Below is the solution:
Let us say that the disk goes through a vertical elevation change of one meter.
The change in potential energy will equal the change in kinetic energy
PE = KEt + KEr
mgh = ½mv² + ½Iω²
for a uniform disk, the moment of inertia is
I = ½mr²
and
ω = v/r
mgh = ½mv² + ½(½mr²)(v/r)²
mgh = ½mv² + ¼mv²
gh = ¾v²
v² = 4gh/3
v² = u² + 2as
if we assume initial velocity is zero
v² = 2as
a = v² / 2s
s(sinθ) = h
s = h/sinθ
a = 4gh/3 / 2(h/sinθ)
a = ⅔gsinθ
a = ⅔(9.8)sin25
a = 2.8 m/s²
