Question

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

Answer 1

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