Rotational Dynamics with sphere?

Question

A uniform solid disk rolls without slipping down a 15.0° inclined plane. What is the acceleration of the disk's center of mass?
The moment of inertia of a uniform solid disk about an axis that passes through its center = ½mr².
The moment of inertia of a uniform solid disk about an axis that is tangent to its surface = 3⁄2mr².

 

I got the answer a=1.69 from coming up with the equation 2/3*g*sinthetha. Did I do this right? 

Answer 1

Start with Newton's 2nd law for the center of mass acceleration.  Static friction f_s opposes the downward acceleration.  Taking down as positive,

 

(i)  mg sinθ - f_s = ma

 

Torque about center of mass due to friction:

 

f_sR = Iα = [(1/2)mR²]α = [(1/2)mR²](a/R) ⇒

 

f_s = ma/2

 

Plug f_s = ma/2 into equation (i) and get

 

mg sinθ - ma/2 = ma

 

g sinθ = (3/2)a

 

a = (2/3)g sinθ = (2/3)(9.8 m/s²)sin15º = 1.69 m/s²

 

You answer is correct.