0
$\omega$
$\frac{\omega}{2}$
$2\omega$

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Let the moment of inertia of the wheel about the axis be $I$ . Initially, the first wheel is rotating at the angular speed $\omega$ about the axle and the second wheel is at rest. Take both the wheels together as the system. The total angular momentum of the system before the coupling is $I\omega$ + 0 = $I\omega$ .

When the second wheel is dropped into the axle, the two wheels slip on each other and exert forces of friction. The forces of friction have torques about the axis of rotation but these are torques of internal forces. No external torque is applied on the two-wheel system and hence the angular momentum of the system remains unchanged. If the common angular speed is $\omega_{1}$ , the total angular momentum of the two-wheel system is $2I\omega_{1}$ after the coupling.

Given angular momentum is conserved,

$2I\omega_{1}$ = $I\omega$

$\implies$ $\omega_{1}$ = $\frac{\omega}{2}$