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I Power Output Given Certain Torque

  1. Apr 23, 2016 #1
    Hi,
    I have a problem that I've been trying to figure out for a while but cannot seem to get on my own. I'm trying to figure out how much power will be generated in a motor given a certain amount of torque. The problem is, to find the amount of power output , I also need to know the rotational velocity, which I cannot seem to find. I don't know how much resistance the motor would give.

    Here's a simple example:
    A man rotates a shaft of radius 5 m, with a constant force of 10 N, on a dc motor. How much power does the motor output?

    I know that the rotational velocity is missing, as well as the number of turns in the motor, but I don't know how much it would be without knowing the specifications of the motor. If you could steer me in the right direction and maybe give me similar examples, it would be much appreciated, thanks!

    Also, how would I know how much resistance the motor provides, in Newtons?
     
  2. jcsd
  3. Apr 24, 2016 #2

    Svein

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    As Power = Torque × Rotational speed, you need to know more.
     
  4. Apr 25, 2016 #3
    As you seem to understand, without the rpm you simply cannot know. However, assuming that as the man applies 10 N (which would be 50 Nm of torque) the shaft does not accelerate then that means that the resistance is also 10 N (50 Nm). If it wasn't, then the shaft would accelerate until the forces balance out, until the man can't keep up, or until something gets destroyed.
     
  5. Apr 25, 2016 #4

    russ_watters

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    Staff: Mentor

    Your best bet with the information given is to estimate how fast a person can get around a 5m radius circle.
     
  6. Apr 25, 2016 #5

    Nidum

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    The rotational speed is going to be impossibly slow for a directly driven dynamo but let's work it out anyway for curiosity :

    4 minute mile : 15 mph or 6.7 M/sec
    Walking speed : 3 to 5 mph or 1.35 to 2.25 M/Sec

    Choose a spot value - say 2 M/sec

    Circumference of circle = 10 Pi = 31.42 M .

    Revs/sec = 2 / 31.42 = 0.064 rev/sec or 3.8 rpm .
     
    Last edited: Apr 25, 2016
  7. Apr 25, 2016 #6

    Merlin3189

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    You boys have beaten me to it - fortunately, as I'm much more long-winded!
    Anyhow, when I got to working out rotational speed, I thought maybe that's not important. 10N at 2m/s is 20W mechanical input power to the dynamo and use whatever gears you need to make that a sensible rotation speed.

    The suggestion I would make to OP is, stop thinking of it as a motor and just Google dynamo or DC generator and find endless pages of explanation at all levels

    But I hate these cryptic questions: they intrigue me, but frustrate me. I just hope he tells us what it is he really wants to know.
     
  8. Apr 27, 2016 #7
    Sorry it took so long to get back, but thanks so much for the explanations! That helps me to understand it better. As for the cryptic question, I'm sorry, but I don't know if I'll be able to get much more specific. I want to be able to understand how much power output I will get if I can be applying a constant force to a motor, but it makes it hard when I don't know how fast that shaft will rotate due to the force I place on it. However, it does seem a little bit clearer now!
     
  9. Apr 27, 2016 #8

    Merlin3189

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    Glad you came back. Sorry I was so tetchy about it!
    This is one point that hasn't really come up yet, unless I missed it before. If you try to rotate a dynamo, the resistance depends on the electrical load attached. A dynamo unconnected is easy to rotate - you are only overcoming friction and magnetic losses. Once a load is connected and current flows, that creates another force (torque), usually the dominant one, opposing the rotation. (If the loss force is large, compared to the load force, then the generator is not efficient. This site suggests 85% is average for PMDC generator.)
    "How fast the shaft will rotate" is not a simple question (without more info) because it is determined by a balance between several variables. If the shaft turns faster, the output emf increases, so the current through the load probably increases, which increases the opposing torque and may tend to slow the shaft. That's the output side. On the input side, you say the man pushes with 10N. Being intelligent he can monitor that and try to keep it constant as the shaft speed varies, but there will be a limit, if it goes too fast and he can't keep up, when that force will fall. Or he may try to keep the speed constant by pushing harder if it slows and less if it speeds up. Again there must be a limit to how hard he can push.

    You see, so much depends on what your real problem is. I find it hard to believe that someone is walking round in a 5m radius circle pushing a long arm with only 10N force, so I assume that is just some random numbers you made up to help us. If we knew what you really wanted to know (or maybe, why you want to know) there's more chance we can say something useful.
    As others have said, that simply is not an answerable question. Power out depends on power in. Force is not power.
    So what I'm asking is, why do you say "constant force"? Where does this constant force come from? Maybe if we know more about it, we can work out what power you have available, then we'd be getting somewhere.
     
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