Physics: Hoop down ramp; time and velocity?

Question

Given:
-height of ramp
-length of ramp
-mass of hoop
-inner and outer radius of hoop
-frictionless

Would my notes acurately predict the time in an ideal-frictionless situation (this is a physics 101 course):


Find Time:

Find α:
Using T=Iα
I=(1/2)M(R^2+R^2)
T=Mg(sinφ)R
Mg(sinφ)R=((1/2)M(R^2+R^2))α
(φ=slope of ramp)
(R's refer to inner
and outer radii)

find θ
d=θR
(d=top of ramp)
(R is outer hoop radius)

find time
θ=(1/2)αt^2

---------
Find ω_final
(ω_f)^2=(ω_i)^2+2αθ

Find final velocity
v_f=(ω_f)*R
(R is outer hoop radius)

Answer 1

If the surface is frictionless, the hoop will slide instead of rolling.
F = mgsinφ = ma
a = g(h^2/(L^2)^(1/2) = gh/L
α = 0
ω = 0
θ = 0

Assuming the hoop starts at the top of the ramp, and reaches the end of the ramp when the point of contact reaches the end of the ramp,
v = √2ghL/L = √2gh