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

Question

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

Moment of inertia for a solid cylinder is I = mr2/2

A) Moments of force F about the center of the disk = I* = F*r


(mr2/2) * = F*r
(m*r / 2) * = F

= 2*F / (m*r)

= 2 * 8.526 / (2.27kg * 0.010 mm) = 751 rad/s^2

b)

the linear acceleration of mass is a

using second law of motion

a = net force/effective mass

a = 0.870* 9.8/(0.870+ 0.5 * 2.27*r^2/r^2)

a = 4.25m/s^2

now, angular acceleratio   = a/r

angular acceleration   = 4.25/(0.01)

angular acceleratio    = 425 rad/s^2

c) use s = ut+0.5at2     ( intial velocity = 0)

s = 0.5at2

Distance travelled in first 0.510 s will be s1 = 0.5*4.25*(0.510)2    =   0.5527125 m

Distance travelled in first 0.710 s will be s2 = 0.5*4.25*(0.710)2   = 1.0712125 m

Distance travelled btw 0.510s to 0.710s will be = 1.0712125 m - 0.5527125 m = 0.5185 m answer

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