Question

M, a solid cylinder (M=2.43 kg, R=0.123 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.770 kg mass, i.e., F = 7.554 N. Calculate the angular acceleration of the cylinder.

If instead of the force F an actual mass m = 0.770 kg is hung from the string, find the angular acceleration of the cylinder. M m 3.09x101 rad/s^2 You are correct. Your receipt no. is 149-2534 Previous Tries How far does m travel downward between 0.470 s and 0.670 s after the motion begins? 4.34x10-1 m You are correct. Your receipt no. is 149-6546 Previous Tries 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 0.418 m in a time of 0.510 s. Find Icm of the new cylinder. You know how to write an equation for the torque from the second part. You are given quantities to calculate the acceleration (review motion in one dimension if this is not obvious), and in this case you need to solve for the moment of inertia. Submit Answer Incorrect. Tries 1/12 Previous Tries

Answer 1

For angular acceleration T= I x RF - MR? xq = 7.554 x 0.123 = 2.43 X0.123X0-123 a = 7.554 x2 = 2.43 X0-123 = 50.5 rad) sepe