Who set the wheels in motion ?!

Consider the center of disc to be the origin of a Cartesian plane with x-axis parallel to the ground.

Now, taking a small element of mass $dm$ at a distance of $r$ from the center with width $dr$ and making an angle of $\theta$ from the x-axis and subtending an angle $d\theta$ at the center.

Thus area covered by the element $=$ length * width $= rd\theta \cdot dr$

(Since radius times the angle gives us the arc length)

Now using unitary method we get: $dm = \left( rd\theta \cdot dr \right) \cdot \left( \dfrac{M}{\pi R^2} \right)$

Now, we can calculate the resultant velocity of his small element by considering two of its velocities: $V$ and $\omega R$ .

Thus the resultant velocity of the element will be:

$v_{net} = \sqrt{v^2 + \omega^2r^2 + 2v\omega rsin\theta}$

(Since angle between the two velocities will be $(90-\theta)$ )

Thus the final kinetic energy of the quarter of the wheel using proper limits will be:

$KE_{net} = \int\limits_0^{\pi/2} \int\limits_0^{R} \left(\frac{1}{2}\right)\cdot dm\cdot v_{net}^2$

Thus, by solving the above expression, we get:

$KE_{net} = \left( \dfrac{9\pi + 16}{48\pi} \right) mV^2$