Acceleration of a rolling body on an inclined plane is given by $$a = \frac{g sin\theta}{1 + I/(mR^2)}$$. Since the acceleration is independent of mass and depends only on the geometry (moment of inertia factor), the sphere has a smaller inertia factor than the disk, meaning the sphere has a greater acceleration and reaches the bottom first.

Acceleration without slipping is $a_1 = \frac{g sin\theta}{1 + I/(mR^2)} = \frac{5}{7}g sin\theta$. Acceleration with pure slipping is $a_2 = g sin\theta$. The ratio $a_1/a_2$ is $5/7$.

Using energy conservation, initial kinetic energy equals potential energy at max height: $\frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) = mgH$. Substituting $H = \frac{3v^2}{4g}$, we get $1 + \frac{I}{mR^2} = 2$, which gives $\frac{I}{mR^2} = 1$. This corresponds to a ring.

The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.

Static friction provides the necessary torque for rolling without slipping, converting translational kinetic energy into rotational kinetic energy without dissipating mechanical energy.

Using conservation of energy, potential energy equals total kinetic energy: $Mgh = \frac{3}{4}Mv^2$. Solving for velocity gives $v = \sqrt{\frac{4}{3}gh}$.

Acceleration of a rolling body is given by $$a = \frac{g \sin\theta}{1 + I/MR^2}$$. Since the solid cylinder has a smaller moment of inertia ratio than the hollow cylinder, its acceleration is greater, so it reaches the bottom first.

The acceleration on an inclined plane is inversely proportional to $1 + I/MR^2$. Solid sphere has the lowest moment of inertia coefficient ($2/5$), giving it maximum acceleration and shortest time to reach the bottom.

Using conservation of mechanical energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a solid sphere ($I = \frac{2}{5}MR^2$), solving yields $v = \sqrt{\frac{10}{7}gh}$.