Question

A solid disk with a c 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 7 Newtons is applied 0.34 meters directly to the right of center in the +x direction, a force of 23 Newtons is applied 0.77 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, if the disk starts from rest, what is the angular velocity of the disk after these forces are applied for 2 seconds, in rad/s? If clockwise, include a negative sign.

Answer 1

given
M = 3 kg
R = 1.5 m
moment of inertia of the disk about its center,

I = (1/2)*M*R^2

= (1/2)*3*1.5^2

= 3.375 kg.m^2

Net torque, Tnet = F1*r1*sin(0) + F2*r2*sin(90) - F3*R*sin(90)

= 0 + 23*0.77 - 7*1.5

= 7.21 N.m

angular acceleration, alfa = Tnet/I

= 7.21/3.375

= 2.14 rad/s^2

angular velocity of the disk at t = 2s,

w = wo + alfa*t

= 0 + 2.14*2

= 4.28 rad/s <<<<<<<<<<--------------------Answer