Question

A body is suspended on a spring balance in a ship sailing along the equator with a speed $$v'$$. If $$\omega$$ is the angular speed of the earth and $$\omega_0$$ is the scale reading when the ship is at rest, the scale reading when the ship is sailing, will be very close to :

Answer 1

Correct option is C. $$w_0\left(1\mp \dfrac{2\omega v'}{g}\right)$$
When the ship is at rest: $$g_e=g-\omega^2R$$ ... (1)

When the ship is moving with velocity v' from west to east

$$\omega'=\omega_e+\omega_b$$

When the ship is moving east to west

$$\omega'=\omega_e-\omega_b$$

On combing the equations we can write:

$$\omega'=\omega_e\pm\omega_b$$

$$\Rightarrow \omega'=\omega\pm\dfrac{v'}{R}$$

g effective of the ship then becomes:

$$g_e'=g-R(\omega\pm\dfrac{v'}{R})^2$$

$$\Rightarrow g_e'=g-R(\omega^2\pm\dfrac{2\omega v'}{R}+\dfrac{v'^2}{R^2})$$

The last term is very small and can be neglected

$$\Rightarrow g_e'=g-R(\omega^2\pm\dfrac{2\omega v'}{R})$$

$$g_e'=g-R\omega^2\mp2\omega v'$$

From Eq 1:

$$g_e'=g_e\mp2\omega v'$$

$$g_e'=g_e(1\mp\dfrac{2\omega v'}{g_e})$$

multiplying boths sides by the mass m and using weight w=mg, we get,

$$\Rightarrow W'=W_o(1\mp\dfrac{2\omega v'}{g_e})$$