Question

A train of mass $m$ moves with a velocity $v$ on the equator from east to west. If $\omega$ is the angular speed of earth about its axis and $R$ is the radius of the earth then the normal reaction acting on the train is

• A. $mg[1-\frac{(\omega R-2v)\omega }{g}-\frac{v^2}{Rg}]$
• B. $mg[1-2\frac{(\omega R-v)\omega }{g}-\frac{v^2}{Rg}]$
• C. $mg[1-\frac{(\omega R-2v)\omega }{g}+\frac{v^2}{Rg}]$
• D. $mg[1-2\frac{(\omega R-v)\omega }{g}+\frac{v^2}{Rg}]$

Answer 1

The correct option is A $mg[1-\frac{(\omega R-2v)\omega }{g}-\frac{v^2}{Rg}]$

The Earth rotates from west to east.


So, the net velocity of train,

$v_{net}=\omega R-v$

Balancing force for the circular motion of the train,

$mg-N=\frac{m{v}_{net}^2}{R}$

$\Rightarrow N=mg-\frac{m(\omega R-v{)}^2}{R}$

$\Rightarrow N=mg-\frac{m}{R}({\omega }^2{R}^2+v^2-2\omega Rv)$

$\Rightarrow N=mg-m({\omega }^{2}R+\frac{v^2}{R}-2\omega v)$

$\Rightarrow N=mg[1-\frac{(\omega R-2v)\omega }{g}-\frac{v^2}{Rg}]$

Hence, option (a) is the correct answer.