lays a crucial role in this problem.

First, we know that the angular momentum (L) is conserved when there is no external torque acting on the system. Angular momentum can be defined as the product of the moment of inertia (I) and the angular velocity (ω), given by the formula:

$$L = I\omega$$

Given in the question, when the child stretches his arms out, the moment of inertia of the system becomes double. If we let the initial moment of inertia be $I$, then the final moment of inertia after the child stretches his arms becomes $2I$.

Since angular momentum is conserved and initially, the system had a certain angular momentum $L = I\omega$, after the change, it still maintains the same angular momentum, now with the new moment of inertia, leading to a new angular velocity $\omega'$ such that:

$$L = I\omega = 2I\omega'$$

From this equation, we solve for $\omega'$ to find that:

$$\omega' = \frac{\omega}{2}$$

This indicates that the angular velocity is halved when the moment of inertia is doubled.

The kinetic energy (K) of a rotating body is given by:

$$K = \frac{1}{2}I\omega^2$$

Initially, the kinetic energy is:

$$K = \frac{1}{2}I\omega^2$$

After the moment of inertia is doubled and the angular velocity is halved:

$$K' = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2$$

$$K' = \frac{1}{2} \cdot 2I \cdot \frac{\omega^2}{4}$$

$$K' = \frac{1}{2}I\omega^2 \cdot \frac{1}{2}$$

$$K' = \frac{1}{2}K$$

Therefore, the kinetic energy of the system after the child stretches his arms out, resulting in the moment of inertia becoming double, is $\frac{K}{2}$.

The correct answer is Option A $$\frac{K}{2}$$.

Rotational Motion

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