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Homework Help: On finding Torque & RPM

  1. Apr 6, 2014 #1
    Presently, three things I need clarity on, please.

    Thing 1:

    I'll be using the following equation to arrive at fair approximations of torque and rpm, given various motor rotors of different radii, masses, and with different stator coils (electromagnets) providing the repelling force....

    T = I x alpha = F x r x sin(theta):

    By example...
    A rotor Mass of 100 kg,
    'r' of 2 meters,
    A force of 75 Newtons,
    point of force at a 90 degree angle.

    First filling in T = F x r x sin(theta), as such:
    T = 75 N x 2 m x sin(90) = 150 N*m
    T = 150 N*m

    Next, add the other part of the equation, T = I x alpha; I = 1/2Mr^2, hence:
    T = 150 N*m = 1/2M x r^2 x alpha
    Resolved progressively...
    T = 150 N*m = 50 kg x 2^2 x alpha
    T = 150 N*m = 50 kg x 4 x alpha
    T = 150 N*m = 200 x alpha

    Next, for alpha:
    alpha = 150 divided by 200 = .75 rad/s^2

    Next, convert .75 rad/s^2 to RPM using this equation:
    ([rad/s]/6.28318530718) x 60 = RPM
    *** Notice, rad/s is not squared in this equation. That is correct? ***
    (Note: 6.28318530718 was Pi x 2, I just filled it in....)
    Solved progressively...
    ([.75 rad/s]/6.28318530718) x 60 = RPM
    .11936620731891365 x 60 = RPM
    .11936620731891365 x 60 = 7.161972439134819 RPM
    .75 rad/s^2 = 7.161972439134819 RPM


    Thing 2: Please refer to attachment "thing 2".

    With the angle known, and the distance between A & B known, can the length of line 3 be found; line 3 being perpendicular to line 2 (as needed).

    I think I read somewhere that if the hypotenuse is multiplied by sin(theta), it will give the length of the 'opposite'.

    Thing 3: Some clarity on 'r', please.
    In finding moment of inertia, using I=1/2Mr^2, 'r' is strictly the radius of the mass in question; never in any way having to do with where the point of action (of the repelling force) is? Not 100% clear on that. The obvious answer seems 'yes', but uncertainty creeps in when, in 'thing 1', converting rad/s to RPM. I'm thinking that if it's the circumference of the rotor that we are using to arrive at rad/s and/or RPM, the 'r' in this case might be the distance from the axis to the point of force, which is slightly outside the tangible rotor, where the actual repelling force created by the interaction of the rotor magnet and the stator coil occurs; and hence a slightly greater radius, circumference, distance...., see what I'm getting at? Leave it be if not. My capacity for these things is only so forgiving here at this point....

    My apologies if the questions could have been clearer.

    Sincerely grateful for your guidance.

    Attached Files:

    Last edited: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2

    Simon Bridge

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    Thing 1.
    - if there is a torque, there will not be a constant rpm.
    - so it is not correct to turn alpha into rpm like that.

    Calculations for the torque on an electric motor rotor depend on the geometry of the rotor, and the details of the generating fields. I don't see that information in your calculations.

    Thing 2.
    You know angles ∠ABC and ∠BCA and length AB and you want to know length AC ... yes, that can be done. Use the sine rule in general or just normal trigonometry where one of the angles is 90 deg.

    Thing 3.
    r is the distance from the pivot axis to the mass-element in question.
    for a point mass a distance r from the pivot: I=mr^2

    and object has a moment of inertia without being acted on by a force, so the r has nothing to do with the point of action etc. thus not to be confused with the moment arm in the torque equation.
  4. Apr 6, 2014 #3
    Hi Simon Bridge.

    Thing 1:

    Would you not agree that a motor should have constant rpm if torque & load is unchanging?
    The parameters of the electric motor, regarding how the force is created, are beside the point here, as concerns this equation. What we have is a set force acting upon a rotor. Elements of internal resistance accounted for -- CEMF, hysteresis, temperature coefficient, surface area, air gap, saturation, commutation, phases, so forth, as well as dimensions, material densities..., of course.

    Our "frequency" on the matter seems off. I found a few other forums focused more on motor design and electrical engineering. More populated, so the equation is finding useful feedback there.....

    Thing 2:

    Perhaps I should have been clearer.

    The only known variables:
    1) One angle (at the location depicted in the diagram).
    2) Distance between A & B.

    Objective: Find line 3 (the dotted line) given the two known variables; line 3 needs to be perpendicular to line 2. Note: The one known angle is not 90 degrees.

    This is simply an inquiry on how, alternatively, one might go about finding the "moment arm", 'r', to be applied to the equation T=rFsin(theta), with only the two known variables mentioned. As mentioned, somewhere down the road I found that multiplying the hypotenuse (line 1) by sin(theta) gives the length of the "opposite" (line 3).... Then curiosity set in, yerp.....

    If you could write solutions out simply enough, that would be helpful.

