Question

A uniform disk with mass m = 8.63 kg and radius R = 1.38 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 347 N at the edge of the disk on the +x-axis, 2) a force 347 N at the edge of the disk on the

Answer 1

All forces are in the direction of y+ as given

R = 1.38 m

Torque due to F1 = F1*R = 347*1.38 = 478.86

F2 is passing through the center of disc..

Torque = ZERO

F3 is at a perpendicular distance of Rcos(33) from origin

Torque = 347*1.38*cos(33) = 401.6

since forces are in the xy plane only

there will be only torque in Z direction

the x-component of the net torque about the z axis on the disk = ZERO

the y-component of the net torque about the z axis on the disk = ZERO

the z-component of the net torque about the z axis on the disk = 478.86 - 401.6 = 77.26

Moment of inertia I = mr^2 /2 = 8.63 * 1.38^2 /2 = 8.22

alpha = 77.26 / 8.22 = 9.4 rad/s^2

angular velocity at t=1.8 w = 9.4*1.8 = 16.92 rad/s

energy = .5 Iw^2 = 1176.64