Answer
$\frac{1}{3}$
Work Step by Step
Let's assume,
mass of the solid disk = M
Radius of the solid disk = R
Speed of the solid disk = V = $r\omega$
We can write,
Translational kinetic energy $(K_{1})= \frac{1}{2}MV^{2}$
Rotational kinetic energy $(K_{2})= \frac{1}{2}I\omega^{2}-(1)$
We know that, for a solid disk,
$I=\frac{1}{2}MR^{2}-(2)$
$(2)=\gt (1)$
$K_{2}=\frac{1}{2}\times\frac{1}{2}MR^{2}\times\frac{V^{2}}{R^{2}}=\frac{1}{4}MV^{2}$
Total kinetic energy $(K)=K_{1}+K_{2}$
$K=\frac{1}{2}MV^{2}+\frac{1}{4}MV^{2}=\frac{3}{4}MV^{2}$
Now we can find the fraction as follows,
$\frac{K_{2}}{K_{1}+K_{2}}=\frac{\frac{1}{4}MV^{2}}{\frac{3}{4}MV^{2}}=\frac{1}{3}$
