Answer
$$
K_{2}=9.71 \times 10^{3} \mathrm{J}
$$
Work Step by Step
The rotational inertia of a rigid body depends on how its mass is distributed.
since the rotational inertia of a cylinder is } $I=\frac{1}{2} M R^{2} $ (Table 10-2($\mathrm{c}$) its rotational kinetic energy is
$$
K=\frac{1}{2} I \omega^{2}=\frac{1}{4} M R^{2} \omega^{2}
$$
For the larger cylinder, we obtain
$$
K_{2}=\frac{1}{4}(1.25 \mathrm{kg})(0.75 \mathrm{m})^{2}(235 \mathrm{rad} / \mathrm{s})^{2}=9.71 \times 10^{3} \mathrm{J}
$$
