Question

A solid disk with a c of 1/2, mass of 2 kg, and radius of 1.5 meters lies on a horizontally (so the normal force and weight can be ignored). A force of 8 Newtons is applied 0.2 meters directly to the right of center in the +x direction, a force of 20 Newtons is applied 0.7 meters directly below of center in the +x direction, and a force of 7 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 acceleration by these three forces about the center of the disk in rad/s2? If clockwise, include a negative sign.

Answer 1

here,

mass of disk , m = 2 kg

radius , r = 1.5 m

the force 1 , F1 = 8 N is at r1 = 0.2 m

the force 2 , F2 = 20 N is at r2 = 0.7 m

the force 3 , F3 = 7 N is at r3 = r = 1.5 m

let the angular acceleration be alpha

equating the torques

net torque = I * alpha

(F1 * r1 * sin(0) + F2 * r2 * sin(90) - F3 * r3 * sin(90)) = (0.5 * m * r^2) * alpha

( 8 * 0.2 * 0 + 20 * 0.7 - 7 * 1.5) = ( 0.5 * 2 * 1.5^2) * alpha

solving for alpha

alpha = 1.56 rad/s^2

the angular acceleration is 1.56 rad/s^2