M, a solid cylinder (mass = 2.27 kg, r = 0.010 m) pivots on a thin, fixed, frict

Question

M, a solid cylinder (mass = 2.27 kg, r = 0.010 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with à 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 Submit Answer Tries 0/20 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. Submit Answer Tries 0/20 How far does m travel downward between 0.510 s and 0.710 s after the motion begins? Submit Answer Tries 0/20

Explanation / Answer

When the force F = 8.526 is acting down, this force provides the linear Acceleration. And an of the mass exerts a torque on cylinder giving an angular
accceleration .

So,

a = R

For the mass M:

F = I /R = MxR^2/2

/R= Ma/2

For the mass m:

ma = mg -F = mg - Ma/2

a = mg/ (M/2 +m) = 8.526 / (1.135+0.870) = 4.25 m/sec^2

= a/R = 4.25 /0.01 = 425.23 rad/sec^2

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