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 T.
a. Find the speed of the ship due to rotation of the earth about its axis.
b. Find the difference between T, and the earth’s attraction on the bob.
c. If the ship sails at speed u, what is the tension

Answer 1

Recall the concept of tension. When two or more objects are in contact with each other then they exert forces on each other. Tension is also one such force. It is a pair of action-reaction forces that act at the end points along the length of the object.

Complete step by step answer:
Step I:
It is given that the bob has a mass $m$ and the ship sails from east to west. Suppose the angular speed of the rotation of earth is $\omega$ and the radius of Earth is R, then the speed of the ship due to the rotation of the Earth about its axis is given by
$\Rightarrow v = \omega R$

Step II:
The tension in the string of the bob is $T$ and the weight of the bob due to Earth’s attraction is $mg$. The centripetal force will be responsible for the inward motion of the bob. This force is given by $m{\omega ^2}R$
The tension force is then written as
${T_0} = mg + m{\omega ^2}R$---(i)
${T_0} - mg = m{\omega ^2}R$

Step III:
The rate of change of angular displacement when the object is rotating is known as angular velocity. An object having angular velocity will also have linear velocity. Their relation is written as
$v = {\omega _1}R$---(ii)
Where v is the linear velocity
And R is the radius
When the ship is stationary, the sailing speed is $v = \omega R$
From equation (i)
The tension in the string is written as
$T = mg + m{({\omega _1} + \omega )^2}R$---(iii)
The tension due to rotation of Earth is
${T_0} = mg + m{\omega ^2}R$---(iv)

Step IV:
Subtracting (iii) and (iv),
$T - {T_0} = mR[{({\omega _1} + \omega )^2} - {\omega ^2}]$
$T - {T_0} = mR(\omega _1^2 + {\omega ^2} + 2\omega {\omega _1} - {\omega ^2})$
$\Rightarrow T - {T_0} = mR(\omega _1^2 + 2\omega {\omega _1})$---(v)

Step V:
From equation (ii),
$\Rightarrow {\omega _1} = \dfrac{v}{R}$
Here the radius of Earth is very large and the velocity of the ship will be small. Therefore, ${\omega ^2}$ will have infinitely small values. So, it can be neglected.
Therefore equation (v) becomes,
$T - {T_0} = mR(2\omega {\omega _1})$
Substitute the value of ${\omega _1}R$ from equation (ii) in the above equation,
$T - {T_0} = 2mv\omega$
$\Rightarrow T = {T_0} + 2mv\omega$

Note: It is important to note that the force of tension will always point away from the string along the direction of the bob. But it always acts in an opposite direction to that of gravitational force. Since the bob is hanging, the force of tension acting should be balanced otherwise the bob will accelerate down due to the effect of gravitational force.