which will reach the bottom first, a hallow cylider or a solid cylider wth the same mass in an inclined plane?

Answer 1

For each cylinder
mgh=(mv^2)/2+(Iω^2)/2 =(mv^2)/2+(Iv^2)/2R^2.
For hallow cylinder the moment of inertia is I=mR^2, and for solid cylinder I=(mR^2)/2.
Therefore, the velocities at the bottom are
- for hallow cylinder v=sqroot(gh),
- for solid cylinder v=sqroot(1.33gh).
The solid cylinder will be first.

Answer 2

A solid cylinder rolls faster, because a smaller fraction of the available potential energy gets converted to rotational kinetic energy. That leaves more energy available for forward motion.

This is true even if the masses and radii are different.

For a solid cylinder that has rolled doownhill a vertical distance H,
M g H = (M/2)V^2 + (1/2) I w^2
= (M/2)V^2 + (1/2)(M*R^2/2)(V/R)^2
= (M/2)V^2 + (M/4)V^2
V^2 = (4/3) g H

For a hollow thin-walled cylinder,
M g H = (M/2) V^2 + (M/2)V^2 = M V^2
V^2 = g H