Question

A solid disk with a c value of 1/2, mass of 3 kg, and radius of 1.5 meters lies on a horizontally (so the normal force and weight can be ignored). A force of 9 Newtons is applied 0.22 meters directly to the right of the center in the +x direction, a force of 24 Newtons is applied at 0.81 meters directly below of the center in the +x direction and a force of 8 Newtons is applied at the edge of the disk directly above the center in the +x direction. Again, ignoring the weight and normal force, what is the angular velocity of the disk after these forces are applied for 3 sec, in rad/s? If clockwise, include a negative sign.

Answer #1

here,

mass of disk , m = 3 kg

radius , r = 1.5 m

F1 = 9 N is at r1 = 0.22 m

F2 = 24 N is at r2 = 0.81 m

F3 = 8 N is at r3 = r = 1.5 m

let the angular acceleration be alpha

equating the torques

T = I * alpha

(F1 * r1 * sin(90) + F2 * r2 * sin(90) - F3 * r3 * sin(90)) = (0.5 * m * r^2) * alpha

(0 + 24 * 0.81 - 8 * 1.5) = 0.5 * 3 * 1.5^2 * alpha

solving for alpha

alpha = 2.2 rad/s^2

t = 3 s

the angular velocity , w = 0 + alpha * t

w = 0 + 2.2 * 3 rad/s

w = 6.6 rad/s

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