A) M, a solid cylinder (M=1.67 kg, R=0.133 m) pivots on a thin, fixed, frictionl

Question

A) M, a solid cylinder (M=1.67 kg, R=0.133 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 0.750 kg mass, i.e., F = 7.357 N. Calculate the angular acceleration of the cylinder.

B)If instead of the force F an actual massm= 0.750 kg is hung from the string, find the angular acceleration of the cylinder.

C)How far does m travel downward between 0.430 s and 0.630 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 0.464 m in a time of 0.470 s. Find Icmof the new cylinder.

Explanation / Answer

A) I = ½mR² = .01477 kg·m² a = Q/I = F*R/I = 66.24 rad/sec² B) F = k*m*g/(m+k) = 3.9613 N (k = ½) a = F*R/I = 28.2 rad/sec² C) a = a*R = 3.7506 m/s²; dy = ½a*(.69² - .49²) = .4426 m D) a = 2y/t² = 3.054 m/s² ? a = 22.966 rad/sec² ? F = m(g - a) = 14.23 N and Icm = F*R/a = .0824 kg·m²

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