The angular momentum of a flywheel having a rotational inertia of 0.884 kg middo

Question

The angular momentum of a flywheel having a rotational inertia of 0.884 kg middot m^2 about its central axis decreases from 9.90 to 0.750 kg m^2/s in 1.50 s. (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the magnitude of the average power done on the flywheel? (a) Number Units (b) Number Units (c) Number Units (d) Number Units By accessing this Question Assistance, you will learn while you earn points based on the Point Potential policy set by your instructor.

Explanation / Answer

a) we know, Torque = the rate of change of angular momentum

= dL/dt

= (9.9 - 0.75)/1.5

= 6.1 N.m

b) angular acceleration, alfa = -T/I

= -6.1/0.884

= -6.9 rad/s^2

initial angular velocity, wi = 9.9/0.884 = 11.2 rad/s

final angular velocity, wf = 0.75/0.884 = 0.848 rad/s

angular displacement, theta = wi*t + (1/2)*alfa*t^2

= 11.2*1.5 + (1/2)*(-6.9)*1.5^2

= 9.04 radians

c) Workdone = change in kinetic energy

= (1/2)*I*(wf^2 - wi^2)

= (1/2)*0.884*(0.848^2 - 11.2^2)

= -55.1 J

d) AVerage power = |Workdone|/t

= 55.1/1.5

= 36.7 W

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