Answer:
aCM = (2/3)*g*Sin θ
Explanation:
Consider a uniform solid disk having mass M, radius R and rotational inertia I about its center of mass, rolling without slipping down an inclined plane.
In order to get the linear acceleration of the object’s center of mass, aCM ,
down the incline, we analyze this as follows:
The force of gravity (W = Mg) acting straight down is resolved into components parallel and perpendicular to the incline.
Since the object rolls without slipping there is a force of friction (Ff) acting on the object, at it’s point of contact with the incline, in the direction up the incline.
Newton’s 2nd Law gives then for acceleration down the incline
∑Fx' = m*aCM ⇒ m*g*Sin θ - Ff = m*aCM
The force of friction also causes a torque around the center of mass
having lever arm R so we can also write
τ = R*Ff = I*α
Solving for the friction, Ff = I*α / R
This is used in the expression derived from the 2nd Law:
m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM
The objects angular acceleration is related to the linear acceleration of the edge that contacts the incline by
a = R*α
Since the object rolls without slipping this has the same magnitude as aCM so we have that
α = aCM / R
Using this in
m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM
⇒ aCM = (m*g*Sin θ*R²) / (I + m*R²)
if I = (1/2)*m*R² (for a uniform solid disk)
we get
aCM = (2/3)*g*Sin θ
⇒ 
