Answer
We can list the shapes in the order they arrive at the finish line:
frictionless cube, solid sphere, solid cylinder, hollow sphere, hollow cylinder
Work Step by Step
Let $h$ be the vertical height of the starting position. We can use conservation of energy to find the speed $v$ of each object at the bottom of the slope.
solid sphere:
$U_g = KE_{tr}+KE_{rot}$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$Mgh = \frac{7}{10}Mv^2$
$v = \sqrt{\frac{10~gh}{7}}$
hollow sphere:
$U_g = KE_{tr}+KE_{rot}$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{3}MR^2)~(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{3}Mv^2$
$Mgh = \frac{5}{6}Mv^2$
$v = \sqrt{\frac{6~gh}{5}}$
solid cylinder:
$U_g = KE_{tr}+KE_{rot}$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)~(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$
$Mgh = \frac{3}{4}Mv^2$
$v = \sqrt{\frac{4~gh}{3}}$
hollow cylinder:
$U_g = KE_{tr}+KE_{rot}$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)~(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$
$Mgh = Mv^2$
$v = \sqrt{gh}$
frictionless cube:
$U_g = KE_{tr}$
$Mgh = \frac{1}{2}Mv^2$
$Mgh = \frac{1}{2}Mv^2$
$v = \sqrt{2gh}$
If a shape has a larger speed at the bottom of the ramp, then it arrives to the finish line earlier since it has a larger acceleration. We can list the shapes in the order they arrive at the finish line:
frictionless cube, solid sphere, solid cylinder, hollow sphere, hollow cylinder
