What happens when the centripetal force is equal and opposite to the centrifugal force?












1












$begingroup$


We say that centrifugal force is fictitious, yet we still use it in some problems. If the centrifugal force is equal and opposite to the centripetal force wouldn't that make the net force zero?










share|cite|improve this question









New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    It would make the net force zero. Which is the problem. Because the body in question is accelerating.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    But that will be for the st. Line only isn't it
    $endgroup$
    – Santosh Khatri
    4 hours ago






  • 1




    $begingroup$
    Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
    $endgroup$
    – Dvij Mankad
    4 hours ago


















1












$begingroup$


We say that centrifugal force is fictitious, yet we still use it in some problems. If the centrifugal force is equal and opposite to the centripetal force wouldn't that make the net force zero?










share|cite|improve this question









New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    It would make the net force zero. Which is the problem. Because the body in question is accelerating.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    But that will be for the st. Line only isn't it
    $endgroup$
    – Santosh Khatri
    4 hours ago






  • 1




    $begingroup$
    Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
    $endgroup$
    – Dvij Mankad
    4 hours ago
















1












1








1





$begingroup$


We say that centrifugal force is fictitious, yet we still use it in some problems. If the centrifugal force is equal and opposite to the centripetal force wouldn't that make the net force zero?










share|cite|improve this question









New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




We say that centrifugal force is fictitious, yet we still use it in some problems. If the centrifugal force is equal and opposite to the centripetal force wouldn't that make the net force zero?







newtonian-mechanics rotational-dynamics reference-frames centripetal-force centrifugal-force






share|cite|improve this question









New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Qmechanic

106k121961222




106k121961222






New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









Santosh KhatriSantosh Khatri

122




122




New contributor




Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Santosh Khatri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    It would make the net force zero. Which is the problem. Because the body in question is accelerating.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    But that will be for the st. Line only isn't it
    $endgroup$
    – Santosh Khatri
    4 hours ago






  • 1




    $begingroup$
    Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
    $endgroup$
    – Dvij Mankad
    4 hours ago
















  • 1




    $begingroup$
    It would make the net force zero. Which is the problem. Because the body in question is accelerating.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    But that will be for the st. Line only isn't it
    $endgroup$
    – Santosh Khatri
    4 hours ago






  • 1




    $begingroup$
    Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
    $endgroup$
    – dmckee
    4 hours ago










  • $begingroup$
    The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
    $endgroup$
    – Dvij Mankad
    4 hours ago










1




1




$begingroup$
It would make the net force zero. Which is the problem. Because the body in question is accelerating.
$endgroup$
– dmckee
4 hours ago




$begingroup$
It would make the net force zero. Which is the problem. Because the body in question is accelerating.
$endgroup$
– dmckee
4 hours ago












$begingroup$
But that will be for the st. Line only isn't it
$endgroup$
– Santosh Khatri
4 hours ago




$begingroup$
But that will be for the st. Line only isn't it
$endgroup$
– Santosh Khatri
4 hours ago




1




1




$begingroup$
Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
$endgroup$
– dmckee
4 hours ago




$begingroup$
Newton's 2nd law is not restricted to straight lines (if that is what "st. Line" is meant to convey). I know that the experience of interaction when riding a merry-go-round or carnival ride is visceral and intense, but you have to frame your thinking about it carefully. When you are going around in a circle you are not in equilibrium: you are accelerating toward the center of that circle. Believe that—really take it into account—and you can make sense of it all. The force that is applied to you is always toward the center.
$endgroup$
– dmckee
4 hours ago












$begingroup$
The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
$endgroup$
– Dvij Mankad
4 hours ago






$begingroup$
The title of the question should serve as a title to the whole question and one should probably avoid writing the text of the question in literal continuation of the phrase/sentence in the title. In other words, the body of the question should be such that it can convey some meaning on its own and need not be read in literal continuation of the title.
$endgroup$
– Dvij Mankad
4 hours ago












1 Answer
1






active

oldest

votes


















3












$begingroup$

First, it must be stated that Newton's laws only hold in inertial frames. What this means is that accelerations must arise from forces (Second Law), and these forces arise from interactions (Third Law). The issue with rotating frames is that accelerations arise when forces of interactions are not present, so Newton's laws do not hold in rotating frames.



