A cylinder of radius 20cm is mounted on an axle coincident with its axis so as to be free to rotate. A cord is wound on it and a 50-g mass is hung from it. If after being released, the mass drops 100cm in 12s, find the moment of inertia of the cylinder? Take down as positive. Torque=I*(alpha), i.e., b*T=I*(alpha) => T=I*(alpha)/b ...(i) m*g - T=m*a ...(ii), on adding i and ii, we get I=m*b(g-a)/(alpha); now alpha=a/b, where a=acceleration, b=radius, T=tension, I=moment of inertia and s=1m. v_avg=(v+u)/2=.04166 m/s: v^2=u^2+2*a*s therefore a=-u^2/2/s=-8.681*10^-4. Hence I=225.78 kg.m^2. I also did the calculation from the energy point of view and got the same answer. Yet the answer in the book is 1.4kg.m^2. Can anyone show me where I went off track? Thanks very much.