A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.08

Question

A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.08 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR^2.) Attached to the disk are four low-mass rods of radius b = 0.16 m, each with a small mass m = 0.5 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 32 N. At the instant when the center of the disk has moved a distance d = 0.035 m, a length w = 0.013 m of string has unwound off the disk. At this instant, what is the angular speed of the apparatus? At this instant, what is the speed of the center of the apparatus? You keep pulling with constant force 32 N for an additional 0.023 s. Now what is the angular speed of the apparatus?

Explanation / Answer

b)

torque = F*R

net torque = I*alpha

F*R = I*alpha

F*R = ((1/2)*M*R^2 + 4*m*b^2)*a/R

32*0.08 = ((1/2)*1.2*0.08^2)+(4*0.5*0.16^2))*a/0.08

a = 3.72 m/s^2 <<_----answer

v = sqrt(2*a*d) = sqrt(2*3.72*0.035)= 0.51 m/s

++++++++++++++

alpha = a/R

angle rotated theta = w/R = 0.013/0.08 = 0.1625 rad

w = sqrt(2*alpha*theta)

w = sqrt(2*3.72/0.08*0.1625)

w = 3.88 rad/s

___________

c)

w1 = w1 + alpha*t

w1 = 3.88 + (3.72/0.08*0.023) = 4.95 rad/s

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