Can you answer a b c and Also calculate numerically the angle through which the

Question

Can you answer a b c and

Also calculate numerically the angle through which the apparatus turns, in radians and degrees.

??radians = rad ??degrees = ° 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)MR2.) Attached to the disk are four low-mass rods of radius b-0.15 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-22 N. At the instant when the center of the disk has moved a distance d = 0.048 m, a length w = 0.018 m of string has unwound off the disk. 1m M, R w+ d (a) At this instant, what is the speed of the center of the apparatus? x m/s (b) At this instant, what is the angular speed of the apparatus? radians/s (c) You keep pulling with constant force 22 N for an additional 0.026 s. Now what is the angular speed of the apparatus? 2 = radians/s

Explanation / Answer

a)

we know that from the kinetic equation we get

v2 =2as

s =0.048m,

a =F/(1.2+4*0.8) =22N/(1.2+4*0.8) =5m/s2

Then the speed at the center of the apparatus is v =Sqrt(2*5*0.048) =0.692m/s

b)

we know that

t =I*alpha

22*0.08m =(MR2/2+4mb2)alpha

=[(1.2)(0.08)2/2+4*0.8kg*(0.15)2]alpha

=[0.0758]alpha

then alpha =1.76/0.0758 =23.218rad/sec2

then the angular speed of the apparatus is theta =w/R =0.018m/0.08 =0.225rad

wo =Sqrt(2*alpha*theta) =Sqrt(2*23.218rad/sec2*0.225rad) =3.232rad/sec

c)

Now the final angular speed of the apparatus is

wf =wo+alpha*t =3.232rad/sec+(23.218rad/sec2)(0.026s) =3.835rad/sec

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