Auniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. what is its acceleration? (enter the magnitude. use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

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