Calculate the time????

Question

i only have one question.Need help with #7 Use the others for information.

Thanks!!!!!!

1) The armature of a shunt motor has a mass of 250 kg and a diameter of 30 cm. Assume the mass of the armature is evenly distributed in the armature. Find the rotational inertia of the armature.

Ans. 2.81 Kg-m^2

2) How fast will the unloaded armature accelerate in comming up to speed if the armature current is so regulated that the armature developes a constant torque of 120 N-m?

Ans. 42.7 rad/sec^2

3) Assume the torque stays constant at 120 N-m. How long will it take for the unloaded armature to accelerate from zero up to full speed of 3000 rpm?

ans: 7.35 sec

4) What horsepower is developed by the motor while running at 3000 rpm and delivering a torque of 120 N-m to a load?

Ans. 50.5 hp

5) If the power to the motor is shut off while the motor is running at 3000rpm and a brake is applied, how much heat in joules will be dissipated in the brake in order to bring the motor to a stop?

Ans. 138,527 J

6) Suppose the motor has a bearing shaft diameter of 3.2 cm and the coefficient of friction of each bearing is 0.008. find the torque of friction of the bearings.

Ans. 0.314

7)referring to problem 5, suppose that, instead of the brake being applied, the motor with the power cut off is allowed to run until it is brought to a stop by the torque of friction of the bearings. how long will the armature take to come to a stop?assume all other losses (such as windage) are zero.

a. 420 sec

b. 1260 sec

c. 2810 sec

tHanks!!!^_^

Answer 1

Let:

T be the torque,

I be the moment of inertia,

a be the angular deceleration

t be the time to stop,

w be the initial angular speed.

T = Ia

0 = w - at

Therefore:

t = w / a

t = wI / T

w = 3000 * 2pi / 60 rad/s

I = 2.81 kg m^2

T = 0.314 Nm

t = (3000 * 2pi * 2.81) / (60 * 0.314)

= 2811.4

= 2810 sec. to 3sf.

Answer (c).