A question about two objects going down an inclined plane.?

cylinder

Iα=mgsinθr

α=mgsinθr/I

α=gsinθ/r

a=αr

a=gsinθ

d=1/2at^2

tc=sqrt(2d/a)

tc=sqrt[(2d/gsinθ)]

sphere

Iα=mgsinθr

α=mgsinθr/I

α=3gsinθ/(2r)

a=αr

a=3/2gsinθ

d=1/2at^2

ts=sqrt(2d/a)

ts=sqrt[4/3(d/gsinθ)]

then

ts=[sqrt(2/3)]tc

ts=0.8tc

sphere wins

Yep, the sphere wins every time, all other things equal.

Why?

Because both start and end with the same total energy (TE) assuming their masses are the same (which you did not specify). Their TE = PE = mgH where PE is the potential (starting) energy at the top of the ramp H high for each one. That's why their total energies must be the same at the bottom of the roll out...the conservation of energy.

And at the bottom, TE = KE = 1/2 mv^2(1 + k) assuming no slippage; all the PE is converted into kinetic energy from the conservation of energy law. v is the linear speed (the speed of the hub) and k = 1 for the cylinder and 2/3 for the sphere. It's the linear speed that determines how fast each object gets to the bottom of the ramp.

As you can see TE = 1/2 mv^2(1 + 1 ) = 1/2 mV^2 (1 + 2/3) = TE so that V/v = sqrt((1 + 1)/(1 + 2/3)) = 1.095 and V = 1.095 v, showing that V > v to keep the total energies the same value as each rolls off the bottom of the ramp. The hollow sphere wins every time. QED.