M, a solid cylinder (M=1.99 kg, R=0.133 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force of F which equals the weight of a 0.830 kg mass, i.e, F=8.142 N. Calculate the angular acceleration of the cylinder. I tried drawing a free body diagram and setting the forces equal to ma. F_n- mg-F_t= Ma_y since a= 0, F_n= Mg + F_t torque= RF_t sin 90= Ia RF_t=Ia I=(1/2)MR^2 F_t= (1/2)MRa mg-ma=(1/2)MR(a/R) (1/2M +m)a= mg a=mg/ (1/2 M=m) I plugged in my numbers, but I think the equation is wrong. Help?