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Constant Angular Velocity

  1. Nov 5, 2006 #1
    I have a question from an old exam, asking for the torque required to rotate a triangular shape about one of its sides. Please find the question attached, and any help is greatly appreciated.

    Until the picture gets approved:

    Find the torque needed to rotate a traingular plate of sides a, b and c (a^2 + b^2=c^2) about the side c with constant angular velocity w. The mass of the plate is M.

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    Last edited: Nov 5, 2006
  2. jcsd
  3. Nov 5, 2006 #2


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    What is the moment of inertia of a triangle, relative to one of its sides?
  4. Nov 5, 2006 #3
    Hi SGT

    Thank you for your reply. I am working with the original poster on this problem. The question was asked to see if anyone was aware of a reason for why the torque should not simply be zero if the object is rotating with constant angular velocity. No other forces act on this triangle (gravity, air resistance, friction, etc)

    To answer your question:
    The moment of inertia will be I=(b.h^3)/12 for b=base, h=height
    which in this case is b=c and h=a.b/c resulting in I=a^3.b^3/12.c^2
    for rotation about side c.

    I have since found that Euler's Eqn of rotational motion may be the key, but am still trying to work through the math. It appears that 9 moments of inertia may be required to construct an inertia matrix. Any help is still very much requested and appreciated.
  5. Nov 5, 2006 #4


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    I'm not sure about the Euler's Equations approach. I have not looked at that for a long time, but I think it would be more easily applicable if the plate were rotating without constraint about its CM along some arbitrary axis. In this problem there is a constraint forcing the plate to rotate about the long side of the triangle. I have an idea on this that does not require the moment of inertia tensor.

    I visualized the triangle rotating about a vertical axis with the b side toward the bottom and the a side toward the top. The CM for a triangle is 1/3 of each leg from the right angle. This puts the CM some distance from the axis of rotation that I will call d, and some distance above the lower vertex of the triangle I will call e. After some messy algebra I found values for d and e that certainly need to be verified, but I found

    d = ab/3c

    e = (b^2 + c^2)/3c

    As the triangle rotates, each bit of mas dm of the triangle is in circular motion with centripetal force

    dmv^2/r = dm(ω^2)r

    where r is the distance from the axis. Since this is linear in r, the total force is equivalent to all the mass being concentrated at the CM at distance e above the lower vertex and distance d from the axis. So there is a force acting on the axis of M(ω^2)d at a distance e from the lower vertex.

    torque = M(ω^2)d*e

    If the constraint were an axle with bearings at the top and bottom of the triangle, you could calculate the magnitude of the forces required at both ends to keep the triangle vertical. If it were a shaft at the lower end, the equation above should be the moment at the tip of the triangle.
    Last edited: Nov 6, 2006
  6. Nov 5, 2006 #5
    I suspect that Euler is the key to solving the question, as it was a major part of the course, and seems to be the method they want us to take.

    Still, initial thoughts led me to believe the answer was 0 (no time dependency).

    Also calculated the moment of inertia the same as above:
    The moment of inertia is about the axis is (bh^3)/12, with base in this case being =c, and the perpendicular height can be found to be (ab)/c.

    So the matrice you find is:
    | Ixx Ixy 0 |
    | Iyx Iyy 0 |
    | 0 0 Izz|

    then you put that into Euler's equation, only the k component will a non-zero answer, and solve for moment? Vague framewoirk?
    Last edited: Nov 6, 2006
  7. Nov 6, 2006 #6
    Possible answer

    All that goes to
    Possible answer of torque=
    /Mw^2ab(b^2-a^2)\ -->
    |-------------------| k
    \ 12(a^2+b^2) /

    That is the torque is in the k direction.

  8. Nov 6, 2006 #7
    Might have something

    Here is the way I have come up with a solution (by no means the correct one)

    Define the origin of my coordinate frame as being at the intersection of sides a and c. Or at the top of the triange as visualized by Dan. Define x direction as being in the same direction as side a and y direction as same direction as b.

    To solve the question need to find Hx and Hy and use them in

    T = w X H ---- X meaning cross

    Resolve W into x and y directions and get
    Wx = -w.b/c
    Wy = w.a/c

    Now find H
    From eulers equations
    Hx = Ixx.Wx + Ixy.Wy
    Hy = Iyy.Wy + Iyx.Wy

    Therefore need to find inertia matrix:
    Ixx Ixy
    Iyx Iyy

    To do that I find Ixx and Iyy as if the coordinate from was at the CoM. Then using paralell axis theorem find the new I at the point I mentioned previously as my origin.

    Once I have Hx and Hy I found
    T = w X H

    this result came as

    T=Mw^2.ab/c^2 . (-b^2/3+a^2/4)
  9. Nov 6, 2006 #8
    shankur, are are you at Uni atm?

