M, a solid cylinder (M=1.91 kg, R=0.115 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.91 kg, R=0.115 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 0.850 kg mass, i.e., F = 8.338 N. Calculate the angular acceleration of the cylinder. b.) If instead of the force F an actual mass m = 0.850 kg is hung from the string, find the angular acceleration of the cylinder. c.) How far does m travel downward between 0.630 s and 0.830 s after the motion begins? d.) 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 0.440 m in a time of 0.470 s. Find Icm of the new cylinder.

Explanation / Answer

(a) = rF sin
also, = I
rF sin = I

In the pulley-mass system, the string always acts at right angles to the pulley,
so, = 90o and sin = 1
rF = I

= rF/I

For a cylinder of solid mass, the moment of inertia is dened to be 0.5Mr2.

= 2F/Mr = (2*8.338)/(1.91*0.115) = 75.92 rad/sec2

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