    Thing 3:

    Looking over notes, searching, searching....
    Anything more?

    thanks for your input, Simon
    Last edited: Apr 6, 2014
  5. Apr 6, 2014 #4

    Simon Bridge

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    The net torque is zero for a constant angular velocity.
    The motor would need to supply a constant torque to oppose the torque supplied by the unchanging load.
    A different torque results in a non-zero angular acceleration.

    Well done then.

    ##\sum\tau = I\alpha## ... the ##I## term depends on the geometry of the setup.
    You'll only get away from this by making assumptions and approximations about the geometry.

    You also said that:
    ... which means you know two angles and the other dimensions can be found by trigonometry.
    Note: it's needed!

    ... "perpendicular to" is not "90 degrees"? ... or it is not always 90deg?

    So no - only knowing one length and one angle, you cannot find more.

    You can't.

    You only have a hypotenuse if one of the angles is 90degrees!
    Make up your mind!

    Maybe in that specific case, the angle was close to 90 deg?

    For an arbitrary triangle with angles A B C, and lengths a,b,c defined the usual way (length a opposite angle A etc) then we can say:

    If you know angles A and B then C=180-A-B, if you also know any one of the sides, then the sine rule will find the other two.

    If a is a "hypotenuse", then A is 90 degrees, and the sine rule collapses into the usual trig relations.

    If you know two sides and the angle between them, then the cosine rule will find the other side.

    It is easy to show geometrically that if you only know one angle and one side, there is no way to get any of the other dimensions ... put another way: there is no unique triangle that has that side and that angle. You need more information.
  6. Apr 7, 2014 #5
    Thanks for hanging out with me (and tolerating me haha), Simon...

    I managed to find the youtube vid where I got the equation from. Searched all day for it..... Should say, it may be quite boring to you at your level (I even get restless watching it) and so if you'd want to click mid-way into the vid, it will soon get more into the equation at hand. Basically the same thing I initially posted. Not sure how this might change things. May be well enough to hear that the dude is full of it haha. Anywho, at the outset I simply needed some validation of the equation itself. You have outright challenged it; as a result I need to alter things on my end -- I have a LOT of intermingling equations over here....., but at least with this one I'll have direction away from this one particularly.... Anywho, here it is. I posted it on another forum as well. It will be good to hear feedback....

    ok...Thing 1:

    "The net torque is zero for a constant angular velocity. The motor would need to supply a constant torque to oppose the torque supplied by the unchanging load. A different torque results in a non-zero angular acceleration."

    I probably would agree, textbook-wise (if I actually read the text myself lol). Though, with motors and my altogether intuitive approach at times, I still tend to think of it as a constant torque overcoming a constant load; resulting in a constant rpm. Not that I'm stubbornly dismissing any of your input (for which I am grateful); I'm simply giving you my perspective that may or may not hold with time. The continuous excitation of the stator coils by continuous commutation, in constant interaction with the rotor magnets, pulse after pulse after pulse..., it is a constant overcoming of resistance; a constant production of force/torque.... Howsoever termed, to me there is constant Torque being produced, constant Work and constant RPM. Finding RPM, given known factors stated, being the only crux of the matter at present, just the same....

    "...The I term depends on the geometry of the setup. You'll only get away from this by making assumptions and approximations about the geometry."

    The dimensions are always articulated beforehand, to set into the equations proper; there is no approximation with the dimensions. I don't know what to add. Feeling challenged with no point here...

    Thing 2:

    Hmmm. For the sake of our sanity haha, this one is not terribly important. I know how to find torque, just the same. As mentioned, there seemed to be a shortcut in there to be found. I had needlessly confused the issue by relating the "line of action" to the "adjacent", relating the distance from A to B (axis to point of force) as being the "hypotenuse", relating the unknown "moment arm" ('r') to the unknown "opposite". What I had not accounted for was that the hypotenuse, itself, involves a 90 degree angle. Thanks for waking me up to that. The correlation still intrigues me and has me curious, though :P

    Thing 3: A bit clearer now. Not terribly difficult -- just my over-curious nature questioning things....

    Cheers :)
    Last edited by a moderator: Sep 25, 2014
  7. Apr 7, 2014 #6

    Simon Bridge

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    Thanks - maybe later.

    This is not how torque works - an unbalanced torque produces an angular acceleration.
    An object will keep spinning at a uniform angular velocity unless acted on by an unbalanced torque.

    "constant work" does no mean anything.
    You can do work at a constant rate ... that would be a constant power?

    You will need to revise some of your ideas about engineering and physics in order to be able to get the benefit out of asking questions of engineers and physicists here.

    Good luck.
    Last edited by a moderator: Sep 25, 2014
  8. Apr 7, 2014 #7
    This boat has too many holes. Dream it away on high water, however far, it will not float.
    I guess the art of patching holes takes many years of experience. I guess it all starts with being sure the patches are without leaks, themselves. Quite risky at high waters, don't you think?

    My majors are Philosophy & English. It's late, I write weird things, excuse me.

    Interesting discourse, exchanging riddles.

    Thanks for the [however elusive] bits of direction I gathered along the way, just the same.
    Too many fingers, not enough hands....

    Good luck, yourself.
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