A pendulum having a bob of mas 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`. a. Find the speed of the ship due to rotation of the earth about its axis. b. find the difference between `T_0` 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.

Explanation

When the ship is at rest on the equator the scale reading= mg - centrifugal force
`W_(0) = mg - m omega^(2) R`
The resultant centrigul force acting on the ship due to its linear speed `v_(0)` is `m ( omega +-(v_(0))/(R ))^(2)` R when `omega` and `v_(0)` are along the same drection ( eastward motion ) `+` ve sign is applied, when they are in the opposite direction - ve sign is applied.
The scale reading when the ship is in motion is,
`W = mg - m ( omega - ( v_(0))/( R ))^(2) R`
When `(v_(0))/( R) lt lt omega`, we have
`( omega - ( v_(0))/( R ))^(2) = omega^(2) - ( 2 omega v_(0))/( R )`, neglecting higher power terms.
`W = mg - m ( omega^(2) +- ( 2 omegav_(0))/( R )) R`
`= mg - m omega^(2) R +- R m omega v_(0)`
`= W_(0) +- 2 m omega v_(0) = W_(0)( 1+- ( 2 m omegav_(0))/( W_(0)))`
In this equation position sign refers to the westward motion and negative sign refers to the eastward direction.
Let `W_(0) = mg ` then W is given by
`W = W_(0) ( 1+ ( 2 omega v_(0))/( g ))`