Question

A body weighs ${W_r}$ 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 is this train will now weight as
A. ${W_r} \times \dfrac{{2v\omega }}{g}$
B. ${W_r}\left( {1 - \dfrac{{2v\omega }}{g}} \right)$
C. ${W_r}\left( {\dfrac{{g - v\omega }}{g}} \right)$
D. ${W_r}\left( {1\dfrac{{3v\omega }}{g}} \right)$

Answer 1

Weight is the product of mass and acceleration. Due to motion of the train, relative acceleration of the body with respect to ground with change and hence the relative weight will change. Calculate the relative acceleration towards the center of earth to calculate the relative weight.

Complete step by step answer:
Acceleration due to gravity acts towards the center of the earth. Now, when the train moves along the equator of the earth, radial acceleration will act on the train which will act along the center of the earth, but away from the center of the earth.
Thus, the net acceleration acting along the center of the earth will be $(g - a)$.
Now, we know that weight is equal to the product of mass and acceleration acting towards the center of the earth.
Therefore, apparent weight, when the train is moving is given by
${W_a} = m(g - a)$ . . . (1)
Now, radial acceleration is given by
${a_r} = \dfrac{{{v^2}}}{r}$
Where,
$v$ is velocity and
$r$ is the radius of the circle on which the motion is being done.
Since, the earth rotates from west to east and the train is also moving in the direction west to east, the relative velocity ${v_r}$ of earth will be the sum of linear velocity of the train and linear velocity of earth.
$\Rightarrow {v_r} = v + R\omega$
Where,
$R$ is the radius of earth
$\omega$ is the angular velocity
$v$ is linear velocity.
$\therefore {a_r} = \dfrac{{{v_r}^2}}{R}$
$\Rightarrow {a_r} = \dfrac{{{{(v + R\omega )}^2}}}{R}$
Therefore, from equation (1), we get
${W_a} = m(g - {a_r})$
$\Rightarrow m\left( {g - \dfrac{{{{(v + R\omega )}^2}}}{R}} \right)$
On simplifying it, we get
$\Rightarrow m\left[ {g - \dfrac{{{v^2}}}{R} - \dfrac{{2vR\omega }}{R} - \dfrac{{{R^2}{\omega ^2}}}{R}} \right]$
$\Rightarrow m\left[ {g - \dfrac{{{v^2}}}{R} - 2v\omega - R{\omega ^2}} \right]$
Multiplying and dividing the above equation by $g$, we get
${W_a} = mg\left[ {1 - \dfrac{{{v^2}}}{{Rg}} - \dfrac{{2v\omega }}{g} - \dfrac{{R{\omega ^2}}}{g}} \right]$ . . . (2)
Now observe
$R$ is much greater than $v$. Therefore, $\dfrac{{{v^2}}}{{Rg}}$ is nearly equal to zero.
Thus we can neglect the term $\dfrac{{{v^2}}}{{Rg}}$ from equation (2)
Also, we know that the earth rotates in such a slow pace that we don’t even feel it. That means $\omega$ is very small. That means ${\omega ^2}$ is much smaller than $\omega$. We also know that when a large quantity multiplied with a very small quantity then the result is small too.
For example: $5000 \times 0.001 = 5$
Therefore, though $R$ is large, $R{\omega ^2}$ is very small.
Therefore, we can also neglect her term $\dfrac{{R{\omega ^2}}}{g}$ from equation (2).
Therefore, equation (2) becomes
${W_a} = mg\left[ {1 - \dfrac{{2v\omega }}{g}} \right]$
Therefore, the same body in the train will weight as ${W_r}\left[ {1 - \dfrac{{2v\omega }}{g}} \right]$

Therefore, from the above explanation, the correct answer is, option (B) ${W_r}\left[ {1 - \dfrac{{2v\omega }}{g}} \right]$

Note: To solve this question, you need to know the concept of relative motion. Also, in this question you need to understand how to neglect terms which do not differ our solution much. Neglecting appropriate terms helps in simplifying the complex equations without much effort and time.