Question

M, a solid cylinder (M=1.79 kg, R=0.127 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 0.670 kg mass, i.e., F = 6.573 N. How far does m travel downward between 0.370 s and 0.570 s after the motion begins?

Answer 1

mass m = 0.670 kg and \[\omega\] is 33.1 rad/s

Answer 2

First thing you have to find is the angular acceleration of the cylinder, under the torque created by the force.

Answer 3

57.8 rad/s

Answer 4

then what should i do? There's also a second part to this question that i can't find, 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 0.5823 m in a time of 0.570 s. Find Icm of the new cylinder.

Answer 5

57.83 rad/s² is the correct angular acceleration, provided the force F is F = 6.573 N as you stated.
Now convert this angular acceleration to a linear acceleration for the moving mass and work out its velocity, then its position, from there.