A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is:

For a body rolling without slipping, the ratio of rotational kinetic energy to translational kinetic energy is $$\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2}$$. Using $$v = R\omega$$ and $$I = kmR^2$$ (where $$k$$ is the moment of inertia factor), this ratio equals $$k$$.

Given that $$KE_{rot} = 50\%$$ of $$KE_{trans}$$, we have $$\frac{KE_{rot}}{KE_{trans}} = \frac{1}{2}$$, so $$k = \frac{1}{2}$$.

Checking known bodies: a solid sphere has $$I = \frac{2}{5}mR^2$$ ($$k = 2/5$$), a solid cylinder has $$I = \frac{1}{2}mR^2$$ ($$k = 1/2$$), a hollow cylinder has $$I = mR^2$$ ($$k = 1$$), and a ring also has $$I = mR^2$$ ($$k = 1$$).

Since $$k = 1/2$$ corresponds to a solid cylinder, the body is a solid cylinder.