A 5.0 kg, 60-cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. What is the cylinder's initial angular acceleration? ok so this is what i think i should do: Torque= I [tex]\alpha[/tex] and i need to find the F(weight)= mg = (5)(9.8) = 49 N radius = .5(.6)= .3 Torque= Fd = 49(.3) = 14.7 14.7 = I [tex]\alpha[/tex] this is the part which i am not sure on, does I = .5 m r^2 = .5(5)(.3^2) = .225 14.7 .225 [tex]\alpha[/tex] [tex]\alpha[/tex] = 65.33 but this answer is incorrect, and i do not know where i am making the mistake.
Try using the Parallel Axis theorem to find the moment of inertia. This is needed because the axis of rotation is not through the centre of mass of the disk. [tex]I = I_{CM} + MD^2[/tex] where [tex]I_{CM}[/tex] is the moment of inertia for a solid disk, M is the mass of the disk and D is the distance from the centre of mass to the new axis.