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 T₀. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T₀ and the earth's attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string? Angular speed of earth's rotation is ω and radius of the earth is R.

Answer 1

(a) Speed of the ship due to rotation of the Earth is v = ωR, where R is the radius of the Earth and ω is its angular speed.

(b) The tension in the string is given by
T₀ = mg − mω²R
∴ T₀ − mg = mω²R

(c) Let the ship move with a speed v.
Then the tension in the string is given by

$$T=mg-m{\omega }_1^{2}R$$

$$=T_0-\frac{{(v-\omega R)}^2}{R^2}R$$

$$=T_0-\frac{(v^2+{\omega }^2{R}^2-2\omega Rv)}{R}R$$

$$\therefore T=T_0+2\omega vm$$