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Calculating the horsepower for a motor

  • #1

Homework Statement


This is not really a homework question, it is a question for a project I am working on. So I have a heavy rod that is 1.4 m long and I want to get a motor to rotate it (I would like to see if this is even plausible). I calculated the moment of inertia for this rod to be 75.2 kgm2. Now I want the rod to rotate approximately 42 degrees in around 0.4 seconds. Thus using the kinematic equations of motions I have:

θ = 0.5αt2
0.746 = 0.5(α)(0.4)2
α = 9.323 rad2/s​

Not to find the torque I need for this to occur:
Iα = τ
τ = 75.2(9.323)
τ = 701.1 Nm = 571 ft lb​

Finally to figure out the horsepower I can do:
HP = τ(RPM)/5252​

The issue I have here is how do I know what RPM to select? Or do I base it on the standard for a 60 Hz motor (ie. 1200 RPM or 1800 RPM). For example, if I choose 1 RPM, the torque suggests that it should move 42 degrees in 0.4 seconds but at 1 RPM that just seems unreasonable. Also, how do I account for the winding up of the motor? I would imagine it takes a little bit of time just to wind up to a certain speed. I guess I should choose a DC motor for the best results since there is less of a wind up?
 

Answers and Replies

  • #2
SteamKing
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42 degrees = 42/360 = 0.1167 rev
t = 0.4 sec
n = 0.1167 rev/0.4s = 0.2917 rev/sec
N = 60 * n = 17.5 rev/min = 17.5 RPM

P = 571*17.5/5252 = 1.90 hp

Calculations are easy once you stop confusing yourself.
 
  • #3
42 degrees = 42/360 = 0.1167 rev
t = 0.4 sec
n = 0.1167 rev/0.4s = 0.2917 rev/sec
N = 60 * n = 17.5 rev/min = 17.5 RPM

P = 571*17.5/5252 = 1.90 hp

Calculations are easy once you stop confusing yourself.
So I would need a gearbox that can get me to 17.5 RPM right? When choosing this gearbox wouldn't my torque values change as well so that would affect the horsepower value? Finally, doesn't it take time to wind up so it won't be 17.5 RPM right at the start? The rod in total weighs 250 lbs.
 
  • #4
SteamKing
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You wanted the piece to move 42 deg. in 0.4 sec. That's equivalent to 17.5 RPM. Starting and stopping this piece, for whatever reason, I figure, is your job. After all, you know the details.
 
  • #5
CWatters
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I get a different answer...

The final angular velocity v is given by..

v = u + a*t

Where

u = Initial angular velocity = 0
α = The angular acceleration = 9.323 rad/s2

So

v = = 9.323 * 0.4
= 3.73 rad/s

Converting that to rpm gives 35.6 rpm after 0.4 seconds.

So you need a motor and gearbox that can deliver the required torque from zero rpm upto to at least 35.6 rpm.

The power will vary during this time but will be at a maximium at max angular velocity.

Power = Torque * angular velocity
= 701 * 3.73
= 2614W
= 3.49 horsepower.

The use of a gearbox will effect the torque and rpm required from the motor but not the power. A gearbox reduces one and increases the other so the product (the power) remains the same. (Can also be implied from conservation of energy). Frictional losses in the gearbox would need to be compensated for by increased power.

It's late here so someone check my figures!
 

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