Uniform solid disk rolls without slipping down a 19.0° inclined plane. what is the acceleration of the disk's center of mass?

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