Angular momentum in different points

Clash Royale CLAN TAG#URR8PPP
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
add a comment |
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
add a comment |
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
newtonian-mechanics angular-momentum rotational-dynamics reference-frames conservation-laws
edited Dec 28 '18 at 12:55
Qmechanic♦
102k121831161
102k121831161
asked Dec 27 '18 at 21:12
FrogfireFrogfire
162
162
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
|
show 7 more comments
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
add a comment |
Angular momentum relative to an origin $mathcal O_1$
$$ mathbfL_mathcal O_1 = mathbfr_mathcal O_1 times p_mathcal O_1$$
where $mathbf r_mathcal O_1$ is the position vector to the particle relative to some origin $mathcal O_1$.
Now suppose that angular momentum is conserved in $mathcal O_1$. Then
$$ fracd mathbf L_1dt = mathbfdotr_1 times p_1 + mathbfr_1 times dotp_1 = frac1m mathbfp_1 times p_1 + mathbfr_1 times dotp_1 =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbfp_1 = mathbfp$). It then follows that
$$ mathbfr_1 times F_1 =0 . $$
Now, let's look at some other origin $mathcalO_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbfr_2 times F_2 stackrel?=0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450722%2fangular-momentum-in-different-points%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
|
show 7 more comments
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
|
show 7 more comments
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.
answered Dec 27 '18 at 21:23
Michael SeifertMichael Seifert
14.6k22752
14.6k22752
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
|
show 7 more comments
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
1
1
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law.
– Bill N
Dec 27 '18 at 23:10
2
2
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
@BillN: If you're defining angular momentum to be conserved whenever $Delta mathbfL = int pmbtau , dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological.
– Michael Seifert
Dec 27 '18 at 23:39
1
1
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
@BillN How do you define "conserved"?
– FGSUZ
Dec 28 '18 at 0:07
1
1
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
@BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand.
– FGSUZ
Dec 28 '18 at 21:29
1
1
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
@BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion.
– Aaron Stevens
Dec 29 '18 at 2:20
|
show 7 more comments
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
add a comment |
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
add a comment |
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.
answered Dec 27 '18 at 21:54
Mozibur UllahMozibur Ullah
4,65022249
4,65022249
add a comment |
add a comment |
Angular momentum relative to an origin $mathcal O_1$
$$ mathbfL_mathcal O_1 = mathbfr_mathcal O_1 times p_mathcal O_1$$
where $mathbf r_mathcal O_1$ is the position vector to the particle relative to some origin $mathcal O_1$.
Now suppose that angular momentum is conserved in $mathcal O_1$. Then
$$ fracd mathbf L_1dt = mathbfdotr_1 times p_1 + mathbfr_1 times dotp_1 = frac1m mathbfp_1 times p_1 + mathbfr_1 times dotp_1 =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbfp_1 = mathbfp$). It then follows that
$$ mathbfr_1 times F_1 =0 . $$
Now, let's look at some other origin $mathcalO_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbfr_2 times F_2 stackrel?=0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
add a comment |
Angular momentum relative to an origin $mathcal O_1$
$$ mathbfL_mathcal O_1 = mathbfr_mathcal O_1 times p_mathcal O_1$$
where $mathbf r_mathcal O_1$ is the position vector to the particle relative to some origin $mathcal O_1$.
Now suppose that angular momentum is conserved in $mathcal O_1$. Then
$$ fracd mathbf L_1dt = mathbfdotr_1 times p_1 + mathbfr_1 times dotp_1 = frac1m mathbfp_1 times p_1 + mathbfr_1 times dotp_1 =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbfp_1 = mathbfp$). It then follows that
$$ mathbfr_1 times F_1 =0 . $$
Now, let's look at some other origin $mathcalO_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbfr_2 times F_2 stackrel?=0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
add a comment |
Angular momentum relative to an origin $mathcal O_1$
$$ mathbfL_mathcal O_1 = mathbfr_mathcal O_1 times p_mathcal O_1$$
where $mathbf r_mathcal O_1$ is the position vector to the particle relative to some origin $mathcal O_1$.
Now suppose that angular momentum is conserved in $mathcal O_1$. Then
$$ fracd mathbf L_1dt = mathbfdotr_1 times p_1 + mathbfr_1 times dotp_1 = frac1m mathbfp_1 times p_1 + mathbfr_1 times dotp_1 =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbfp_1 = mathbfp$). It then follows that
$$ mathbfr_1 times F_1 =0 . $$
Now, let's look at some other origin $mathcalO_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbfr_2 times F_2 stackrel?=0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
Angular momentum relative to an origin $mathcal O_1$
$$ mathbfL_mathcal O_1 = mathbfr_mathcal O_1 times p_mathcal O_1$$
where $mathbf r_mathcal O_1$ is the position vector to the particle relative to some origin $mathcal O_1$.
Now suppose that angular momentum is conserved in $mathcal O_1$. Then
$$ fracd mathbf L_1dt = mathbfdotr_1 times p_1 + mathbfr_1 times dotp_1 = frac1m mathbfp_1 times p_1 + mathbfr_1 times dotp_1 =0 $$
but since the direction of momentum is frame-independent, the first term vanishes (that is, $mathbfp_1 = mathbfp$). It then follows that
$$ mathbfr_1 times F_1 =0 . $$
Now, let's look at some other origin $mathcalO_2$, given that $L$ is conserved in $mathcal O_1$. Well the first term much vanish again, that's fine but what about the second term? Does
$$mathbfr_2 times F_2 stackrel?=0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.
edited Dec 27 '18 at 21:46
answered Dec 27 '18 at 21:32
InertialObserverInertialObserver
2,242623
2,242623
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450722%2fangular-momentum-in-different-points%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown