A solid cylinder (M=2.43 kg, R=0.131 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 0.570 kg mass, i.e., F = 5.592 N. Calculate the angular acceleration of the cylinder.
If instead of the force F an actual mass m = 0.570 kg is hung from the string, find the angular acceleration of the cylinder.
How far does m travel downward between 0.430 s and 0.630 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 0.381 m in a time of 0.510 s. Find Icm (I sub cm) of the new cylinder.
2007-04-02
13:40:04
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2 answers
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asked by
Anonymous