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3d dynamics- determine the couple exerted

  1. Sep 17, 2011 #1
    Please see uploaded image...

    So the question is, what is Hg(dot) ?

    so I know that I need w1 in rad/sec, so 1800*(2(pi)/60)=188 rad/sec

    I want to establish a set of moving axes at the center of the rotating disk on the saw that is parallel to the stationary axis. My w is then,

    w=w1k+w2J : Always

    Now I want to have my w in terms of my moving axes, so

    w=w1k+rw2j : Instantaneously

    Where r is the distance from my moving axes origin to the stationary x-axis.

    w(dot)=α={w1(dot)k+w1k(dot)}+{rw2(dot)J+rw2J(dot)}

    Since w2 is a constant angular velocity, w2(dot)=0

    Jdot=0

    α=w1(dot)k+w1[wxk]

    wxk=(w1k+rw2j)xk=0+rw2i

    α=2w1w2i+w1(dot)k

    Okay- so now I need to determine my Inertial products and moments. This is where I'm getting stuck, though... I know that;

    HG(dot)=(Ixxαx-Ixyαy-Ixzαz)i+(Iyyαy-Ixyαx-Iyzαz)j+(Izzαz-Ixzαx-Iyzαy)k

    and since αy doesn't exist,

    HG(dot)=(Ixxαx-Ixzαz)i+(Ixyαx-Iyzαz)j+(Izzαz-Ixzαx)k

    but I don't understand how to find Ixx, Izz, Ixz, Ixy, or Iyz. I know that Ixy, Ixz, and Iyz are the mass products of intertia and that Izz and Ixx are the mass moments of inertia. But I don't know how to use these equations to solve for the components of I...

    Also- the answer to the whole problem is

    HG(dot)=-0.366i lb-ft
     

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  3. Sep 18, 2011 #2
    So I've been thinking on this one more and I know that the radius of gyration, rg=sqrt(I/m) and that I equals 1/2mr^2, so I can solve both for I and get that r equals 1.06, but this still doesn't help me find Ixx and I don't know if this r is around the y-axis or the radius of the disk...
     
  4. Sep 18, 2011 #3

    rude man

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    You've got some formulas mixed up For one, I = mr^2 where r is radius of gyration. 1/2 m r^2 is kinetic energy. Youre also trying to make this problem look harder than I think it is.

    I say 'think' because I solved it & didn't get the "right" answer, assuming that it is the right answer. It might be because I don't know how to handle the antiquated British units correctly.

    Still, let me give you a couple of suggestions:

    1. What is I, the moment of inertia of the saw plus rotor about the rotor axis?
    2. What is L, the angular momentum of saw plus rotor about the rotor axis?
    3. How about torque = dL/dt?
     
  5. Sep 19, 2011 #4
    The professor gave us the answer from the solution manual for the text. He says it's correct, he wants us to figure out how to get there...

    rg is radius of gyration and r is the distance from the moving axes to the y-axis of the stationary axis.

    okay- new thoughts. If I assume Ixy=Ixz=Iyz=0, then the problem simplifies a bit, but I still don't understand how to get Ixx or Izz... This would make the problem set up more comfortable. The other thought I'm having is that Ixx must be zero, if we want the k direction to vanish. Now, I just need to understand how to find Ixx and presumably the problem is solved....
     
  6. Sep 19, 2011 #5
    Okay! More new thoughts if anyone out there at all can help me even a little bit... Ixx=m*rg^2 where m is the mass of the saw motor and rotor and rg is the radius of gyration.

    rude man- If you could tell me how you solved it maybe I can extrapolate the technique taught in class from there? I need all the help I can get on these problems!
     
  7. Sep 19, 2011 #6

    rude man

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    OK.
    First: the rotational inertia of the saw and rotor is I = mr^2 where
    r = 1.5" or 1.5/12 ft
    m = 2 lbs.
    (This is about the z axis initially.)

    Then, the angular momentum of the saw + rotor is Iw where
    w = 1800 rpm * 1/60 rps/rpm * 2π rad/rot.

    The direction of angular momentum can be given in terms of the cylindrical coordinate system (r,θ,y). It is in the -θ direction. (Don't even try to use cartesian! DesCartes was an s.o.b anyway - he cut his living dog's torso open to see if the dog's soul would pop out! A real jerk).

    Finally, the precession torque T = dL/dt. This is the negative of the answer you're looking for. The word 'couple' sounds a bit funny but I must assume it means torque or moment. At any rate your answer is in ft-lbs. Is this perhaps a UK thing? I had 'couples' only in my statics course.

    But what is dL/dt? If you make a diagram of vector ∆L = (L + ∆L) - L as the saw swings around the y axis, you'll see that ΔL ~ Lsinθ where θ is the angle about the y axis generated in time t. For small t, sinθ = θ and we can say dL = Ldθ or dL/dt = Ldθ/dt. But dθ/dt is just the angular rate w1 at which the saw is rotated about the y axis. The distance of the saw's c.m. to the y axis does not enter the calculations.

    This is exactly analogous with how you derive the formula for acceleration of a mass in a circular orbit: a = w^2 * r, except here directed towards the orbit center.

    So T = w1*L, the direction of dL/dt being tangential to the direction of rotation of the c.m. of the saw, pointing in the -θ direction.

    The direction of dL/dt is tangential to the circular orbit of the saw. So the operator must impart a counteracting turning torque -T to the saw, in the direction of twisting the saw so the part closest to the center of rotation points up (but of course he maintains the attitude of the saw horizontal).

    Sorry about the many edits. I did this in too much of a hurry, had to leave the house in a hurry. Ignore all posts but the last. Let me know how it turns out?)
     
    Last edited: Sep 19, 2011
  8. Sep 20, 2011 #7

    rude man

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    ANOTHER EDIT:!! The torque T due to the rotation about the y axis is directed inward towards the center of the orbit, i.e in the -r direction. So the countervailing torque must be in the +r direction. Which means the torque the man must exert on the saw is a downward pitch motion, i.e. the saw needs to be pushed so as to attempt to push its leading edge downwards.

    I hope and believe this is my final post on this subject! Thanks for your patience.
     
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