 1

Answer:

Answer:

Elevation of outer rail above the inner rail = $\underline{\boxed{\tt \purple{0.01 \:m}}}$

Explanation:

Given:

Radius of curvature (r) = 1 km = 1000 m

Distance between the rails (d) = 1 m

Speed of train (v) = 36 km/h

To Find:

Elevation of outer rail above inner rail (h).

Now, firstly we will convert speed into m/sec,

$\implies{\sf{\dfrac{36\times 5}{18}\;m/sec}}$

$\implies{\bf{10\;m/sec}}$

Now, the angle of elevation would be,

$\implies{\sf{\theta = \tan^{-1}\bigg(\dfrac{v^{2}}{rg}\bigg)}}$

Let, g = 10 m/s

$\implies{\sf{\theta = \tan^{-1}\bigg(\dfrac{10^{2}}{1000\times 10}\bigg)}}$

$\implies{\sf{\theta = \tan^{-1}\bigg(\dfrac{100}{10000}\bigg)}}$

$\implies{\sf{\theta = \tan^{-1}\bigg(\dfrac{1}{100}\bigg)}}$

$\implies{\boxed{\sf{\theta =0.01\;rad}}}$

Now, the elevation of outer rail will be,

$\implies{\sf{h=d\sin \theta}}$

$\implies{\sf{h=1\times \sin\times 0.01}}$

$\implies{\boxed{\tt{h=0.01\;m}}}$