The angular momentum of a flywheel having a rotational inertiaof 0.200 kg·m 2 ab

Question

The angular momentum of a flywheel having a rotational inertiaof 0.200 kg·m2 about itscentral axis decreases from 3.20 to 1.600 kg·m2/sin 1.60 s. (a) What is the magnitude of the average torque acting on theflywheel about its central axis during this period?
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 torque acting on theflywheel about its central axis during this period?
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 = 0.2 kg m^ 2

Initial angular momentum L = 3.2 kg m^ 2/ s

Final angular momentum L ‘ = 1.6 kg m^ 2/ s

Time t= 1.6 s
(a) .  the magnitude of the average torque acting on theflywheel about its central axis during this period T = dL / t

from this angular acceleration = T / I

                                                = 5 rad / s ^ 2

let the angle does the flywheel turn be thenfrom the relation = w't - ( 1/ 2) t ^ 2

                 = 0 -(0.5 * 5 * 1.6 ^ 2) Since final angularspeed w ' = 0

                   = 6.4 rad
(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

                             = 8 rad / s

So, W = 6.4 J

(d).Power P = W / t= 4 watt

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