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

Explanation:

It is given that the pendulum with a bob of mass m hanging in a ship is sailing along the equator from east to west and the tension in the string is $T_0$ when the ship is stationary with respect to water.  

(a) The speed of the ship due to rotation of the earth about its axis:  

$\mathrm{V}=\omega R$

(b) The difference between $T_0$ and the earth's attraction on the bob:$T_{0}=m g r=m g-m \omega^{2} R$ Therefore, the difference is $T_{0}-m g=m \omega^{2} R$.

(c) If the ship sails at speed v, with angular speed of earth's rotation ω and radius of the earth R, the tension in the string:

The tension can be defined as $T=m g-m \omega_{1}^{2} R.$Thus, $T=T_{0}+2 \omega v m$.

Answer 2

a)v=Rw (formula)

b)earth attraction= GMm/R^2

=mg

according to figure

t0+mRw^2=mg

t0=mg-mRw^2

hence

=mg-t0

=mRw^2

c)t0+2wvm