# Conservation of energy in rotational motion [duplicate]

Suppose a boy is standing on a platform which is free to rotate about an axis passing through its center. The Kinetic energy of the boy and the platform is K. If the boy stretches his hands so that the moment of inertia of the sytem(boy + platform) gets doubled. Then I have to find the new Kinetic energy of the system. As there is no torque so angular momentum remains conserved. Since M.O.I gets doubled so angular velocity gets halved. Therefore the new kinetic energy is:-

$$K'= \frac{1}{2}I'×(w')^2$$

$$K'= \frac{1}{2}×2I×(\frac{w}{2})^2$$

$$K'= \frac{1}{2}×\frac{1}{2}×I×w^2$$

$$K'= K/2$$

Thus the kinetic energy gets halved. So is the law of conservation of energy being violated here? Is the total energy not conserved?