Solve the following :


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 To. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between To 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 $\omega$ and radius of the earth is $R$.


(a) Since, ship is at rest with respect to water.

So, Angular speed of ship=Angular speed of earth



$V_{s}=R \omega$

(b) The tension in string is given by

$T_{0}=m g^{\prime}-m R \omega^{2}$

$m g^{\prime}-T_{0}=m R \omega^{2}$

(c) If ship sails at speed v, then tension

$T=m g^{\prime}-m R \omega^{z^{2}}$

$T=m g^{\prime}-m R\left(\frac{V-R \omega}{R}\right)^{2}$

$T=m g^{\prime}-\frac{m\left(V^{2}+R^{2} \omega^{2}-2 V R \omega\right)}{R}$

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