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rotational dynamics
flywheel battery can be used in place of lead-acid batteries to provide a short-term electrical power supply when mains power fails. Energy is stored as rotational kinetic energy in a rapidly spinning rotor, which is driven up to speed by a mains-powered motor. To recover the energy, the motor is operated in reverse as a generator driven by the spinning rotor.
(a) The rotor of a flywheel battery is a thin-walled circular cylinder of mass 160 kg and mean radius 0.34 m, which can be rotated at a maximum safe angular speed of 44 000 rev min–1.
Calculate
(i) the moment of inertia of the rotor about its own axis,
.................................................................................................................
(ii) the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev min–1.
.................................................................................................................
.................................................................................................................
.................................................................................................................
(3)
(b) The manufacturer claims that the power loss due to friction when the rotor is running at its maximum safe angular speed is 2.0 W and that the mean power loss over the range of speeds from rest up to the maximum safe angular speed is 0.60 W.
Use the manufacturer’s data to calculate
(i) the frictional torque acting on the rotor when spinning at its maximum safe angular speed,
.................................................................................................................
.................................................................................................................
(ii) the time it takes to come to rest from this angular speed under the action of the frictional torque alone.
.................................................................................................................
.................................................................................................................
(2)
(c) When it is required to supply electrical power during a mains failure, the flywheel battery can supply a constant load of 2.5 kW for 17 hours. Calculate the rotor’s angular speed at the end of this period if it starts at 44 000 rev min–1.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(3)
(Total 8 marks)
(a) (i) (use of I = Smr2 gives) I = 160 × 0.342 = 18.(5) kg m2 (1)
(ii) w = = 4.6(1) × 103 rad s–1 (1)
(use of Ek = ½Iw2 gives) Ek = 0.5 × 18.5 × (4.61 × 103)2 = 1.9(6) × 10^8J (1)
(allow C.E. from (a)(i) and (a)(ii)) 3
(b) (i) (use of P = Tw gives) T = (1)
(allow C.E. from (a)(ii))
(ii) Ek = Pt t = = 3.2(7) × 108 s (1)
(allow C.E. from (a)(ii)) 2
(c) energy supplied = 2.5 × 103 × 17 × 60 × 60 = 1.5(3) × 10^8J (1)
½Iw22 – ½Iw12 = 1.5 (3) × 10^8(J) (1)
w2 = 2.1(7) × 103 rad s–1 (20, 700 rev min–1) (1)
(allow C.E. from (a)(ii)) 3
[8]
could somebody explain part c) please? i've got all the answer correct up till then
(a) The rotor of a flywheel battery is a thin-walled circular cylinder of mass 160 kg and mean radius 0.34 m, which can be rotated at a maximum safe angular speed of 44 000 rev min–1.
Calculate
(i) the moment of inertia of the rotor about its own axis,
.................................................................................................................
(ii) the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev min–1.
.................................................................................................................
.................................................................................................................
.................................................................................................................
(3)
(b) The manufacturer claims that the power loss due to friction when the rotor is running at its maximum safe angular speed is 2.0 W and that the mean power loss over the range of speeds from rest up to the maximum safe angular speed is 0.60 W.
Use the manufacturer’s data to calculate
(i) the frictional torque acting on the rotor when spinning at its maximum safe angular speed,
.................................................................................................................
.................................................................................................................
(ii) the time it takes to come to rest from this angular speed under the action of the frictional torque alone.
.................................................................................................................
.................................................................................................................
(2)
(c) When it is required to supply electrical power during a mains failure, the flywheel battery can supply a constant load of 2.5 kW for 17 hours. Calculate the rotor’s angular speed at the end of this period if it starts at 44 000 rev min–1.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(3)
(Total 8 marks)
(a) (i) (use of I = Smr2 gives) I = 160 × 0.342 = 18.(5) kg m2 (1)
(ii) w = = 4.6(1) × 103 rad s–1 (1)
(use of Ek = ½Iw2 gives) Ek = 0.5 × 18.5 × (4.61 × 103)2 = 1.9(6) × 10^8J (1)
(allow C.E. from (a)(i) and (a)(ii)) 3
(b) (i) (use of P = Tw gives) T = (1)
(allow C.E. from (a)(ii))
(ii) Ek = Pt t = = 3.2(7) × 108 s (1)
(allow C.E. from (a)(ii)) 2
(c) energy supplied = 2.5 × 103 × 17 × 60 × 60 = 1.5(3) × 10^8J (1)
½Iw22 – ½Iw12 = 1.5 (3) × 10^8(J) (1)
w2 = 2.1(7) × 103 rad s–1 (20, 700 rev min–1) (1)
(allow C.E. from (a)(ii)) 3
[8]
could somebody explain part c) please? i've got all the answer correct up till then
You work out the energy that the battery supplies, add that to the kinetic energy of the flywheel to get the new kinetic energy then work out the new angular velocity from the energy.
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