Question

What should be the angular velocity of the rotation of the earth about its own axis so that the weight of the body at the equator reduces to $\dfrac{3}{5}$ of its present value? (Take R as the radius of the earth)
${\text{A}}{\text{. }}\sqrt {\dfrac{g}{{3R}}} \\\ {\text{B}}{\text{. }}\sqrt {\dfrac{{2g}}{{3R}}} \\\ {\text{C}}{\text{. }}\sqrt {\dfrac{{2g}}{{5R}}} \\\ {\text{D}}{\text{. }}\sqrt {\dfrac{{2g}}{{7R}}} \\\$

Answer 1

Hint: If we consider the mass of the man to be m, then at the equator, the mass will reduce by a factor of $\dfrac{3}{5}$. The students must take into consideration the centrifugal force which acts towards the outward direction in a circular motion. Thereafter, equating the forces acting on the man at the equator, the students can find the answer.

Complete step by step solution:

In the figure, the curve with an arrow indicates the direction of rotation of the earth about its axis of rotation. The equator is also indicated in the figure.
Now, when the man is at the equator, he will be experiencing gravitational force towards the centre of the earth because of his weight and also centrifugal force away from the centre because of the circular motion executed by the earth (rotation).
Let us consider that the angular velocity of rotation of the earth is $\omega$ and the mass of the man is m. The radius of the earth is given to be R. Now, as stated in the question that the mass of the man at the equator becomes $\dfrac{3}{5}$ of its original value, so let us equate the forces acting on the man at the equator,
$\dfrac{3}{5}mg = mg - m{\omega ^2}R$
where g is the acceleration due to gravity and the second term $m{\omega ^2}R$ is the centrifugal force. We have got this equation from the fact that at the equator, the weight of the man is reduced due to the centripetal force acting on him.
Now, from the above equation, we get,
$\dfrac{3}{5}g = g - {\omega ^2}R \\\ \Rightarrow \dfrac{2}{5}g = {\omega ^2}R \\\ \Rightarrow {\omega ^2} = \dfrac{{2g}}{{5R}} \\\ \Rightarrow \omega = \sqrt {\dfrac{{2g}}{{5R}}} \\\$
Hence, the correct answer is option C.

Note: We know that the value of acceleration due to gravity changes at different places on earth but in this question, we have related the mass of the man at the equator to his original mass, so we do not need to consider different value of g, but if we are to relate the mass of the man at the equator with the mass of the man at the pole, then we will have different g on both sides of the equation.