A uniform wheel of radius R is made to rotate about its axis at an...

Let the mass of the uniform wheel be $m$. A uniform wheel is considered as a uniform disc, so its moment of inertia about its central axis is $I = \dfrac{1}{2}mR^2$.

When the rotating wheel is placed on the rough horizontal surface, the frictional force acts at the point of contact. The normal reaction and gravitational force pass through the center of mass and are equal and opposite. Therefore, the net torque about any point on the ground along the line of motion is zero, and the angular momentum of the wheel about the point of contact is conserved.

Initial angular momentum about the point of contact:
$L_i = I\omega = \dfrac{1}{2}mR^2\omega$

Let $v$ be the linear speed and $\omega'$ be the angular speed when pure rolling begins. For pure rolling, $v = \omega' R$.

Final angular momentum about the point of contact:
$L_f = I\omega' + m v R = \dfrac{1}{2}mR^2\omega' + m(\omega' R)R = \dfrac{3}{2}mR^2\omega'$

By conservation of angular momentum ($L_i = L_f$):
$\dfrac{1}{2}mR^2\omega = \dfrac{3}{2}mR^2\omega'$

$\omega' = \dfrac{\omega}{3}$

The linear speed of the wheel is:
$v = \omega' R = \dfrac{\omega R}{3}$