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Question: A pendulum having a bob mass \(m\) is hanging in a ship sailing along the equator from east to west....

A pendulum having a bob mass mm is hanging in a ship sailing along the equator from east to west. When the strip is stationary with respect to water, the tension in the string is T0T_{0}. The difference between T0T_{0} and earth attraction on the bob is:
A.mg+mω2R2 B.mω2R3 C.mω2R2 D.mω2R \begin{aligned} & A.\dfrac{mg+m{{\omega }^{2}}R}{2} \\\ & B.\dfrac{m{{\omega }^{2}}R}{3} \\\ & C.\dfrac{m{{\omega }^{2}}R}{2} \\\ & D.m{{\omega }^{2}}R \\\ \end{aligned}

Explanation

Solution

We know that the time period T=2πLgT=2\pi \sqrt{\dfrac{L}{g}} where LL is the length of the simple pendulum and gg is the acceleration due to gravity. Due to the rotation of the earth, the pendulum experiences a centripetal force, pulling the pendulum towards the earth.
Formula used: T0=mg+mω2RT_{0}=mg+m\omega^{2}R

Complete step-by-step solution:
Let us assume that the earth rotates along its axis with an angular velocity ω\omega. Given that the mass of the pendulum is mm. Let the LL is the length of the simple pendulum. Letθ\theta be the angle made by the pendulum at the point of contact.
Now the pendulum experiences tension T0T_{0}, which acts radially to the pendulum, along the lengths of the pendulum.
The pendulum also experiences a normal force that is given by mgmg and a centripetal force which is given by CF=mω2RCF=m\omega^{2}R where RR is the radius of the earth. Taking the components we get mgsinθmgsin\theta and mgcosθmgcos\theta

We now form the free body diagram that the net force acting along the positive y-axis is equal to the net force acting along the negative y-axis. Then we can say the tension T0=mg+mω2RT_{0}=mg+m\omega^{2}R.
Then the difference between the T0T_{0} and earth attraction on the bob will be T0mg=mω2RT_{0}-mg=m\omega^{2}R.
Hence the difference between the T0T_{0} and earth attraction on the bob will be mω2Rm\omega^{2}R
Thus the answer is D. mω2Rm\omega^{2}R.

Note: This might seem like a complex question but it can be solved easily if the concept of centripetal force and the formulas are known. This question is asked frequently. To understand the question, students are suggested to draw the free body diagram of the pendulum.