The combination of an applied force and a friction force produces a constant tot
ID: 1439174 • Letter: T
Question
The combination of an applied force and a friction force produces a constant total torque of 35.0 N. m on a wheel rotating about a fixed axis. The applied force acts for 6.10 s. During this time, the angular speed of the wheel increases from 0 to 10.3 rad/s. The applied force is then removed, and the wheel comes to rest in 59.9 s. Find the moment of Inertia of the wheel, Find the magnitude of the torque due to friction. Find the total number of revolutions of the wheel during the entire Interval of 66.0 s.Explanation / Answer
a) net torque, Tnet = I*alfa
==> I = Tnet/alfa
= Tnet/(wf-wi)/t
= Tnet*t/(wf - wi)
= 35*6.1/(10.3 - 0)
= 20.7 kg.m^2
b) friction due to torque, T_friction = I*alfa_friction
= 20.7*(0 - 10.3)/59.9
= -3.56 N.m
,agnitude of toruw due to friction = 3.56 N.m
c) angular displacement = (10.3^2 - 0^2)/(2*(10.3 - 0)/6.1) + (0^2 - 10.3^2)/(2*(0 - 10.3)/59.9)
= 340 rad
= 34*/(2*pi)
= 54.1 revolutions