Question

A disk with a c of 1/2 about its center, has a mass of 4 kg, and a radius of 0.7 meters. It is originally at rest and a unknown force is applied at the edge of the disk in the -x direction directly above the center. Another force of 13 N is applied directly to the left of the center, halfway between the center edge, in the +y direction. Another force of 6 Newtons is applied directly below the center at the edge of the disk in the +x direction. Lastly, a force of 6 Newtons is applied directly to the right of the center in +x direction at the edge of the disk. Ignoring the normal force and weight force, after these forces are applied for 2 seconds, the rotational kinetic energy of the disk about its center is 251 Joules. How much, in Newtons, is the unknown force?

Answer 1

F1 = ?

F2 = 13 N

F3 = 6 N

F4 = 6 N

kinetic energy K = (1/2)*I*w^2

K = (1/2)*(1/2)*M*R^2*w^2

251 = (1/2)*(1/2)*4*0.7^2*w^2

W = angular speed = 22.6 rad/s

angular acceleration alpha = (w - wo)/t

alpha = (22.6-0)/2 = 11.3 rad/s^2

torque due to F1 , T1 = +F1*R*sin90 = +F1*R

torque due to F2 , T2 = -F2*R/2*sin90 = -6.5*R

torque due to F3 , T3 = +F1*R*sin90 = +6*R

torque due to F4 , T4 = -F1*R*sin180 = 0

net torque = F1*R - 6.5*R + 6*R + 0

from newtons second law

net torque = I*lapha

F1*R - 0.5*R = (1/2)*M*R^2*alpha

F1 - 0.65 = (1/2)*4*0.7*11.3

F1 = 16.47 N