The rotor of an electric motor has a mass 250 kg and a diameter of 48 cm. The mo

Question

The rotor of an electric motor has a mass 250 kg and a diameter of 48 cm. The motor is switched on and the rotor set spinning. Calculate the power required to accelerate the rotor from rest to 1,500 rev/min in 6s. (Assume the radius of gyration of a cylinder is 0-707r and zero frictional forces in the motor.) A shaft is coupled lo the motor and the combined Moment of inertia of shaft and motor is now 22 kg m2. The speed of the motor now drops to 900 rev/min and it is accelerated again to bring it back to 1500 rev/min during 10 revolutions of the shaft. In the friction of the couple is 80 Nm, calculate i. The change in kinetic energy of rotation of the combined shaft and rotor to bring the speed back to 1500 rev/min. ii. The average torque required to produce the acceleration.

Explanation / Answer

b) I = 22 kgm2, initial angular velocity 0 = 900 rev/min = 900*2/60 rad/s = 94.25 rad/s

final angular velocity = 1500 rev/min = 1500*2/60 rad/s = 157.08 rad/s

angular displacement = 10 rev = 10*2 rad

f = 80 Nm

i) E = I2/2 - I02/2 = 1.75*105 J

ii) ( - f)* = E

= 2845 Nm

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