Question

A disk with a c of 1/2 about its center, has a mass of 3 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 11 N is applied directly to the left of the center, halfway between the center edge, in the +y direction. Another force of 8 Newtons is applied directly below the center at the edge of the disk in the +x direction. Lastly, a force of 8 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 2.5 seconds, the rotational kinetic energy of the disk about its center is 329 Joules. How much, in Newtons, is the unknown force?

Answer 1

F1 = ?

F2 = 11 N

F3 = 8 N

F4 = 8 N

radius R = 0.7 m

c = 1/2

mass M = 3 kg

moment of inertia I = c*M*R^2

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

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

329 = (1/2)*(1/2)*3*0.7^2*w^2

W = angular speed = 30 rad/s

angular acceleration alpha = (w - wo)/t

alpha = (30-0)/2.5 = 12 rad/s^2

torque due to F1 , T1 = +F1*R*sin90 = +F1*R ( counterclockwise torque)

torque due to F2 , T2 = -F2*R/2*sin90 = -11*R/2*sin90 = -5.5*R ( clockwise)

torque due to F3 , T3 = +F3*R*sin90 = +8*R   ( counterclockwise torque)

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

net torque = F1*R - 5.5*R + 8*R + 0 = F1*R + 2.5R

from newtons second law

net torque = I*lapha

F1*R + 2.5*R = (1/2)*M*R^2*alpha

F1 + 2.5 = (1/2)*M*R*alpha

F1 + 2.5 = (1/2)*3*0.7*12

F1 = 10.1 N