Answer
The rotational kinetic energy is $~~0.03125~J$
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 rotational kinetic energy is $\frac{1}{3}$
We can find the rotational kinetic energy:
$K_{rot} = (\frac{1}{3})(0.09375~J)$
$K_{rot} = 0.03125~J$
The rotational kinetic energy is $~~0.03125~J$
