Question

A body situated on earth’s surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by \( \left( M \text{ and } R \text{ be the mass and radius of earth respectively} \right) \)

• A. \( \frac{MgR}{4} \)
• B. \( \frac{MgR}{5} \)
• C. \( \frac{MgR}{2} \)
• D. \( \frac{MgR}{3} \)

Hint:

In problems involving rotational motion and weightlessness, equate the centrifugal force with gravitational force to find the critical value of the energy.

Answer 1

The Correct Option is B

Step 1: Understanding the concept.
The body becomes weightless when the gravitational force is completely balanced by the centrifugal force due to the rotation of the Earth. The centrifugal force is given by: \[ F_{\text{centrifugal}} = M \omega^2 R \] where \( \omega \) is the angular velocity of Earth. The condition for weightlessness is when the centrifugal force equals the gravitational force: \[ M g = M \omega^2 R \] Step 2: Calculating the angular velocity.
The rotational kinetic energy of the Earth is: \[ KE = \frac{1}{2} I \omega^2 \] where \( I = \frac{2}{5} M R^2 \) is the moment of inertia of the Earth. Setting this equal to the required condition, we solve for \( \omega^2 \), and find that the critical value of the rotational kinetic energy is \( \frac{MgR}{5} \).
Step 3: Conclusion.
Thus, the critical rotational kinetic energy is \( \frac{MgR}{5} \), corresponding to option (B).