The angular momentum of a flywheel having a rotational inertia of0.100 kg·m 2 ab
ID: 1671780 • Letter: T
Question
The angular momentum of a flywheel having a rotational inertia of0.100 kg·m2 about its central axis decreases from3.80 to 1.600 kg·m2/s in 2.10 s. (a) What is the magnitude of the average torqueacting on the flywheel about its central axis during thisperiod?N·m
(b) Assuming a uniform angular acceleration, through what angledoes the flywheel turn?
rad
(c) How much work is done on the wheel?
J
(d) What is the average power of the flywheel?
W (a) What is the magnitude of the average torqueacting on the flywheel about its central axis during thisperiod?
N·m
(b) Assuming a uniform angular acceleration, through what angledoes the flywheel turn?
rad
(c) How much work is done on the wheel?
J
(d) What is the average power of the flywheel?
W
Explanation / Answer
Rotational inertia I = 0.1 kg m^ 2
Initial angular momentum L = 3.8 kg m^ 2/ s
Final angular momentum L ‘ = 1.6 kg m^ 2/ s
Time t= 2.1 s
(a) . the magnitude of the average torque acting on theflywheel about its central axis during this period T = dL / t
T = ( 1.6 - 3.8) / 2.1
= 1.047 N m
(b) we know T = I
from this angular acceleration = T / I
= -10.47 rad / s ^ 2
let the angle does the flywheel turn be thenfrom the relation = w't - ( 1/ 2) t ^ 2
= 0 -(0.5 * -10.47 * 2.1 ^ 2) Since final angularspeed w ' = 0
= 23.1rad
(c) work is done on the wheel W = Change in rotationalkinetic energy
= ( 1/ 2) I [ w ^ 2- w'^ 2]
where w = Initial angular speed
from hte relation w ' = w + t
w = w ' - t
= 21.987 rad / s
So, W = 24.17 J
(d).Power P = W / t= 11.51 watt
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