PLease Help with this question

Question

VSuppose that the height of the incline is h = 15.2 m. Find the speed at the bottom for each of the following objects.

solid sphere	I know this answer
spherical shell	My original answer was 1337	Your response is off by a multiple of ten. m/s
hoop	I know this answer
cylinder	Need Help with this answer I got 12.2 which is wrong	Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

PLease help with the "cylinder" and the spherical shell answer i got those two wrong

Not sure if this equation is needed thank you in advance

vCM =	2gh	1 +	ICM/MR2		1/2

Answer 1

Let

m (unknown) be the mass of the object

r (unknown) be the radius of the object,

g = 9.81 m/s² be gravitational acceleration,

v (to be computed) be the translational velocity at the bottom,

ω = v/r be the angular velocity at the bottom,

I = kmr² be the moment of inertia of the object,

where

k = 2/5 for a solid sphere

k = 2/3 for a spherical shell

k = 1 for a hoop

k = 1 for an empty cylinder

k = 1/2 for a solid cylinder

The potential energy -- mgh -- at the top is converted at the bottom into

translational kinetic energy -- mv²/2 -- and

rotational kinetic energy -- Iω²/2.

That gives the equation

mgh = mv²/2 + Iω²/2

This can be rewritten and simplified into

mgh = mv²/2 + kmr²ω²/2

mgh = mv²/2 + kmv²/2

2gh = v² + kv²

v = √(2gh/(1 + k))

Spherical shell:

v = √(2×9.81×15.2)/(1 + 2/3)) = 13.37

Empty cylinder:

v = √(2×9.81×15.2)/(1 + 1)) = 12.2

Solid cylinder:

v = √(2×9.81×15.2)/(1 + 1/2)) = 14.1