Pure Rolling

@Steven Chase has already posted a detailed solution. I am sharing an alternate approach to this problem.

Consider an elementary ring of radius $r$ , thickness $dr$ with height $t_o$ into the plane of the paper. The mass of this ring is:

$dM = \rho(r) 2 \pi r dr t_o$

The total mass of the disk is:

$M = \int_{0}^{R}\rho(r) 2 \pi r t_o (dr)$

The moment of inertia of this elementary ring is about the central axis of the ring (axis of rotation) is:

$dI = (dM) r^2$

And therefore:

$I = \int_{0}^{R}\left(\rho(r) 2 \pi r t_o (dr)\right) r^2$

The speed of the disk is $v$ and since it is purely rolling, its angular velocity is $\omega = v/R$ . Now the total kinetic energy of the disk is:

$T = T_{translation} + T_{rotational}$ $T = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$

Plugging in the required expressions of mass and inertia, and simplifying, gives the required answer.

Consider the velocity of an infinitesimal patch on the disk:

$\dot{x} = v + r \omega \, \sin\theta \\ \dot{y} = -r \omega \, \cos \theta$

Square of the speed of the patch:

$u^2 = \dot{x}^2 + \dot{y}^2 = v^2 + 2v \, \omega \, r \, \sin\theta + r^2 \omega^2$

Area, volume, and mass of patch:

$dA = r \, dr \, d\theta \\ dV = t_0 \, dA = t_0 \, r \, dr \, d\theta \\ dm = \rho \, dV = r^2 \, dr \, d\theta$

Kinetic energy of patch:

$dE = \frac{1}{2} \, dm \, u^2 = \frac{1}{2} \, (r^2 \, dr \, d\theta) \, (v^2 + 2v \, \omega \, r \, \sin\theta + r^2 \omega^2) \\ = \frac{1}{2} \, (v^2 \, r^2 + 2v \, \, \omega \, r^3 \, \sin\theta + r^4 \omega^2) \, dr \, d\theta$

The kinetic energy is equal to the following integral:

$E = \int_0^{2 \pi} \int_0^R \frac{1}{2} \, (v^2 \, r^2 + 2v \, \, \omega \, r^3 \, \sin\theta + r^4 \omega^2) \, dr \, d\theta$

Recall the relationship between the angular speed and the other quantities:

$\omega = \frac{v}{R}$

Plugging in and evaluating the integral results in:

$E = \frac{8 \pi \, v^2 \, R^3}{15}$