A 8 kg disk with a radius of 2 meters has several forces applied to it as shown

Question

A 8 kg disk with a radius of 2 meters has several forces applied to it as shown in the sketch. F_1 is 38 N and is applied at the edge of the disk, F_2 is 13 N is applied at 1/4 the radius from the center of the disk, F_3 is 6 N is applied at 3/4 the radius from the center, F_4 is 21 N is applied at the edge of the disk, and, lastly, F_5 is 19 N and is applied at the edge of the disk at an angle of 25 degrees. If the moment of inertia about the center for the disk is I = 1/2 M R^2, and the disk is initially spinning at 32 rad/s counter-clockwise its center, how long, in seconds, does it take the disk to be spinning at 9 rad/s counter-clockwise about its center?

Explanation / Answer

Mass of disc = 8kg ; Radius of disc = 2m

let us determine the net torque acting in the disc about X

Torque due to F1 = F1r = 38*2 = 76Nm(clockwise)

Torque due to F2 = F2r/4 = 13*2/4 = 6.5Nm(anticlockwise)

Torque due to F3 = F3(3/4)r = 6*(3/4)*2 = 9Nm(anticlockwise)

Torque due to F4 = F4 *0 = 0

Torque due to F5 = F5 cos 25o r = 19cos25o *2 = 34.44Nm(clockwise)

Net torque = 76-6.5-9+34.44 = 94.94Nm(clockwise)

mass moement of inertia of disc =I= (1/2)mr2 = (1/2)*8*22 = 16kgm2

Let angular acceleration be =a rad/s2

T=Ia

a=T/I = 94.94/16 = 5.934 rad/s2 (clockwise)

Initial angular speed = 32rad/s (anticlockwise)

Final angular speed = 9rad/s(anticlockwise)

Time taken to reduce speed from 32 to 9 is = (32-9)/5.934 = 3.88 seconds

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