The combination of an applied force and a friction force produces a constant tot
ID: 1496457 • Letter: T
Question
The combination of an applied force and a friction force produces a constant total torque of 36.3 N middot m on a wheel rotating about a fixed axis. The applied force acts for 5.80 s. During this time, the angular speed of the wheel increases from 0 to 9.8 rad/s. The applied force is then removed, and the wheel comes to rest in 59.0 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 64.8 s.Explanation / Answer
from the rotational kinematic equation
wf = wi + alpha t
wi =0
alpha = wf/ t =9.8/5.5 = 1.68 rad/s^2
T = I alpha
I = T/ alpha = 36.3 Nm/ 1.68 = 21.48 kg m^2
(b)
wf = wi + alpha t
0 = 9.8 + apha ( 59)
alpha = -0.166 rad/s^2
T = I alpha = 21.48 kg m^2( 0.166) = 3.56 Nm
(c)
theta _f = 1/2 (1.68) ( 5.8)^2 = 28.25 rad
for t = 64.8 s
thet_f = 9.8(64.8) - 1/2 ( 0.166 rad/s^2) ( 64.8)^2 = 286.51968 rad
total revolution is 286.51968+ 64.8 =351.3 rev