Science > Physics > Translational Kinetic Energy and Rotational Kinetic Energy.
| Topic: |
Science > Physics |
| User: |
"Starblade Darksquall" |
| Date: |
01 Sep 2003 12:08:36 AM |
| Object: |
Translational Kinetic Energy and Rotational Kinetic Energy. |
What is the conservation rule governing transitions between
Translational Kinetic Energy (TKE) and Rotational Kinetic Energy
(RKE)? Suppose that a system of molecules bounces against eachother.
We know that the net linear momentum and the net angular momentum
remain the same, but what rules govern the net changes in TKE and RKE?
The first thought on my mind is TKE + RKE = constant, but then I got
to thinking. Isn't TKE slightly different from RKE?
TKE doesn't invole any radian units, but RKE does. Wouldn't this cause
some difficulties? Or is it assumed that 1 radian can be considered
simply as the number 1?
Also, I am sure that TKE is scalar, but what about RKE? I'd imagine
that it's also scalar. But am I wrong about this? Am I wrong about
these things? What if they're not scalar? What if neither of them are?
Then how do the conservation laws work?
These things bother me because just because we think that things
should work a certain way doesn't mean that they actually do work that
way, as is often found out. So... how does the whole TKE and RKE thing
pan out? Are they compatible or incompatible units of energy?
(...Starblade Riven Darksquall...)
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| User: "Starblade Darksquall" |
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| Title: Re: Translational Kinetic Energy and Rotational Kinetic Energy. |
01 Sep 2003 10:28:04 PM |
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"Old Man" <nomail@nomail.net> wrote in message news:<3f5241ff_2@newsfeed>...
Starblade Darksquall <Starblade13@Yahoo.com> wrote in message
news:4aa861fb.0308312108.1e6d41ba@posting.google.com...
What is the conservation rule governing transitions between
Translational Kinetic Energy (TKE) and Rotational Kinetic Energy
(RKE)? Suppose that a system of molecules bounces against eachother.
We know that the net linear momentum and the net angular momentum
remain the same, but what rules govern the net changes in TKE and RKE?
The first thought on my mind is TKE + RKE = constant, but then I got
to thinking. Isn't TKE slightly different from RKE?
Conservation of energy:
E = [(m / 2)(v_r)^2] + [L^2 / 2mr^2] + V(r)
Where radial velocity, v_r = d|r| / dt, and L is angular momentum.
E - V(r) = [KE of radial motion] + [KE of angular momentum]
[Old Man]
I'm not looking for the KE of radial motion vs the KE of angular
momentum. I'm looking for the KE of LINEAR motion plus the KE of
angular motion. What exactly are the equations for that?
TKE doesn't invole any radian units, but RKE does. Wouldn't this cause
some difficulties? Or is it assumed that 1 radian can be considered
simply as the number 1?
Also, I am sure that TKE is scalar, but what about RKE? I'd imagine
that it's also scalar. But am I wrong about this? Am I wrong about
these things? What if they're not scalar? What if neither of them are?
Then how do the conservation laws work?
These things bother me because just because we think that things
should work a certain way doesn't mean that they actually do work that
way, as is often found out. So... how does the whole TKE and RKE thing
pan out? Are they compatible or incompatible units of energy?
(...Starblade Riven Darksquall...)
.
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| User: "Old Man" |
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| Title: Re: Translational Kinetic Energy and Rotational Kinetic Energy. |
02 Sep 2003 12:11:44 PM |
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Starblade Darksquall <Starblade13@Yahoo.com> wrote in message
news:4aa861fb.0309011928.1c54dd58@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<3f5241ff_2@newsfeed>...
Starblade Darksquall <Starblade13@Yahoo.com> wrote in message
news:4aa861fb.0308312108.1e6d41ba@posting.google.com...
What is the conservation rule governing transitions between
Translational Kinetic Energy (TKE) and Rotational Kinetic Energy
(RKE)? Suppose that a system of molecules bounces against eachother.
We know that the net linear momentum and the net angular momentum
remain the same, but what rules govern the net changes in TKE and RKE?
The first thought on my mind is TKE + RKE = constant, but then I got
to thinking. Isn't TKE slightly different from RKE?
Conservation of energy:
E = [(m / 2)(v_r)^2] + [L^2 / 2mr^2] + V(r)
Where radial velocity, v_r = d|r| / dt, and L is angular momentum.
E - V(r) = [KE of radial motion] + [KE of angular momentum]
[Old Man]
I'm not looking for the KE of radial motion vs the KE of angular
momentum. I'm looking for the KE of LINEAR motion plus the KE of
angular motion. What exactly are the equations for that?
Silly! Even though angular momentum is constant,
KE(rotational) => not constant
KE(linear) + KE(rotational) => not consant
KE(linear) = KE(rotational) + KE(radial) = constant
Radial motion is a component of linear motion. A particle with
constant linear momentum, wherein its straight line trajectory does
not intersect the origin, also has constant angular momentum:
L = r x p = r*p*sin(theta) = b*p = constant
where b is the impact parameter (distance of clsest approach).
KE(linear) = p^2 / 2m = KE(rotational) + KE(radial) = constant
note that, even though angular momentum is constant, KE(rotational)
is mot constant
KE(rotational) = L^2 / 2*m*r^2 => not constant
[Old Man]
(...Starblade Riven Darksquall...)
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| User: "Sam Wormley" |
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| Title: Re: Translational Kinetic Energy and Rotational Kinetic Energy. |
01 Sep 2003 10:51:00 PM |
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Starblade Darksquall wrote:
I'm not looking for the KE of radial motion vs the KE of angular
momentum. I'm looking for the KE of LINEAR motion plus the KE of
angular motion. What exactly are the equations for that?
Kinetic Energy of a rigid rotating body about a fixed axis x
E_rot = 1/2 J_x omega^2
where J_x in the moment of inertia and omega the angular velocity.
So the total KE would be something like
3
1/2 mv^2 + sum J_ik omega_i omega_k where i,k = x,y,z
i,k=1
where m is the total mass, J_ik: components of the tensor of inertia J, omega_i:
components of the angular velocity vector omega.
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| User: "Sam Wormley" |
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| Title: Re: Translational Kinetic Energy and Rotational Kinetic Energy. |
01 Sep 2003 11:04:31 PM |
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Sam Wormley wrote:
Starblade Darksquall wrote:
I'm not looking for the KE of radial motion vs the KE of angular
momentum. I'm looking for the KE of LINEAR motion plus the KE of
angular motion. What exactly are the equations for that?
Kinetic Energy of a rigid rotating body about a fixed axis x
E_rot = 1/2 J_x omega^2
where J_x in the moment of inertia and omega the angular velocity.
So the total KE would be something like
3
1/2 mv^2 + sum J_ik omega_i omega_k where i,k = x,y,z
i,k=1
where m is the total mass, J_ik: components of the tensor of inertia J, omega_i:
components of the angular velocity vector omega.
Kinetic Energy
http://scienceworld.wolfram.com/physics/KineticEnergy.html
Equations (11) and (12) and there derivations
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| User: "Uncle Al" |
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| Title: Re: Translational Kinetic Energy and Rotational Kinetic Energy. |
01 Sep 2003 11:19:21 AM |
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Starblade Darksquall wrote:
What is the conservation rule governing transitions between
Translational Kinetic Energy (TKE) and Rotational Kinetic Energy
(RKE)?[snip]
We all know your game, schmuck. Nobody is playing.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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