A string is wrapped around a uniform disk of mass M = 2.1 kg and radius R = 0.06

Question

A string is wrapped around a uniform disk of mass M = 2.1 kg and radius R = 0.06 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.10 m, each with a small mass m = 0.7 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 = 0.039 m, a length w = 0.022 m of string has unwound off the disk.

(a) At this instant, what is the speed of the center of the apparatus?
v =  m/s

(b) At this instant, what is the angular speed of the apparatus?
1 =  radians/s

(c) You keep pulling with constant force 32 N for an additional 0.040 s. Now what is the angular speed of the apparatus?
2 =  radians/s

Explanation / Answer

a) k trans = 1/2mtot*vcm^2

vcm = sqrt(2k/mtot)

vcm = sqrt(2f*d/M+4m) = 0.713 m/s

b) fd+L^2/2I=f(w+d)

L = sqrt(2Ifw)

L = Iw

w = sqrt(2fw/I) = sqrt(4fw/MR^2+8mb^2) = sqrt(2.816/0.00756+0.056)

w = 6.656 rad/s

c) Lf = (frt+sqrt(2Iwf))

wf = (0.0768+0.21)/0.06356 = 4.51 rad/s

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