Solving Motor Rotation: 0 to 1800rpm in 0.49s?

  • Thread starter Thread starter fahraynk
  • Start date Start date
  • Tags Tags
    Motor Rotation
AI Thread Summary
The discussion revolves around calculating the time required for an electric motor's rotor to accelerate from 0 to 1800 rpm with a constant torque of 20 in-lb. The initial calculations incorrectly treated weight as mass, leading to an erroneous time estimate of 188.5 seconds instead of the correct 0.49 seconds. Participants emphasized the importance of using consistent units, noting that weight must be converted to mass by dividing by gravity. Additionally, torque should be converted from inch-pounds to foot-pounds for accurate calculations. Ultimately, the correct answer is confirmed to be 0.49 seconds.
fahraynk
Messages
185
Reaction score
6

Homework Statement


The rotor of an electric motor weighs 10 pounds and is 4 inches in diameter. What is the length of time required for the motor speed to increase from 0 to 1800rpm, assuming a constant electrical torque of 20 in-lb and zero external load during this period? Assume the rotor is a homogeneous cylinder.

Homework Equations

The Attempt at a Solution


Moment of inertia (I) of cylinder according to google is $$I=1/2 MR^2 = 1/2*10*2^2=20\\\\\
1800 rpm = \frac{1800}{60} \frac{rotations}{second} * \frac{2\pi rads}{rotation} = 188.5\\\\
20*188.5=20t\\\\
t=188.5$$
The book says the answer is 0.49 seconds. I got 188.5 seconds. Which is correct?
 
Physics news on Phys.org
0.49 s is correct.

Your overall approach is fine. But you are not giving enough consideration to the units of the various quantities. For example, the rotor weighs 10 pounds. But does M in the formula I = (1/2)MR2 represent weight? You should rework the problem plugging in units for each quantity in the calculation. If the units in your calculation do not reduce to seconds when you solve for t, you know you made an error.
 
  • Like
Likes fahraynk
TSny said:
0.49 s is correct.

Your overall approach is fine. But you are not giving enough consideration to the units of the various quantities. For example, the rotor weighs 10 pounds. But does M in the formula I = (1/2)MR2 represent weight? You should rework the problem plugging in units for each quantity in the calculation. If the units in your calculation do not reduce to seconds when you solve for t, you know you made an error.
A slightly different take on the error... Not sure which is more helpful.
fahraynk said:
20 in-lb
The lb there does not represent mass. What does it represent?
 
  • Like
Likes fahraynk
I got it. There were 2 problems. They gave me weight but I thought it was mass, so I had to divide by gravity. The other was they gave me torque in inch pounds, I had to convert it to foot pounds. Thank you both.
 
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