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Problem finding the total kinetic energy of bicycle system

  1. May 9, 2017 #1
    I've been working on this problem for a while and here comes the problem: I'm building a balancing system for a bicycle using Gyroscopic effect (take a look at Figure 2 for more details). My system consists of 2 parts:
    • the bicycle itself
    • a flywheel on top of it
    All parameters are given. The question is I have to derive formula for calculating total kinetic energy of the system. Figure 1 is the formula given in the article but I'm a bit confused about how to get it. I can understand the first 3 terms in the formula like this:

    (1) translational kinetic energy of the bike's center of mass

    (2) translational kinetic energy of the flywheel's center of mass

    (3) rotational kinetic energy of the bike about its center of mass

    Fig. 1

    Since the last 3 terms in the given expression involve moment of inertia, I guess these are rotational kinetic energy of the flywheel about some reference which I don't know yet.

    Please tell me how to get these last 3 terms. I would be very appreciated for your help.

    Fig. 2

    Reference: see attached PDF file

    Attached Files:

  2. jcsd
  3. May 10, 2017 #2
    I'm thinking about rotating frame of reference. Any ideas everyone ?
  4. May 15, 2017 #3


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    Sorry I deleted my original post because I was totally wrong. I miss read as φ as ω in a key place.

  5. May 15, 2017 #4
    In the now deleted post from CWatters, there was a quotation from a text (apparently) regarding the meaning if Ip and Ir. I would interpret these in this way:

    Ir = Mass moment of inertia with respect to a radial line (the same for all radial lines by symmetry)
    Ip = Mass moment of inertia with respect to the axis of rotation

    Does that help at all?
  6. May 16, 2017 #5
    Yes and thank you for your reply. Then expression (4) seems like the rotational kinetic energy of the flywheel about gimbal axis which relates to the derivative of φ. But I'm still stuck at finding the meaning of expressions (5) and (6). Any ideas?
  7. May 16, 2017 #6
    In the XYZ coordinate system, the roll angular velocity of the entire assembly is simply theta-dot*k, where k is a unit vector in the Z direction. Now, resolve that velocity component on the body coordinate system of the flywheel. There is an axial component theta-dot*sin(phi) and a component about a radial line in amount theta-dot*cos(phi). These are the two velocity components that go into making up terms 5 and 6.
  8. May 18, 2017 #7
    The expression for T in post #1 seems to be lacking a term representing the spin velocity of the flywheel. Is it not spinning?

    Assuming that the flywheel is spinning with angular velocity w relative to the frame, there should be a term of the form
    (1/2)*Ip*(w +/- theta-dot*sin(phi))^2
    where the +/- choice depends on the direction chosen positive for w. This is because w and theta-dot*sin(phi) are colinear angular velocties which should be added before squaring. At least, that is how it looks to me.
  9. May 18, 2017 #8
    First of all, can you show me the way to resolve this roll angular velocity into 2 components above: axial component and radial component

    Second of all, the flywheel is spinning with angular velocity w, indeed. But I think this missing term should be (1/2)*Ip*(w^2), which is the rotational kinetic energy of the flywheel about its center of mass.
  10. May 19, 2017 #9
    In the global XYZ fram, the bicycle roll is described by the vector theta-dot*k where k is a unit vector in the Z-direction.

    Now, in the vertical plane, resolve this vector into components along the spin axis and perpendicular to the spin axis to get, respectively
    theta-dot*sin(phi) and theta-dot*cos(phi)

    They are both in the vertical plane, but one is aligned with the spin axis will the other is normal to it and parallel to a radial line on the flywheel.

    The components along the spin axis add together to get the net rotational velocity of the flywheel = (w +/- theta-dot*sin(phi)). They must be added before expressing the kinetic energy which depends on the net rotational velocity (the sum/difference).

    If this is hard to follow, I suggest that you draw the appropriate vectors on your original diagram. Then I think it will be evident.
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