susus Posts: 138, Reputation: 2 Junior Member   #1 Dec 25, 2010, 01:29 AM

Rotating Mass M, a solid cylinder (mass = 1.39 kg, r = 0.100 m) pivots on a thin, fix Rotating Mass M, a solid cylinder (mass = 1.39 kg, r = 0.100 m) pivots on a thin, fixed, frictionless beari? Rotating Mass M, a solid cylinder (mass = 1.39 kg, r = 0.100 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 0.650 kg mass, i.e. F = 6.370 N. Calculate the angular acceleration of the cylinder. http://img577.imageshack.us/img577/4850/prob28a1007fcyl2.gif (in rad/s^2) If instead of the force F an actual mass m = 0.650 kg is hung from the string, what is the angular acceleration of the cylinder. (in rad/s^2) http://img163.imageshack.us/img163/6202/prob28b1007mcyl2.gif How far does m travel downward between 0.650 s and 0.850 s after the motion begins? (in m) I tried this , but it's wrong :( The inertial momentum of solid cylinder is I=M*R^2/2, where M and R are the mass and the radius. (This is yield with integration of the point momentums over the hall cylinder). Then F*R = I * Ang, F*R is the momentum of the force, Ang is the angular acceleration we want to find. Thus Ang = F*R/ (M*R^2/2) = 2*F/ M*R, in your case 0.92rad/s^2 Obviously the origin of the force doesn't matter, so that is the same when hanging the actual mass. The linear acceleration a of the periphery, or of the string is a = Ang * R = 2*F/ M (see- it doesn't depend on the radius). which is about 9.2m/s^2. In 0.850s the mass will travel a*0.85^2/2, in 0.650s a*0.65^2/2, so the distance between is a*0.85^2/2 - a*0.65^2/2 = 1.38m

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