To understand rolling without slipping we need to understand the kinetic energy falls into two modes; linear kinetic energy and rotational kinetic energy. For a rigid body if the velocity of the center of mass $m$ is $v$ then $\frac{1}{2}mv^2$ gives the linear KE. On the other hand, analyzing the motion of the little bits of mass rotating around the center of mass allows us to capture that collective energy by $\frac{1}{2}I \omega^2$ where $I$ is the moment of inertia and $\omega$ is the angular velocity about the center of mass. Generally it is more complicated and you need an inertia tensor to capture the story, but when the rotational motion is just around one axis then we get the usual Freshman stuff.
Ok, so,
$$KE = \frac{1}{2}mv^2+ \frac{1}{2}I \omega^2 $$
Conservation of energy for a rigid body starting from rest gives us that the initial gravitational potential energy $PE = mgh$ is matched by the final $KE$.
Rolling without slipping means that $\omega$ and $v$ are connected; $v = \omega R$ hence we find:
$$ mgh = \frac{1}{2}mv^2+ \frac{1}{2}I (v/R)^2 $$
hence
$$ mgh = \frac{1}{2}(m+ I/R^2)v^2 $$
and so,
$$ v = \sqrt{\frac{2mgh}{m+I/R^2}} $$
We find larger $I$ iimplies smaller $v$. Calculus shows $I = \frac{1}{2}mR^2$ for a solid cylinder whereas $I = mR^2$ for a cylindrical shell. Another common example somewhere between these extremes is the solid sphere with $I = \frac{2}{5}mR^2$.
I think the explanation is largely algebra provided we have already developed the basic rules for rotational kinematics. That said, calculus is naturally employed to derive the moment of intertia formulas.
