| Topic: |
Science > Physics |
| User: |
"Jason Stanidge" |
| Date: |
15 Feb 2006 12:15:13 PM |
| Object: |
Centre of mass theorem for rotaing bodies? |
For a distributed mass, F = M dv/dt where F is the total external force, M
the total mass, and v the velocity of the centre of mass. But take the case
of a gyroscope, where the force acting on the centre of mass is gravity,
assuming it is uniform over the body. The gyroscope doesn't topple over, but
precesses, which imples that the centre of mass theorem doesn't apply to
rotational bodies. Why not? The Theorem seems to be correct for any system
of particles, whether it is rotaing or not, if one looks at the derivation
of the Centre of Mass Theorem:-
For each particle in a body, f = md^2r/dt^2.
If I sum all the forces, including the external forces applied to the body,
and make use
of Newton's third law, then the internal forces all cancel, so I'm left with
just the external force F.
F = sum(k=1, k=n)[mkd^2rk/dt^2] where mk and rk is the mass and radius
vector
of the kth particle.
So F = d^2/dt^2[sum(k=1, k=n)[mk x rk]] . If the centre of mass is defined
as R
= (sum(k=1, k=)[mk x rk] )/M
Then external F = Md^2R/dt^2.
Which of the above lines is false for a rotating body?
Thanks
.
|
|
| User: "JMA" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
16 Feb 2006 06:33:47 PM |
|
|
"Jason Stanidge" <Jameson@nospam.com> escreveu na mensagem
news:dt2f4l$427$1@newsg2.svr.pol.co.uk...
(snip)
Then external F = Md^2R/dt^2.
Which of the above lines is false for a rotating body?
You have a question for which Physics have no answer.
Mass rotation is a pain in the ***** for Physics.
Nor Newton, Nor SR, nor GR, nor whatever solves the
gyroscopic momentum and precession issue.
The best I have seen so far is:
http://nuclear.ucdavis.edu/~cherney/dvd/memorize/fforces.html
http://nuclear.ucdavis.edu/~cherney/dvd/memorize/erigid.html
.
|
|
|
| User: "Jason Stanidge" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
16 Feb 2006 12:38:11 PM |
|
|
"JMA" <jmDOTa@sapo.pt> wrote in message
news:43f51b58$0$30010$a729d347@news.telepac.pt...
"Jason Stanidge" <Jameson@nospam.com> escreveu na mensagem
news:dt2f4l$427$1@newsg2.svr.pol.co.uk...
(snip)
Then external F = Md^2R/dt^2.
Which of the above lines is false for a rotating body?
You have a question for which Physics have no answer.
Mass rotation is a pain in the ***** for Physics.
Nor Newton, Nor SR, nor GR, nor whatever solves the
gyroscopic momentum and precession issue.
The best I have seen so far is:
http://nuclear.ucdavis.edu/~cherney/dvd/memorize/fforces.html
http://nuclear.ucdavis.edu/~cherney/dvd/memorize/erigid.html
Thanks for the link. It's curious that linear momentum can be considered
angular momentum with the reference point at infinity. Is it correct to use
the foundations of linear mechanics to construct rotational mechanics or
does something get left out? ;)
.
|
|
|
|
|
| User: "" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
16 Feb 2006 08:52:16 PM |
|
|
Jason Stanidge wrote:
For a distributed mass, F = M dv/dt where F is the total external force, M
the total mass, and v the velocity of the centre of mass. But take the case
of a gyroscope, where the force acting on the centre of mass is gravity,
assuming it is uniform over the body. The gyroscope doesn't topple over, but
precesses, which imples that the centre of mass theorem doesn't apply to
rotational bodies. Why not? The Theorem seems to be correct for any system
of particles, whether it is rotaing or not
The center of mass theorem does hold for rotational bodies, but there
are additional laws that must also hold for rigid bodies. In
particular, the roational analog of the center of mass theorem, namely
that the net torque equals the moment of inertia times the rate of
change of the angular velocity. This latter law is derived for a rigid
body using the force acting on each body in a similar way to your
derivation, so there is no contradiction.
(Don't forget that there is also an upward force on the gyroscope at
the point of contact of the gyroscope and its platform.)
.
|
|
|
| User: "Jason Stanidge" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
15 Feb 2006 09:40:14 PM |
|
|
<pixel_a_ted@yahoo.com> wrote in message
news:1140144736.890792.84650@z14g2000cwz.googlegroups.com...
Jason Stanidge wrote:
For a distributed mass, F = M dv/dt where F is the total external force,
M
the total mass, and v the velocity of the centre of mass. But take the
case
of a gyroscope, where the force acting on the centre of mass is gravity,
assuming it is uniform over the body. The gyroscope doesn't topple over,
but
precesses, which imples that the centre of mass theorem doesn't apply to
rotational bodies. Why not? The Theorem seems to be correct for any
system
of particles, whether it is rotaing or not
The center of mass theorem does hold for rotational bodies, but there
are additional laws that must also hold for rigid bodies. In
particular, the roational analog of the center of mass theorem, namely
that the net torque equals the moment of inertia times the rate of
change of the angular velocity. This latter law is derived for a rigid
body using the force acting on each body in a similar way to your
derivation, so there is no contradiction.
(Don't forget that there is also an upward force on the gyroscope at
the point of contact of the gyroscope and its platform.)
OK, so what step in the following derivation of the Centre of Mass Theorem
is false for a rotating body then?
For each particle in a body, f = md^2r/dt^2.
If I sum all the forces, including the external forces applied to the body,
and make use of Newton's third law, then the internal forces all cancel, so
I'm left with
just the external force F.
F = sum(k=1, k=n)[mkd^2rk/dt^2] where mk and rk is the mass and radius
vector of the kth particle.
So F = d^2/dt^2[sum(k=1, k=n)[mk x rk]] . If the centre of mass is defined
as R = (sum(k=1, k=)[mk x rk] )/M
Then external F = Md^2R/dt^2.
.
|
|
|
| User: "" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
18 Feb 2006 03:02:29 PM |
|
|
Check out http://science.howstuffworks.com/gyroscope2.htm for a
non-mathematical discussion of precession. The torque due to gravity
and the upward force at the point of contact are not being ignored. The
precession of a gryoscope, i.e. it's not falling down, does not negate
the center of mass theorem.
.
|
|
|
| User: "Jason Stanidge" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
18 Feb 2006 10:30:06 PM |
|
|
<pixel_a_ted@yahoo.com> wrote in message
news:1140296549.327857.148390@g43g2000cwa.googlegroups.com...
Check out http://science.howstuffworks.com/gyroscope2.htm for a
non-mathematical discussion of precession. The torque due to gravity
and the upward force at the point of contact are not being ignored. The
precession of a gryoscope, i.e. it's not falling down, does not negate
the center of mass theorem.
The link is a good explanation of precession. But it doesn't explain why the
COM theorem is still OK for rotating bodies. I'm beginning to think the
problem lies in equating the sum of Fs to the sum of the mometum of the
particles at time t, and so assuming that the effects of the past do not
need to be taken into account, but I'll have to think about this further;
that a rotating body is a well behaved, oscillating mechanical system with
feedback. Sounds crackpot, I know ;)
.
|
|
|
|
|
|
|
| User: "dedanoe" |
|
| Title: Re: Centre of mass theorem for rotaing bodies? |
18 Feb 2006 04:33:40 AM |
|
|
no buddy that's completely wrong approach. nothing deals better with
center of mass like the law of lever (the static form). there's more on
it on http://dedanoe.tripod.com/nukebook
.
|
|
|
|
|
Related Articles |
|
|
