A Ring on the incline

Clearly, normal reaction here N = Mgcosθ where M=6kg

Now,

since pure rolling takes place,

v=ωr or a=αr

Mgsinθ - f = Ma [f is force of friction] ... (i)

\tau = I\alpha ... (ii)

where I = m₁R² +m₂R² [ m₁+m₂ = 6]

since torque applied is by force of friction,

f=Ia/r² ... (iii)

Put (iii) in (i),

a=mgsinθ/(m+I/r²)

α=a/r

and f = Ia/r²

I don't think it's coming. However I may be wrong in Putting " I " Values.

The original answers are -

α = 6.7 rad/s²

N = 46.656 N

f = 9.302 N

friction=10N
angular acc=20/3 rad/s²

i took g=10m/s²

btw i just realized tht i forgot to write the Normal reaction
=30√3 N

You got these answers from Post #2? I check again..