Question

A body suspended on a spring balance in a ship weighs ${}^{\prime }{W}_0{}^{\prime }$ when the ship is at rest. When the ship begins to move along the equator with a speed ${}^{\prime }v{}^{\prime }$, the scale reading is very close to $W_0(1\pm 2\omega v/g)$, where ${}^{\prime }\omega {}^{\prime }$ is the angular speed of the earth.

Answer 1

When at rest, $W_o=mg-m{\omega }^{2}R\approx mg$, as $g>>{\omega }^{2}R$ where $\omega =7.3\times {10}^{-5}radians/second$ is angular speed of earth about its axis, and $R=6.4\times {10}^{6}m$ is radius of earth.

now when ship moves with speed $v$, the effective angular speed of body is $\omega \pm \frac{v}{R}$,

so new weight is

$W=m(g-(\omega \pm \frac{v}{R}{)}^{2}R)=mg-m{\omega }^{2}R+m{v}^2/R\pm 2\omega v=mg\pm 2m\omega v$, ($\because g>>{\omega }^{2}R,v^2/R\approx 0$)

$\Rightarrow W=mg(1\pm 2\omega v/g)=W_o(1\pm 2\omega v/g)$