Question

A disk with a hole cut out of the middle (only changes its c value) has a c value of 0.87 about its center, a mass of 8 kg and a radius of 0.7 meters. One force of 38 N is applied directly above the center on the edge of the disk in the -x direction. Another force of 23 N is applied directly to the right of the center, halfway between the center and edge, in the -y direction. Lastly, a force of 39 N is applied directly below the center of the disk at the edge in the - y direction. Ignoring weight and the normal force (only looking at given forces), if the disk started from rest, how many revolutions does the disk make if these forces are applied for 1.9 seconds?

Answer 1

using equillibrium condtion

summation of all torques on disk = 0

here it is

F1 R1 - F2 R1/2 = Tnet

38 R1   - 23 R1/2 = Tnet

Tnet   = 26.5 *R1

Tnet = 26.5* 0.7

Tnet = 18.55 Nm

Moment of inertia I = c mR^2

I = 0.87 * 8* 0.7^2

I = 3.41 kgm^2

also torque T = I * alpha

I is MOI and alpha is angular accleration

Alpha = Tnet/I

alpha = 18.55/3.41 = 5.44 rad/s^2

using theta (ang Diplacement )   = Wo t + 1/2 *alpha *t^2

theta = 0.5* 5.44* 1.9^2

theta = 9.81 rad

n = theta/2pi

n = no. of revolutions = 9.81/2pi = 1.56 revolutons