Angular speed of rotation of earth is $ω_{0}$ . A train is running along the equator at a speed ‘v’ from west to east with respect to surface of earth. A very sensitive balance inside the train shows the weight of the object as $W_{1}$ . During the return journey when the train is running at same speed from east to west (with respect to surface of earth) the balance shows the weight of the object to be $W_{2}$ . Weight of the object shown by the balance at the poles is $W_{0}$ . The value of $W_{2} - W_{1}$ was found out to be $\frac{p W_{0} ω_{0} v}{g}$ (g is the acceleration due to gravity on the surface of the earth). Find the value of ‘p’?

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Detailed Solution

When train is at rest $W_{0} = m g (V_{0} = W_{0} R)$
When train moves from west to east $W_{1} = m g - \frac{m {(v_{0} + v)}^{2}}{R}$
When train moves from east to west $W_{2} = m g - \frac{m {(v_{0} - v)}^{2}}{R}$
$\Rightarrow W_{2} - W_{1} = \frac{m}{R} [{(v_{0} + v)}^{2} + {(v_{0} + v)}^{2}] = \frac{m}{R} 4 v_{0} v = \frac{4 m v_{0} v}{R} = \frac{4 m v_{0} v}{R} = 4 \frac{W_{0}}{g} W_{0} v$