Question

If a man at the equator would weigh $\dfrac{3}{5}th$ of his weight, the angular speed of the earth is
A.$\sqrt {\dfrac{2}{5}\dfrac{g}{R}}$
B.$\sqrt {gR}$
C.$\sqrt {Rg}$
D.$\sqrt {\dfrac{2}{5}\dfrac{R}{g}}$

Answer 1

Here we have to find the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh $\dfrac{3}{5}$ as much as the present. The weight we feel at the equator of the earth’s surface can be stated in terms of the original weight and the centripetal force undergone by the body due to the rotation of the earth. Find the mathematical expression and put the given values to obtain the required answer.

Complete answer:
Earth is rotating on its own axis. Thus, every object in the equator of the earth will undergo a centripetal force. This centripetal force is because of the gravitational force of the earth. At the equator, the mass will decrease by a factor of $\dfrac{3}{5}$ . We must take into account the centrifugal force which acts in the outward direction in a circular motion.
The centrifugal force is formulated by, $F = mr{\omega ^2}$ , $\omega$ is the angular velocity, $r$ is the radius of the earth, and $m$ is the mass.
The cosine component of the force that holds the weight of the object is,
$\lambda$ is the angle at which the force is divided into the components.
Now, the actual weight on the equator $= W = mg$
Here and $'g$' is the acceleration due to gravity.
According to the specified condition,
The weight on the equator of the earth $= W' = mg'$
Weight on the equator $= W' = \dfrac{3}{5}W = \dfrac{3}{5}{\text{ }}mg\;{\text{ }}\;...........{\text{ }}\left( 1 \right)$
We know that, $\lambda = 0$ at the equator.
Now,
$mg' = mg - mR{\omega ^2}cos\;\lambda$
$\dfrac{3}{5}mg = mg - mR{\omega ^2}cos0\;{\text{ }}(\because \lambda = {\text{ }}0{\text{ }})$
$(\because cos{\text{ }}0{\text{ }} = 1)$
$\Rightarrow \dfrac{3}{5}mg = mg - mR{\omega ^2}$
$\Rightarrow mR{\omega ^2} = mg - \dfrac{3}{5}mg$
$\Rightarrow mR{\omega ^2} = \left( {1 - \dfrac{3}{5}} \right)mg$
$\Rightarrow mR{\omega ^2} = \dfrac{2}{5}mg$
$\Rightarrow R{\omega ^2} = \dfrac{2}{5}g$
$\Rightarrow {\omega ^2} = \dfrac{{2g}}{5}R$
$\Rightarrow \omega = \sqrt {\dfrac{2}{5}\dfrac{g}{R}}$.

Hence option A is correct.

Note:
Centripetal force is a force acting on the body traveling in a circular path and the force is focused towards the center of rotation. For a body at the equator of the earth, the centripetal force is specified by the gravitational force of the earth. As this centripetal force is dependent on the velocity of the rotation, the apparent weight of objects will vary depending on the velocity.