Must show all work and correct answer to get points. Thank you! Do not round int

Question

Must show all work and correct answer to get points.  Thank you!

Do not round intermediate calculations; however, for display purposes, report intermediate steps rounded to four significant figures. Give your final answer(s) to four significant figures. A spool of mass ms = 130 kg and inner and outer radii p = 0.7 m and R = 1.2 m, respectively, is connected to a counterweight A of mass m A = 50 kg via a pulley system whose cord, at one end, is wound around the inner hub of the spool. The center G and the center of mass of the spool coincide, and the radius of gyration of the spool is KG = 1 m. The system is at rest when the counterweight is released causing the spool to move to the right. The spool rolls without slip and the cord unwinds from the spool without slip. Neglecting the inertia of the pulley system, use the impulse-momentum principles to determine the angular speed of the spool 2 s after release.

Explanation / Answer

inertia of spool = ms * kg^2 = 130*1^2 = 130 kg m^2
Let the tension in the rope be T ..

let the friciton between the spool and ground be f
Let the angular acceleration be alpha

The rope and the spool will move with different linear accelerations..

so.. for spool ..Let the linear accelerations be as
so.. as = R*alpha

T - f = ms*as
T - f = 130*1.2*alpha

so.. f = T - 156*alpha

also ..
T*0.7 + f*1.2 = 130 * alpha
T*0.7 + T*1.2 - 187.2alpha = 130*alpha
so.. T = 166.94737 *alpha


for the mass...
acceleration = acceleration of rope / 3 = alpha * 0.7 / 3

mA*g - 3T = mA*a
50*9.8 - 3*166.947368 *alpha = 50alpha * 0.7/3

so.. alpha = 0.9560812 rad/sec2

so.. angular accelerion = 0.9560812 rad/sec2

so... angular velocity after 2 secs = 0.9560812 * 2 = 1.9121624 rad/sec

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