Answer
The translational kinetic energy is $~~0.0625~J$
Work Step by Step
We can find the energy in the system:
$E = \frac{1}{2}kx^2$
$E = \frac{1}{2}(3.00~N/m)(0.250~m)^2$
$E = 0.09375~J$
We can find an expression for the total kinetic energy at the equilibrium position:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I \omega^2$
$K = \frac{1}{2}Mv^2+(\frac{1}{2})(\frac{1}{2}MR^2) (\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$
$K = \frac{3}{4}Mv^2$
Note that the ratio of the kinetic energy that is translational kinetic energy is $\frac{2}{3}$
We can find the translational kinetic energy:
$K_{trans} = (\frac{2}{3})(0.09375~J)$
$K_{trans} = 0.0625~J$
The translational kinetic energy is $~~0.0625~J$
