M, a solid cylinder (M=1.71 kg, R=0.133 m) pivots on a thin, fixed, frictionless bearing. A...

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Physics

M, a solid cylinder (M=1.71 kg, R=0.133 m) pivots on a thin,fixed, frictionless bearing. A string wrapped around the cylinderpulls downward with a force F which equals the weight of a 0.690 kgmass, i.e., F = 6.769 N. Calculate the angular acceleration of thecylinder.

5.95Ã—10¹ rad/s^2
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Your receipt no. is 161-2131 Previous Tries
If instead of the force F an actual mass m = 0.690 kgis hung from the string, find the angular acceleration of thecylinder.
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How far does m travel downward between 0.730 s and 0.930 s afterthe motion begins?
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The cylinder is changed to one with the same mass and radius,but a different moment of inertia. Starting from rest, the mass nowmoves a distance 0.379 m in a time of 0.470 s. FindI_cm of the new cylinder.

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3.7 Ratings (294 Votes)

Torque FR I where I MR22 and is the angualar accceleration FR MR22 2FMR 267691710133 59526 10 rads2 When the force mg 6769 is acting down this force provides the See Answer

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