M, a solid cylinder (mass = 1.39 kg, r = 0.090 m) pivots on a thin, fixed, frict
ID: 2064606 • Letter: M
Question
M, a solid cylinder (mass = 1.39 kg, r = 0.090 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 0.630 kg mass, i.e., F = 6.174 N. If instead of the force F an actual mass m = 0.630 kg is hung from the string, what is the angular acceleration of the cylinder.Image address: http://loncapa2.physics.sc.edu/res/msu/physicslib/msuphysicslib/21_Rot3_AngMom_Roll/graphics/prob28b_1007mcyl2.gif
b. How far does m travel downward between 0.690 s and 0.890 s after the motion begins?
Explanation / Answer
Torque = 6.174*0.09 = 0.555 Nm
Moment of inertia = 0.5mr^2 = 0.5*1.39*0.09^2 = 0.00563Kgm^2
= I
0.555 = 0.005638
= 98.44rad/s^2
acceleration of the mass hung, a = r = 98.44*0.09 = 8.859m/s^2
Distance moved in that time = 0.5a(0.89)^2 - 0.5a(0.69)^2 = 1.4m
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