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Re: Is there more than one type of kinetic energy

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  • Re: Is there more than one type of kinetic energy

    Dear Richard,

    You pose some very interesting points, and ones that I have in some form or
    another discussed with others in the past. I have a simple "thought
    experiment" that I use to satisfy myself that there is meaning in separate
    "terms" (not "components") of kinetic energy calculations.

    I start with a re-statement of the Work/Energy Theorem. The work done on a
    rigid body equals the change in its mechanical energy (with the latter
    including translational kinetic, rotational kinetic, and potential

    Considering translational terms for simplicity, if I know the value over
    time of the net external force applied to a rigid body, and the position of
    its point of application, I can calculate the work done on the rigid body;
    this should equal the change in its translational kinetic energy. But, this
    argument also holds for separate forces applied to the rigid body: the work
    of the net force is equal to the algebraic sum of the works of the
    individual forces.

    If I choose individual forces along three axes of an inertial reference
    frame to replace the net force, I know that "F = ma" applies to each of
    these forces independently of the others. That is, I can calculate the
    body's acceleration along the X-axis independently of its translational
    mechanics along the Y- and Z-axes. If I know the body's initial velocity
    along the X-axis, I can determine its final velocity. From this, I can
    easily determine its change in kinetic energy due to this force. If I do
    this also for the forces chosen along the Y- and Z-axes, I can determine the
    body's change in kinetic energy due to each of these as well. Algebraically
    summing these "change in kinetic energy" terms provides the same result as
    if I were to consider the net force alone:

    Delta TKE = m |V2|^2 - m |V1|^2

    Delta TKE = m [(V2x^2 + V2y^2 + V2z^2) - (V1x^2 + V1y^2 + V1z^2)]

    Delta TKE = m [(V2x^2 - V1x^2) + (V2y^2 - V1y^2) + (V2z^2 - V1z^2)]

    Why is this important? Because the Work/Energy Theorem can be used to
    calculate either work or changes in mechanical energy, if the other is
    known. If I know the work done by a force aligned with the X-axis, I can
    determine the change in velocity along the X-axis of the point of
    application of the force. Practically speaking, I may never choose to do
    this. But, there appears to me to be no violation of the scalar nature of
    "work" in this thought experiment. The individual "terms" of the mechanical
    energy equation can have physical meaning. I deliberately avoided calling
    these quantities "components," and discussed algebraic sums rather than
    vector summations. This also keeps the scalar nature of the work and
    energies intact.

    I have only seen the rotational kinetic energy equation expressed as:

    RKE = [(Ixx)(Wx)(Wx) + (Iyy)(Wy)(Wy) + (Izz)(Wz)(Wz)]

    where W is the angular velocity "omega" (in ASCII text). I believe this
    thought experiment works here as well. However, I have wondered whether
    this equation for RKE is simplified, leaving out products of inertia and
    associated (WiWj) terms that are included in Euler's Equations of Motion.
    As time permits, answering this question will be my next challenge, unless
    other members of BIOMCH-L have an answer for this now.

    Hope this helps,

    Frank L Buczek Jr, PhD
    Director, Motion Analysis Laboratory
    Shriners Hospitals for Children
    1645 West 8th Street, Erie, PA, 16505, USA
    (814) 875-8805 voice, (814) 875-8756 facsimile

    -----Original Message-----
    From: Richard Baker [SMTP:richard.baker@GREENPARK.N-I.NHS.UK]
    Sent: Thursday, March 16, 2000 11:47 AM
    Subject: Is there more than one type of kinetic energy

    In calculating the total amount of work done on the basis of the
    of body segments, during gait for example, it is common (Winter
    Pierrynowski 1980, Viswanath 1999) to consider the potential energy,
    translational kinetic energy and the rotational kinetic energy. In
    calculating total work done the translational and rotational energy
    are treated as separate. Is this justified?

    It is my understanding that there is just one scalar quantity,
    energy, and that the translational and rotational "components" are
    simply a
    means of calculating this total. Given this surely the total kinetic
    for each segment should be treated as a single term for the
    calculation of
    total work.

    Whilst we are on this subject, there is at least one series of
    papers which
    talks of the forward, lateral and vertical components of
    kinetic energy (although the use of the terms does not affect the
    mathematical analysis). Surely kinetic energy is a scalar and there
    is no
    physical meaning to these "components"?

    Am I right? Right, but overly pedantic? Plain wrong?

    I'd be interested in anyone's comments.


    Richard Baker PhD
    Gait Analysis Service Manager
    Musgrave Park Hospital, Stockman's Lane, Belfast, Northern Ireland,
    BT9 7JB
    Tel: +44 2890 669501 ext 2155 or 2849
    Fax: +44 2890 382008

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