Athin spherical shell rolls down an incline without slipping. if the linear acceleration of the center of mass of the shell is 0.23g, what is the angle the incline makes with the horizontal?

Fnet = ma = m*0.23g = mgsinΘ - Ff

where Ff is the friction force upslope.

net torque τ = Ff * r,

but also τ = I*α =(2/3)mr² * a/r = 2mra/3 = 2mr(0.23g)/3 = 0.46mrg/3

Then Ff * r = 0.46mrg / 3

Ff = 0.46mg/3 → put this into net force equation:

m*0.23g = mgsinΘ - 0.46mg/3 → mg cancels; multiply through by 3

0.69 = 3sinΘ - 0.46

3sinΘ = 1.15

sinΘ = 1.15/3

Θ = arctan(1.15/3) = 22.54 º