A string is wrapped around a uniform disk of mass M = 2.2 kg and radius R = 0.11

Question

A string is wrapped around a uniform disk of mass M = 2.2 kg and radius R = 0.11 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 = 0.17 m, each with a small mass m = 0.8 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 = 21 N. At the instant when the center of the disk has moved a distance d = 0.049 m, a length w = 0.030 m of string has unwound off the disk. (a) At this instant, what is the speed of the center of the apparatus? v = 0.617 m/s (b) At this instant, what is the angular speed of the apparatus? omega _1 = 3.598 radians/s (c) You keep pulling with constant force 21 N for an additional 0.022 s. Now what is the angular speed of the apparatus? omega _2 = radians/s

Explanation / Answer

(A) Fnet = m a

21 = (2.2 + 4(0.8)) a

a = 3.89 m/s^2

and vf^2 - vi^2 = 2 a d

v^2 - 0^2= 2(3.89)(0.049)

v = 0.617 m/s

(B) moment of inertia = (2.2 x 0.11^2 /2) + (4 x 0.8 x 0.17^2 )

I = 0.106 kg m^2

Applying torque = I alpha

(0.11 x 21) = 0.106 alpha

alpha = 21.84 rad/s^2

angle revolved = w/r = 0.03 / 0.11 = 0.273 rad

wf^2 - wi^2 = 2(alpha) (theta)

w^2 - 0 = 2(21.84)(0.273)

w = 3.45 rad/s

(c) wf = wi + alpha t

w = 3.45 + (21.84 x 0.022) = 3.93 rad/s

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