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Not sure about finding velocity with this equation

  1. Mar 31, 2014 #1
    I need to find velocity when force, distance, and mass are known.
    I found this equation: Fd = (M/2)V^2

    The process to solve it is put forth by using the metric system. It would be a great help to know if (and how) I can use the imperial system only with this equation.

    I'd like to take you through the process put forth, to hopefully reveal to you where I'm faltering.

    -----------------------------------

    "Step ONE: Weigh the object to find Mass. All other resistance, such as bearing resistance, being negligible, is not accounted for here; and so the WORK done on the object equals its KINETIC ENERGY. If grams, convert to Kg by dividing by 1,000. If pounds, convert to Kg by multiplying by .45"

    > I have a disc that weighs .8 oz, which converts to .02268 Kg

    -----------------------------------

    "Step TWO: Set equations for WORK and KINETIC ENERGY so they are equal.
    WORK = FORCE x DISTANCE; KINETIC ENERGY = 1/2 the MASS of the object x its VELOCITY squared.
    That is: F x D = (M/2) X V^2.
    Enter measurements for FORCE, DISTANCE and MASS.
    Example: FORCE = 2 Newtons, DISTANCE = 5 meters, MASS = 0.7kg.
    Therefore: (2 N) x (5 m) = (0.7kg/2) x v^2 "

    > I converted oz to kg above for "M".
    > Next, I have .3 lbs of force being applied to the disc. (The disc, by the way, is a motor rotor and the force applied is the repelling force via interaction between the rotor magnet and the stator's electromagnet...) So I need to convert .3 lbs to Newtons. 1 lbf = 4.448 N, so .3 lbs = 1.3344 Newtons.
    > Next, for distance I'm assuming it's the circumference of the disc. A 2.365" diameter disc x Pi = 7.428465" circumference. Seems logical now to convert inches to metric, to keep with the metric equation. 1" = .0254m, so 7.428465" x .0254m = .188683011 m

    My numbers so far: (1.3344 N) x (.188683011 m) = (.02268 kg/2) x v^2

    More confusion sets in, whereby this equation is for figuring velocity linearly, not radially (as I need it to be). Therefore, at the end I attach the resolve for converting m/s to RPM, which involves finding the circumference, which I just did in step TWO, and is why I'm not sure about this equation altogether. But I keep on with it....

    -----------------------------------

    "Step THREE: Multiply and divide to simplify the equation.
    Example: (2 N)*(5 m) = (0.7 kg/2)*v^2
    becomes 10 N*m = (0.35 kg)*v^2."

    My numbers: (1.3344 N) x (.188683011 m) = (.02268 kg/2) x v^2
    becomes .251779 N*m = (.01134 kg) x v^2

    -----------------------------------

    "Step FOUR: Divide the left side of the equation by the number on the right side of the equation to isolate v^2.
    Example: 10 N*m = (0.35 kg)*v^2
    becomes 28.6 N*m/kg = v^2."

    My numbers: .251779 N*m = (.01134 kg) x v^2
    becomes 22.2027.. N*m/kg = v^2

    -----------------------------------

    "Step FIVE: Take the square root of the number on the left side of the equation to find the velocity.
    Example: 28.6 N*m/kg = v^2
    The square root of 28.6 equals 5.3, so the velocity is 5.3 m/s."

    My numbers: 22.2027.. N*m/kg = v^2
    Square root of 22.20273.. = 4.71197.. m/s

    -----------------------------------

    Now, for converting meters per second to RPM: RPM = V/(Pi*D).

    RPM = 4.71197.. / 7.428465"

    RPM = .634313..

    ?????????????

    Quite exhausting.
    Any help on the matter, I am truly grateful.
     
    Last edited: Mar 31, 2014
  2. jcsd
  3. Mar 31, 2014 #2
    I am a little confused, you say you are moving the disc 5 meters? or is that just an example?

    what you are doing is rotating the disc about its center axis I assume?

    If so, you cant use the "linear" versions of those equations, basically replace V with theta dot (angular velocity, aka omega (w)) and A with theta double-dot (angular acceleration). Also you cant use the "mass" since its not linear motion, it is rotational motion and therefore you must use the moment of inertia of a disc (you can look that up really easy, i think it's .5mr^2 but dont hold me to that)


    so your energy equation is not 1/2 mv^2 but

    KE = 1/2 Iw^2

    where KE = kinetic energy
    I = moment of inertia kgm^2
    w= omega (angular velocity) rad/sec

    multiplied shoudld give you units of kgm^2/sec^2 = Joules

    If you wish to relate omega with linear velocity, its simple v=rw or w=v/r

    NOTE: this is if you want to find the velocity of the outer-most part of your disc, linear velocity varies with the radius, so the center-most part of the disc is not moving at the same linear velocity as the outermost part.
     
    Last edited: Mar 31, 2014
  4. Mar 31, 2014 #3
    Hey _Bd_

    I'll be investigating your reply. Thanks for setting a clear path.
    As to the '5 meters', that was the example they gave.
    Also, yes, I'll be rotating a disc about its center axis.

    Recently I managed to decipher T=rFsin(theta); comfortable with that equation now.
    About a year ago I got into making motors, so the equations here are all part of the process. You've got a rotor with magnets on it, a stator with electromagnets on it, a means of commutation to excite the e-magnets at a certain point in time, to interact with the rotor magnets, to move the rotor along.... Spent a good part of the year just having too much fun with the easy stuff. Deeper into it now, I have much to figure out.

    The electromagnets on the stator are made to be near-equal in pull force to the rotor magnets. Electricity aside, accounting for the surface area of the two and air gap between them, there is a repelling force shared, of which is the Force. The point of action (of force) changing with angle of rotor magnet and coil orientation, accounted for, and henceforth arriving at your line of action. For reasons I'll not get into, I needed to find 'r' with only angle and the hypotenuse known. Pretty easy: Find sin(theta) with calculator, then multiply by the Hypotenuse = r. Next, I need to find 'velocity when force, distance, and mass are known'. More specifically, to find the RPM, given a rotor circumference and a 'force' upon that rotor --- 'force' being either the resultant Torque or the repelling force mentioned...?? OY!! Well, well... I've been rambling a bit, as usual.... Thought to give a little overview, perhaps of some use for some further direction.

    I'll be analyzing your post _Bd_...
    thanks :)
     
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