THE ANSWER IS NOT 8.5cm M, a solid cylinder (mass = 2.27kg, r = 0.010 m) pivots

Question

THE ANSWER IS NOT 8.5cm

M, a solid cylinder (mass = 2.27kg, r = 0.010 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 0.870 kg mass, i.e., F = 8.526 N. Calculate the angular acceleration of the cylinder. 7.51x102 rad/s 2 You are correct. Your receipt notis 168-49586) Previous Tries If instead of the force F an actual mass m = 0.870 kg is hung from the string, what is the angular acceleration of the cylinder 4.25x102 rad/s 2 You are correct. Your receipt no. is 168-8488Previous Tries How far does 4.25×102 rad/s^2 travel downward between 0.510 s and 0.710 s after the motion begins? You can calculate the linear acceleration from the angular acceleration in the second part. From here, it is just motion in one dimension Submit Answer Incorrect. Tries 3/20 Previous Tries Post Discussion Send Feedback

Explanation / Answer

Acceleration of block = angular acceleration* r
= 4.25*10^2 * 0.01
= 4.25 m/s^2

distance,
d = 0.5*a*(tf^2 -ti^2)
= 0.5*4.25*(0.710^2 - 0.510^2)
= 0.52 m
= 52 cm
Answer: 52 cm

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