However, the second law ($mathbf F=mmathbf a$) is nice to use since it tells us how to determine the position and velocity of a body given initial conditions. Therefore, we define "fictitious" centrifugal and Coriolis forces in order to keep this framework. They are "fictitious" because they are an artifact of the rotating reference frame rather than interactions, but they are not fake (for example, they are very real for anyone going around a sharp turn in a car). Essentially we have opted to abandon the third law in order to keep the second law.



Now, onto your specific inquiry: If you are in a rotating frame, and there is a force equally opposing the centrifugal force, then yes the net force is zero (assuming no Coriolis force either). Therefore in the rotating frame there is no acceleration of the object in question.



Of course, if you looked at the scenario from an inertial frame you would have a non-zero acceleration of the object as there is now a non-zero net force that is the centripetal force.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
    $endgroup$
    – dmckee
    3 hours ago












  • $begingroup$
    @dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
    $endgroup$
    – Aaron Stevens
    2 hours ago











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
});


}
});






Santosh Khatri is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f467294%2fwhat-happens-when-the-centripetal-force-is-equal-and-opposite-to-the-centrifugal%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

First, it must be stated that Newton's laws only hold in inertial frames. What this means is that accelerations must arise from forces (Second Law), and these forces arise from interactions (Third Law). The issue with rotating frames is that accelerations arise when forces of interactions are not present, so Newton's laws do not hold in rotating frames.



However, the second law ($mathbf F=mmathbf a$) is nice to use since it tells us how to determine the position and velocity of a body given initial conditions. Therefore, we define "fictitious" centrifugal and Coriolis forces in order to keep this framework. They are "fictitious" because they are an artifact of the rotating reference frame rather than interactions, but they are not fake (for example, they are very real for anyone going around a sharp turn in a car). Essentially we have opted to abandon the third law in order to keep the second law.



Now, onto your specific inquiry: If you are in a rotating frame, and there is a force equally opposing the centrifugal force, then yes the net force is zero (assuming no Coriolis force either). Therefore in the rotating frame there is no acceleration of the object in question.



Of course, if you looked at the scenario from an inertial frame you would have a non-zero acceleration of the object as there is now a non-zero net force that is the centripetal force.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
    $endgroup$
    – dmckee
    3 hours ago












  • $begingroup$
    @dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
    $endgroup$
    – Aaron Stevens
    2 hours ago
















3












$begingroup$

First, it must be stated that Newton's laws only hold in inertial frames. What this means is that accelerations must arise from forces (Second Law), and these forces arise from interactions (Third Law). The issue with rotating frames is that accelerations arise when forces of interactions are not present, so Newton's laws do not hold in rotating frames.



However, the second law ($mathbf F=mmathbf a$) is nice to use since it tells us how to determine the position and velocity of a body given initial conditions. Therefore, we define "fictitious" centrifugal and Coriolis forces in order to keep this framework. They are "fictitious" because they are an artifact of the rotating reference frame rather than interactions, but they are not fake (for example, they are very real for anyone going around a sharp turn in a car). Essentially we have opted to abandon the third law in order to keep the second law.



Now, onto your specific inquiry: If you are in a rotating frame, and there is a force equally opposing the centrifugal force, then yes the net force is zero (assuming no Coriolis force either). Therefore in the rotating frame there is no acceleration of the object in question.



