Rolling Hoop Down an Incline

Question

A uniform hoop rolls without slipping down a 25° inclined plane. What is the acceleration of the hoop's center of mass?

The moment of inertia of a uniform solid disk about an axis that passes through its center = mr².

The moment of inertia of a uniform solid disk about an axis that is tangent to its surface = 2mr².

Answer 1

T = I*α

Torque is caused by friction between hoop and ramp = F_fr = μ*n = μ*mgcos(25) because normal force is mgcos(θ)

V = R*ω so a = R*α so α = a/R

T = F*R = I*α so plug in: μ*mgcos(25)* R = (mR²)*(a/R) , m cancels out and and so does R and you get μ*g*cos(25) = a but we don't know μ and it wasn't given so we need to cancel it out. Solve to get μ = a/(g*cos(25)) so we can use this in another equation.

Net Force = m*a so force parallel to ramp = mgsin(25) and friction is mgcos(25)*μ so Force of gravity pulling it down ramp minus friction holding it back is the net force = m*a

So let's write this as an equation

m*g*sin(25) - m*g*cos(25)*μ = m*a and plug in μ = a/(g*cos(25)) and cancel out m

g*sin(25) - a = a

which leads to a = (0.5)*g*sin(25) = 4.9sin(25) if g is 9.8

so since we got an equation for μ and plugged it in to cancel it out we see the answer doesn't depend on it (this is because the hoop isn't slipping) and notice how the radius doesn't matter and neither does the mass of the hoop.