Given that the rotational kinetic energy ($K_{rot}$) is $50\%$ of the translational kinetic energy ($K_{trans}$): $K_{rot} = \frac{1}{2} K_{trans}$ $\frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} m v^2\right)$ For pure rolling, $v = R\omega$, so: $\frac{1}{2} I \omega^2 = \frac{1}{4} m (R\omega)^2$ $\frac{1}{2} I \omega^2 = \frac{1}{4} m R^2 \omega^2$ $I = \frac{1}{2} m R^2$ This is the expression for the moment of inertia of a solid cylinder (or a uniform disc) about its central axis. Therefore, the body is a cylinder.