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

energies).

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 = ½ m |V2|^2 - ½ m |V1|^2

Delta TKE = ½ m [(V2x^2 + V2y^2 + V2z^2) - (V1x^2 + V1y^2 + V1z^2)]

Delta TKE = ½ 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 = ½ [(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,

FB

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

fbuczek@shrinenet.org

-----Original Message-----

From: Richard Baker [SMTP:richard.baker@GREENPARK.N-I.NHS.UK]

Sent: Thursday, March 16, 2000 11:47 AM

To: BIOMCH-L@NIC.SURFNET.NL

Subject: Is there more than one type of kinetic energy

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

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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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

energies).

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 = ½ m |V2|^2 - ½ m |V1|^2

Delta TKE = ½ m [(V2x^2 + V2y^2 + V2z^2) - (V1x^2 + V1y^2 + V1z^2)]

Delta TKE = ½ 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 = ½ [(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,

FB

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

fbuczek@shrinenet.org

-----Original Message-----

From: Richard Baker [SMTP:richard.baker@GREENPARK.N-I.NHS.UK]

Sent: Thursday, March 16, 2000 11:47 AM

To: BIOMCH-L@NIC.SURFNET.NL

Subject: Is there more than one type of kinetic energy

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

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

---------------------------------------------------------------

To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

---------------------------------------------------------------

---------------------------------------------------------------

To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

---------------------------------------------------------------