The combination of an applied force and a friction force produces a constant tot
ID: 1468233 • Letter: T
Question
The combination of an applied force and a friction force produces a constant total torque of 35.2 N · m on a wheel rotating about a fixed axis. The applied force acts for 6.20 s. During this time, the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.6 s.
(a) Find the moment of inertia of the wheel.
kg · m2
(b) Find the magnitude of the torque due to friction.
N · m
(c) Find the total number of revolutions of the wheel during the entire interval of 65.8 s.
revolutions
Explanation / Answer
here,
let the moment of inertia of the wheel be I
(a)
final speed after t = 6.2 s , w = 10.2 rad/s
let the accelration be a
w = w0 +a*t
10.2 = 0 a*6.2
a = 1.65 rad/s^2
torque = 35.2 N.m
35.2 = I*a = I*1.65
I = 21.4 kg.m^2
the moment of inertia is 21.4 kg.m^2
(b)
wheel comes to resr in t = 59.6 s
there fore , let the accelration be a'
w = w0 + a'*t'
0 = 10.2 + a*59.6
a = - 0.17 rad/s^2
this deaccelration is due to frictional torque
the torque due to friction = I*a'
= 21.4 * 0.17
= 3.66 N.m
the magnitude of the torque due to friction is 3.66 N.m
(c)
let the total distance travelled be d
d = distance when wheel is accelrating + distance when wheel is deaccelrating
d = ( 0 + 0.5 * 1.65 * 6.2^2) + ( 10.2*59.6 - 0.5 * 0.17 * 59.6^2)
d = 337.7 rad
number of rotation , n = d/2pi = 53.77 rotations