"Spaceman" <Realspace@comcast.not> wrote in message
news:MIKdnW1DTrBolJPZRVn-rQ@comcast.com...

"Jason Stanidge" <Jameson@nospam.com> wrote in message
news:duktti$1pe$1@newsg2.svr.pol.co.uk...

Moment of inertia for a square of side r, mass m, rotating about it's
centre
point with angular velocity w = (mr^2)/6, so KE is (m(wr)^2)/12

But how about calculating the KE the following way?
Spit the square up into 4 squares. Each of these squares has mass m/4

and

moves on a circle of radius r/sqrt2 with angular velocity w, and
rotates
about it's centre with angular velocity w.
KE contributed by 4 squares moving around the circle = 1/2m'v^2 =KE' =
4*(1/2)(m/4)(w^2)(r/sqrt2)^2 with the rest contributed by rotation
about
their centres.
Now split the KE of the 4 rotating squares about their centre as was

done

previously with the whole square:-
Total KE contributed by 4*4 squares moving around a circle of radius
r/(2sqrt2) with mass m/(4*4) KE'' =
4*4*(1/2)(m/(4*4))(w^2)(r/(2*sqrt2))^2
And again splitting into 4:-
Total KE contributed by 4*4*4 squares moving around a circle of radius
r/(2*2sqrt2) with mass m/(4*4*4) KE'''=
4*4*4*(1/2)(m/(4*4*4))(w^2)(r/(2*2sqrt2))^2
etc...

So the total KE of the original rotating square = KE = KE' + KE'' +

KE'''

+...
This is a geometric series with first term a = (1/2)(m)(w^2)(r/sqrt2)^2
and
common ratio n = 1/2
So the sum to infinity = a/(1-n) = ((1/2)(m)(w^2)(r/sqrt2)^2/)(1-1/2) =
1/2
m (wr)^2.

So where have I gone wrong in my calculation?

The center of mass of one of the 4 squares would not be traveling
on the same radius of angular momentum that the larger mass alone
would have.

I don't know what you mean by the radius of angular momentum of the larger
mass. Do you mean the maximum radius of that particle which contributes to
the total momentum of the mass?

So the r would not be r it would be r - ?
It is easy to map but tough to model.
:)

Take a square piece of paper, and draw a cross on it so that it divides it
up into 4 squares and put a dot in the centre of each. Take a pin and
place
it in the centre of the cross, and rotate the sqaure paper around it. You
will see each centre of the four smaller squares moving around a circle
and
at the same time rotating around their centres. Hence the motion of these
four squares is equivalent to the motion of the larger square which they
are
a part of.

"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:osBPf.302397$Ey1.9690156@phobos.telenet-ops.be...

"Jason Stanidge" <Jameson@nospam.com> wrote in message

news:duktti$1pe$1@newsg2.svr.pol.co.uk...

Moment of inertia for a square of side r, mass m, rotating about it's

centre

point with angular velocity w = (mr^2)/6, so KE is (m(wr)^2)/12

Yes, as calculated in the reference frame of the centre in which
the square has angular velocity w.

But how about calculating the KE the following way?
Spit the square up into 4 squares. Each of these squares has mass m/4

and

moves on a circle of radius r/sqrt2 with angular velocity w, and rotates
about it's centre with angular velocity w.
KE contributed by 4 squares moving around the circle = 1/2m'v^2 =KE' =
4*(1/2)(m/4)(w^2)(r/sqrt2)^2 with the rest contributed by rotation about
their centres.
Now split the KE of the 4 rotating squares about their centre as was

done

previously with the whole square:-
Total KE contributed by 4*4 squares moving around a circle of radius
r/(2sqrt2) with mass m/(4*4) KE'' =

4*4*(1/2)(m/(4*4))(w^2)(r/(2*sqrt2))^2

And again splitting into 4:-
Total KE contributed by 4*4*4 squares moving around a circle of radius
r/(2*2sqrt2) with mass m/(4*4*4) KE'''=
4*4*4*(1/2)(m/(4*4*4))(w^2)(r/(2*2sqrt2))^2
etc...

So the total KE of the original rotating square = KE = KE' + KE'' +

KE'''

+...
This is a geometric series with first term a = (1/2)(m)(w^2)(r/sqrt2)^2

and

common ratio n = 1/2
So the sum to infinity = a/(1-n) = ((1/2)(m)(w^2)(r/sqrt2)^2/)(1-1/2) =

1/2

m (wr)^2.

So where have I gone wrong in my calculation?

You have added moments of inertia around different axes
of rotation. Total moment of inertia is to be integrated with
respect to one axis of rotation.

I used the Moment of Inertia to calculate the text-book kinetic energy to
start with. But after that, I'm not using the moment of inertia of at all
when breaking it down into elements. For each element of mass, I'm using
1/2mv^2 which is due to the centre of the mass merely moving aound the
circle. Then to add that contributed by the rotation of that element, I
break it down further into smaller elements and so on.

Furthermore, you have added kinetic energies as calculated
in different reference frames. Kinetic energy is frame
dependent, so you can't just add them together and expect
to get something useful.

Now this is a point that didn't occur to me. I've gone and summed the
kinetic energies of the components of the velocity which isn't equal to the
kinetic energy of the resultant velocity.

Jason

Dirk Vdm

Dirk Van de moortel wrote:

"Jason Stanidge" <Jameson@nospam.com> wrote in message news:duktti$1pe$1@newsg2.svr.pol.co.uk...

So where have I gone wrong in my calculation?

You have added moments of inertia around different axes
of rotation. Total moment of inertia is to be integrated with
respect to one axis of rotation.

Hey, what a surprise, Dirt von don Muton's tutor must be doing a good
job!

Furthermore, you have added kinetic energies as calculated
in different reference frames. Kinetic energy is frame
dependent, so you can't just add them together and expect
to get something useful.

You must have gotten a good tutor finally, you are making real
progress.