Question

A body suspended on a spring balance inside the spaceship weighs $W$ when ship is at rest at the earth's equator. If $W_0$ is the weight of the body in the stationary ship when we assume earth not rotating about its axis and $\omega$ is the angular velocity of rotation of earth then weight $W$ is

• A. $W_0(1+\frac{R{\omega }^2}{g})$
• B. $W_0{(1+\frac{R{\omega }^2}{g})}^{-1}$
• C. $W_0{(1+\frac{R{\omega }^2}{g})}^{1/2}$
• D. $W_0{(1+\frac{R{\omega }^2}{g})}^{-1/2}$

Answer 1

The correct option is B $W_0{(1+\frac{R{\omega }^2}{g})}^{-1}$

$W_0$ is the weight of the body inside the stationary ship at equator when earth is assumed to be not rotating about its own axis and
$W$ is the weight of the body in the stationary ship at equator when earth is rotating with angular speed $\omega$, then the weight of the body is reduced due to reaction of centripetal force by a factor of $\frac{m{v}_2^2}{R}(\because v_e=R\omega )$
$v_e=$ Velocity of stationary ship at the equator of rotating earth.
$\therefore W=W_0-\frac{m{v}_e^2}{R}=W_0-mR{\omega }^2$
But $m=\frac{W}{g}$ (since, earth is actually rotating about its own axis)
$\therefore W=W_0-\frac{WR{\omega }^2}{g}$
$W(1+\frac{R{\omega }^2}{g})=W_0\Rightarrow W=\frac{W_0}{(1+\frac{R{\omega }^2}{g})}$
Hence, the correct answer is option (b).