If instead of the force F an actual mass m = 0.690 kg is hung from the string, find the angular acceleration of the cylinder.
Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion.
How far does m travel downward between 0.630 s and 0.830 s after the motion begins?
The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.373 m in a time of 0.470 s. Find I_cm of the new cylinder.