Topic: Rigid Bodies
Concepts: Energy dissipation, Limiting cases, Rolling down inclined plane

Solution:

Recall the acceleration of an object rolling without slipping down an incline is given by

where the object has moment of inertia $I=\beta mr^2$. For small enough $\theta$, static friction is large enough to provide sufficient torque since

Using $a=r\alpha$ as the object is rolling without slipping,

so we see that the required friction force $f \rightarrow 0$ as $\theta \rightarrow 0$. Thus, the object rolls without slipping (no energy loss) for some range of angles $0<\theta<\theta_c$.

On the other end, if $\theta=90^{\circ}$ then the object drops vertically and there is also no dissipation. In between when $\theta_c<\theta<90^{\circ}$, the object is rolling with slipping so there is energy dissipation. Going a bit past $\theta_c$ leads to a small amount of slipping (and energy loss) since the friction force changes continuously. Hence, the only graph that accounts for all these features is C.