Answer
The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is $~~\frac{4}{3}$
Work Step by Step
We can find an expression for the initial rotational inertia of the system:
$I_i = \frac{1}{2}(4.00m)(R^2)+mR^2$
$I_i = 3.00mR^2$
We can find an expression for the new rotational inertia of the system:
$I_f = \frac{1}{2}(4.00m)(R^2)+m(\frac{R}{2})^2$
$I_f = 2.00mR^2+\frac{mR^2}{4}$
$I_f = \frac{9mR^2}{4}$
$I_f = \frac{3}{4}~I_i$
We can use conservation of angular momentum to find an expression for the new angular velocity of the system:
$L_f = L_i$
$I_f~\omega_f = I_i~\omega_i$
$\frac{3}{4}~I_i~\omega_f = I_i~\omega_i$
$\omega_f = \frac{4~\omega_i}{3}$
We can write an expression for the initial rotational kinetic energy:
$K_i = \frac{1}{2}I_i\omega_i^2$
We can find an expression for the new rotational kinetic energy:
$K_f = \frac{1}{2}I_f\omega_f^2$
$K_f = \frac{1}{2}(\frac{3I_i}{4})(\frac{4~\omega_i}{3})^2$
$K_f = \frac{4}{3}\times \frac{1}{2}I_i\omega_i^2$
$K_f = \frac{4}{3}~K_i$
Therefore:
$\frac{K_f}{K_i} = \frac{4}{3}$
The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is $~~\frac{4}{3}$.