Of course, if you looked at the scenario from an inertial frame you would have a non-zero acceleration of the object as there is now a non-zero net force that is the centripetal force.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
    $endgroup$
    – dmckee
    3 hours ago












  • $begingroup$
    @dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
    $endgroup$
    – Aaron Stevens
    2 hours ago














3












3








3





$begingroup$

First, it must be stated that Newton's laws only hold in inertial frames. What this means is that accelerations must arise from forces (Second Law), and these forces arise from interactions (Third Law). The issue with rotating frames is that accelerations arise when forces of interactions are not present, so Newton's laws do not hold in rotating frames.



However, the second law ($mathbf F=mmathbf a$) is nice to use since it tells us how to determine the position and velocity of a body given initial conditions. Therefore, we define "fictitious" centrifugal and Coriolis forces in order to keep this framework. They are "fictitious" because they are an artifact of the rotating reference frame rather than interactions, but they are not fake (for example, they are very real for anyone going around a sharp turn in a car). Essentially we have opted to abandon the third law in order to keep the second law.



Now, onto your specific inquiry: If you are in a rotating frame, and there is a force equally opposing the centrifugal force, then yes the net force is zero (assuming no Coriolis force either). Therefore in the rotating frame there is no acceleration of the object in question.



Of course, if you looked at the scenario from an inertial frame you would have a non-zero acceleration of the object as there is now a non-zero net force that is the centripetal force.






share|cite|improve this answer











$endgroup$



First, it must be stated that Newton's laws only hold in inertial frames. What this means is that accelerations must arise from forces (Second Law), and these forces arise from interactions (Third Law). The issue with rotating frames is that accelerations arise when forces of interactions are not present, so Newton's laws do not hold in rotating frames.



However, the second law ($mathbf F=mmathbf a$) is nice to use since it tells us how to determine the position and velocity of a body given initial conditions. Therefore, we define "fictitious" centrifugal and Coriolis forces in order to keep this framework. They are "fictitious" because they are an artifact of the rotating reference frame rather than interactions, but they are not fake (for example, they are very real for anyone going around a sharp turn in a car). Essentially we have opted to abandon the third law in order to keep the second law.



Now, onto your specific inquiry: If you are in a rotating frame, and there is a force equally opposing the centrifugal force, then yes the net force is zero (assuming no Coriolis force either). Therefore in the rotating frame there is no acceleration of the object in question.



Of course, if you looked at the scenario from an inertial frame you would have a non-zero acceleration of the object as there is now a non-zero net force that is the centripetal force.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 4 hours ago









Aaron StevensAaron Stevens

13.1k42248




13.1k42248








  • 3




    $begingroup$
    You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
    $endgroup$
    – dmckee
    3 hours ago












  • $begingroup$
    @dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
    $endgroup$
    – Aaron Stevens
    2 hours ago














  • 3




    $begingroup$
    You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
    $endgroup$
    – dmckee
    3 hours ago












  • $begingroup$
    @dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
    $endgroup$
    – Aaron Stevens
    2 hours ago








3




3




$begingroup$
You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
$endgroup$
– dmckee
3 hours ago






$begingroup$
You should emphasize that Newton's laws are defined for inertial frames and that the treatment of non-inertial frame by applying inertial pseudo-forces is a lash-up to lets us apply the machinery of Newtonian mechanics to situations other than those for which in which the subject finds its natural expression. Otherwise you invite misunderstanding.
$endgroup$
– dmckee
3 hours ago














$begingroup$
@dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
$endgroup$
– Aaron Stevens
2 hours ago




$begingroup$
@dmckee I agree those are good points to make. I have added information pertaining to this. Thanks for the suggestion.
$endgroup$
– Aaron Stevens
2 hours ago










Santosh Khatri is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















Santosh Khatri is a new contributor. Be nice, and check out our Code of Conduct.













Santosh Khatri is a new contributor. Be nice, and check out our Code of Conduct.












Santosh Khatri is a new contributor. Be nice, and check out our Code of Conduct.
















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f467294%2fwhat-happens-when-the-centripetal-force-is-equal-and-opposite-to-the-centrifugal%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

CARDNET

Boot-repair Failure: Unable to locate package grub-common:i386

濃尾地震