    We need to compare solutions
  10. Nov 6, 2006 #9
    hahaha, take it u googled too?

    not at uni presently, but will be tomorrow, so will hopefully sort it then. seeing as the exam is the day after would obv wanna sort it 2moro :P
  11. Nov 6, 2006 #10
    I googled a solution to a very similar problem. The one I found is a rectangle not a triangle. But other then that the method should be the same. The method I used works excellently for the rectangle question. The only real differences in the methods for the rectangle as opposed to triangle is the choice of origin and therefore the Inertia matrix you use.

    The problem I have is that I am not sure where the best spot to put the origin. I think it needs to be along side C somewhere, otherwise the origin will be rotating and I think thats bad.
  12. Nov 6, 2006 #11


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    Could you post the link to that problem?
  13. Nov 6, 2006 #12
    the origin has to go either along side c, or at the centre of gravity (where i put it).
  14. Nov 6, 2006 #13
  15. Nov 6, 2006 #14


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    Thanks. I've been calculating the tensor elements about the center of mass. I actually did them about the right angle corner with the x-axis paraller to side b and the y-axis parallel to side a and then translated them to the CM using the parallel axis theorem. I checked one by direct calculation and it agreed. What I've gotten is

    [tex] I_{xxCM} = \frac{{Ma^2 }}{{18}} [/tex]

    [tex] I_{yyCM} = \frac{{Mb^2 }}{{18}} [/tex]

    [tex] I_{zzCM} = \frac{{Mc^2 }}{{18}} [/tex]

    [tex] I_{xyCM} = \frac{{Mab }}{{36}} [/tex]

    How do those look to you?
  16. Nov 7, 2006 #15
    picture finally got approved :D huzzah.

    those moment of inertias look correct
  17. Nov 7, 2006 #16


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    Using these along with [tex] \omega _x = - \omega \frac{b}{c} [/tex] and [tex] \omega _y = \omega \frac{a}{c} [/tex] to calculate the torque about the CM I got

    [tex] \mathbf{N} = \frac{{M\omega ^2 ab}}{{36c^2 }}\left( {a^2 - b^2 } \right) \mathbf{k} [/tex]

    I did the whole thing again about the point that is the base of the triangle altitude on side c using y parallel to side c and x perpendicular to side c in the plane of the triangle and got [itex] \omega _y = \omega [/itex], [itex] \omega _x = 0 [/itex]

    [tex] I_{xx} = \frac{{M\left( {a^6 + b^6 } \right)}}{{6c^2 }} = \frac{{M\left( {c^4 - 3a^2 b^2 } \right)}}{{6c^2 }} [/tex] <== This does not contribute.

    [tex] I_{yy} = \frac{{Ma^2 b^2 }}{{6c^2 }} [/tex]

    [tex] I_{xy} = - \frac{{M\left( {a^2 - b^2 } \right)ab}}{{12c^2 }} [/tex]

    That gives a torque of

    [tex] \mathbf{N} = \frac{{M\omega^2 ab }}{{12c^2 }} \left( {a^2 - b^2 } \right) \mathbf{k} [/tex]

    I think it makes sense for these to be different. They should lead to the same forces at the pivot points. I have not done that yet.
    Last edited: Nov 8, 2006
  18. Nov 7, 2006 #17
    think thats right, except is it missing a factor of ab
  19. Nov 7, 2006 #18


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    Yeah, thanks. I somehow lost that ab fussing with the LaTeX. I edited it to fix it.
  20. Nov 8, 2006 #19
    cheers for all the help.

    just had the exam and the same sort of question popped up.
  21. Nov 10, 2006 #20


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    Some of this was true, and some of it was a fib. The centripetal force for a slice of the triangle perpendicular to the rotation axis (side c) is equivalent to the centripetal force of a mass of that slice located at its center, but the centripetal force for each slice is not proportional to the length of that slice, it is poroprtional to the square of the length of the slice. One cannot use the center of mass of the triangle as the point of application of the net force to compute the torque. One can however sum the forces to find net force, and sum the torques from the individual slices to find the net torque. The net force can still be found as if all the mass were concentrated at the CM distance (d) from the axis. So the net force is still M(ω^2)d. You just cannot assume it is applied at the CM for the torque calculations.

    I did the integral to find the torque about the point at the base of the triangle altitude, and it agrees perfectly with the calculation based on the moment of inertia tensor. That is a good thing. What has me rather perplexed is that I am not getting the same forces on the pivot points using the torque about the CM that I am getting using the torque about the base of the altitude from the two moment of inertia tensor calculations. That cannot be. I'm still looking at it.:yuck:
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