Question

A body weights $W_r$ in a train at rest. The train now begins to move with a velocity of v around the equator from west to east. The angular velocity of the earth is $\omega$. The same body in this train will now weight as

• A. $W_r\times \frac{2v\omega }{g}$
• B. $W_r(1-\frac{2v\omega }{g})$
• C. $W_r\left(\frac{g-v\omega }{g}\right)$
• D. $W_r(1-\frac{3v\omega }{g})$

Answer 1

The correct option is B $W_r(1-\frac{2v\omega }{g})$

Relative acceleration of the train w.r.t
$a=\frac{(v+R\omega {)}^2}{R}$ towards the center of the earth. Thus the apparent weight
$W=m(g-a)=m[g-\frac{v^2+R{V}^2{\omega }^2+2vR\omega }{R}]$
or
$W=mg[1-\frac{v^2}{Rg}-\frac{R{\omega }^2}{g}-\frac{2v\omega }{g}]$
or
$W=v_o[1-\frac{2v\omega }{g}]$ (as $\frac{v^2}{Rg}$ and $\frac{R{\omega }^2}{g}$ are negligibly small)