A pendulum having a bob of mass $m$ is hanging in a ship sailing along the equator from east to west

Question

A pendulum having a bob of mass $m$ is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water, the tension in the string is $T_0$. If the ship sails at speed $v$ (where $v \ll \omega R$), what is the approximate tension in the string? (Angular speed of earth's rotation is $\omega$ and radius of the earth is $R$)

Professor PC Thomas Classes · Answer

The earth rotates west to east with speed $\omega R$. A ship sailing east to west at speed $v$ has an absolute speed of $(\omega R - v)$. Tension $T = mg - m\frac{(\omega R - v)^2}{R} = mg - m\omega^2 R + 2m\omega v - \frac{mv^2}{R}$. Since $T_0 = mg - m\omega^2 R$ and $v^2$ is negligible, $T \approx T_0 + 2m\omega v$.