How Long Does It Take for a Motor to Stop Due to Bearing Friction Alone?

  • Thread starter Thread starter iLuveLfenLied
  • Start date Start date
  • Tags Tags
    Time
AI Thread Summary
The discussion revolves around calculating the time it takes for a motor to stop due to bearing friction. Key calculations include determining the rotational inertia of the armature, which is 2.81 Kg-m^2, and the acceleration of the unloaded armature at a constant torque of 120 N-m, resulting in 42.7 rad/sec^2. Participants are specifically debating the time required for the armature to accelerate to full speed and the stopping time due to bearing friction, with options provided for both scenarios. The stopping time, assuming no other losses, is under scrutiny, with one participant suggesting it takes 420 seconds. Accurate calculations and clarifications on methodology are encouraged for verification.
iLuveLfenLied
Messages
1
Reaction score
0
calculate the time?

Need help with answer #3 and #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?
a. 7.35 sec
b. 34.1
c. 70.2
(my answer is c..am i correct?)

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
(my answer is a..am i correct?)
 
Physics news on Phys.org


It would be easier if you would show your work rather than just giving your answer. As a general rule, it takes less time to check work than it does to do the question oneself and then say whether you are right or wrong.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top