View Full Version : Re: Is there more than one type of kinetic energy

Jon Dingwell
03-16-2000, 12:59 PM
Dear Richard:

You are correct, that energy (kinetic and potential) is a scalar and not a
vector quantity. However, translational and rotational components of
kinetic energy are separate issues, and must both be calculated in order to
obtain the total kinetic energy of a rigid body.

For example, a block sliding across a frictionless table has translational
kinetic energy ("TKE") because its center of mass is moving, but not
rotational kinetic energy ("RKE"). On the other hand, a distributed mass
which is rotating about its center of mass, but not moving (such as a
bicycle wheel spinning in place) has RKE, but not TKE (because the ceter of
mass does not move).

Thus, calculation of TKE accounts for the energy associated with the
movement of the center of mass of an object relative to inertial (i.e.
global) coordinates. Calculation of RKE accounts for the energy associated
with movement of the object with respect to its own center of mass. Thus,
the sum of both gives you the net kinetic energy of the entire system with
respect to global.

As for the notion of "forward, lateral and vertical components" of TKE,
this is slightly more tricky, but it does work out. In general, the TKE
for a rigid body is given by:

TKE = (1/2) M * (V*V)

where M is the mass and V is the TANGENTIAL velocity of the center of mass
(i.e. the instantaneous speed along the direction of motion) and is also a
scalar. The scalar V can be obtained from the magnitudes of the components
of the velocity vector (Vx, Vy, and Vz) along the x, y, and z axes (by the
Pythagorean theorem):

V*V = (Vx*Vx + Vy*Vy + Vz*Vz)

Thus, the equation for TKE then becomes:

TKE = (1/2) M * (Vx*Vx + Vy*Vy + Vz*Vz)

TKE = (1/2)M*(Vx*Vx) + (1/2)M*(Vy*Vy) + (1/2)M*(Vz*Vz)

The only quantity that is of use is the net TKE, taking into account all
three velocity components.

We can then make similar arguments about RKE, and we derive the following

RKE = (1/2) I * (W*W)

where I is the moment of inertia of the rigid body about the instantaneous
axis of rotation, and W is the instantaneous angular velocity of the rigid
body. And so finally, the total kinetic energy is then given by:


KE = (1/2) M * (V*V) + (1/2) I * (W*W)

For a more detailed description, I highly recommend the textbook
"Principles of Dynamics" by Donald Greenwood (Prentice Hall; ISBN #

I hope this helps,

Jon Dingwell

At 04:47 PM 3/16/00 +0000, Richard Baker wrote:
>In calculating the total amount of work done on the basis of the movements
>of body segments, during gait for example, it is common (Winter 1979,
>Pierrynowski 1980, Viswanath 1999) to consider the potential energy, the
>translational kinetic energy and the rotational kinetic energy. In
>calculating total work done the translational and rotational energy terms
>are treated as separate. Is this justified?
>It is my understanding that there is just one scalar quantity, kinetic
>energy, and that the translational and rotational "components" are simply a
>means of calculating this total. Given this surely the total kinetic energy
>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 translational
>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

Jonathan Dingwell, Ph.D.
Postdoctoral Research Associate

Sensory Motor Performance Program
Rehabilitation Institute of Chicago
345 East Superior, Room 1406
Chicago, Illinois, 60611

Phone: (312) 238-1233 [Office] / (312) 238-1232 [Lab]
FAX: (312) 908-2208
E-Mail: j-dingwell@nwu.edu
Web: http://manip.smpp.nwu.edu/dingwell/

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