Question

can someone plz help solve the last one with steps :)

M, a solid cylinder (M-2.39 kg, R-0.131 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls dowmward with a force F which equals the weight of a 0.850 ka mass, 1.e.. F 8.338N. Cálculate the angular acceleration of the cylinder M 5.33x101 rad/s 2 You are correct. Your receipt no, is 154-486 0 Previosus Tries If instead of the force F an actual mass m0.850 kg is hung from the string, find the angular acceleration of the cylinder 3.11x101 rad/s 2 You are correct. Your receipt no. is 154-9996 o Previous Tries 5.33x101 rad/s^2 You are correct. Your receipt no. is 154-486 revious Tries If instead of the force F an actual mass m 0.850 kg is hung from the string, find the angular acceleration of the cylinder. m 3.11x101 rad/s^2 You are correct. Your receipt no. is 154-9996 Previous Tries How far does m travel downward between 0.670 s and 0.870 s after the motion begins? 6.28x10 m You are correct154-5970 Previous Tries TOur receipt no. The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the the new.cylinder mass now moves a distance 0.410 m in a time of 0.510 s. Find lom or Submt Answer Tries 0/20

Answer 1

de0.410m teo.S12 So, detat a2x041 (0.51)2 a3-153mls T- 085(9-8-3-153) ) TE 5.65N ma-Te ma So T Io TT JT = 5-65x0.121XO.131 BIS3 TO-235 kamxo13 D.030185 kam