Mdthbs

Can I travel from Germany to England alone as an unaccompanied minor?

Why won't the ground take my seed?

Convergenge or divergence of series with e

Attempt to de-reference a null object: list initialization

How hard is it to sell a home which is currently mortgaged?

Can Access Fault Exceptions of the MC68040 caused by internal access faults occur in normal situations?

Why did this meteor appear cyan?

How does the Duergar Magic shrink/enlarge ability work with rage?

Different budgets within roommate group

Generate and graph the Recamán Sequence

What is a macro? Difference between macro and function?

Wrong corporate name on employment agreement

Are there any features that help with the roll to avoid the destruction of a Wand of Fireballs when using the last charge?

Is there a category where products don't exist because uniqueness fails?

Is there a way for presidents to legally extend their terms beyond the maximum of four years?

Most importants new papers in computational complexity

What's the easiest way for a whole party to be able to communicate with a creature that doesn't know Common?

Miss Toad and her frogs

Can a US President have someone sent to prison?

Does the Pi 4 resolve the Ethernet+USB bottleneck issue of past versions?

What is the line crossing the Pacific Ocean that is shown on maps?

Is it bad to describe a character long after their introduction?

3D nonogram, beginner's edition

How did researchers use to find articles before the Internet and the computer era?

Acceleration in Circular motion

Physical Variables of Circular MotionTangential acceleration in circular motion?Is this a correct understanding of circular motion?Instantaneous acceleration in non uniform circular motionUniform circular motion dilemmatangential acceleration for uniform circular motionNon Uniform Circular Motion and How External Force Affects The MotionNet acceleration in circular motionCan non-uniform circular motion be called as periodic motion?Intuition for formula of tangential component of acceleration in general curvillinear motion

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

1

$begingroup$

Can motion of a particle be circular if radial acceleration is zero but tangential acceleration is not 0?

newtonian-mechanics forces acceleration rotation

share|cite|improve this question

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What discussion, reading, or thoughts have prompted you to ask that question? Do you understand the meaning of the terms "radial" and "tangential"?
  $endgroup$
  – Bill N
  12 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Can motion of a particle be circular if radial acceleration is zero but tangential acceleration is not 0?

newtonian-mechanics forces acceleration rotation

share|cite|improve this question

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What discussion, reading, or thoughts have prompted you to ask that question? Do you understand the meaning of the terms "radial" and "tangential"?
  $endgroup$
  – Bill N
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Can motion of a particle be circular if radial acceleration is zero but tangential acceleration is not 0?

newtonian-mechanics forces acceleration rotation

share|cite|improve this question

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

$endgroup$

share|cite|improve this question

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

share|cite|improve this question

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

edited 11 hours ago

Aaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

asked 12 hours ago

RiyaRiya

1611 bronze badge

asked 12 hours ago

RiyaRiya

1611 bronze badge

5 Answers
5

active

oldest

votes

5

$begingroup$

No. Radial acceleration is necessary in order to change the direction of the velocity.

share|cite|improve this answer

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Let's look at the general acceleration vector in polar coordinates$^*$:

$$mathbf a=left(ddot r-rdottheta^2right)hat r+left(rddottheta+2dot rdotthetaright)hattheta$$

If we want our object to remain on the same circle, we must have $dot r=0$ and $ddot r=0$. This means our acceleration must have the form:
$$mathbf a=-rdottheta^2hat r+rddotthetahattheta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force of
$$F_r=-mrdottheta^2$$
and a tangential force of
$$F_theta=mrddottheta$$

The radial force is responsible for changing the direction of the particle. The tangential force is responsible for changing the speed of the particle as it moves around the circle.

If we additionally require uniform circular motion, then $ddottheta=0$, and we can only have a radial force without a tangential force. i.e.
$$mathbf F=-mrdottheta^2hat r$$

Therefore, the answer to your question is no. You always need a radial force (technically this specific radial force. Not any radial force will do) for circular motion (uniform or not). However, you can have the tangential force as zero (uniform circular motion) or non zero (non-uniform circular motion).

---

$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

In circular motion, $$vec{a}_{text{radial}}neq0,$$ but $vec{a}_{text{tangential}}$ may or may not be equal to zero.

share|cite|improve this answer

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

I think the premise of your question is false, because the accelerated tangential motion will lead to a change in the centrifugal force on the particle, which in turn will cause radial acceleration. You can't have one kind of acceleration without causing the other in the first place.

But of course, when the acceleration and the resulting adjustment of the radius are finished, you can end up with a new circular orbit at a the new radius. Perhaps this is what you meant?

share|cite|improve this answer

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

$endgroup$

add a comment | 

-2

$begingroup$

For circular motion, we should have
$$v_r=0,qquad v_{phi}=const.$$
This means that both the radial as well as tangential acceleration should be zero. If the tangential acceleration is non-zero, this means that the tangential velocity changes which would make the orbit spiral depending on the sign of the acceleration.

share|cite|improve this answer

answered 12 hours ago

RichardRichard

16499 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is for uniform circular motion. Non-uniform circular motion is also possible. Your speed can change while still remaining on the same circle.
  $endgroup$
  – Aaron Stevens
  12 hours ago
  

  
  
  
  

add a comment | 

Your Answer

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f487829%2facceleration-in-circular-motion%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

No. Radial acceleration is necessary in order to change the direction of the velocity.

share|cite|improve this answer

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

No. Radial acceleration is necessary in order to change the direction of the velocity.

share|cite|improve this answer

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

$endgroup$

add a comment | 

$begingroup$

No. Radial acceleration is necessary in order to change the direction of the velocity.

share|cite|improve this answer

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

$endgroup$

share|cite|improve this answer

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

answered 12 hours ago

garypgaryp

17.3k11 gold badge3131 silver badges6565 bronze badges

5

$begingroup$

Let's look at the general acceleration vector in polar coordinates$^*$:

$$mathbf a=left(ddot r-rdottheta^2right)hat r+left(rddottheta+2dot rdotthetaright)hattheta$$

If we want our object to remain on the same circle, we must have $dot r=0$ and $ddot r=0$. This means our acceleration must have the form:
$$mathbf a=-rdottheta^2hat r+rddotthetahattheta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force of
$$F_r=-mrdottheta^2$$
and a tangential force of
$$F_theta=mrddottheta$$

The radial force is responsible for changing the direction of the particle. The tangential force is responsible for changing the speed of the particle as it moves around the circle.

If we additionally require uniform circular motion, then $ddottheta=0$, and we can only have a radial force without a tangential force. i.e.
$$mathbf F=-mrdottheta^2hat r$$

Therefore, the answer to your question is no. You always need a radial force (technically this specific radial force. Not any radial force will do) for circular motion (uniform or not). However, you can have the tangential force as zero (uniform circular motion) or non zero (non-uniform circular motion).

---

$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Let's look at the general acceleration vector in polar coordinates$^*$:

$$mathbf a=left(ddot r-rdottheta^2right)hat r+left(rddottheta+2dot rdotthetaright)hattheta$$

If we want our object to remain on the same circle, we must have $dot r=0$ and $ddot r=0$. This means our acceleration must have the form:
$$mathbf a=-rdottheta^2hat r+rddotthetahattheta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force of
$$F_r=-mrdottheta^2$$
and a tangential force of
$$F_theta=mrddottheta$$

The radial force is responsible for changing the direction of the particle. The tangential force is responsible for changing the speed of the particle as it moves around the circle.

If we additionally require uniform circular motion, then $ddottheta=0$, and we can only have a radial force without a tangential force. i.e.
$$mathbf F=-mrdottheta^2hat r$$

Therefore, the answer to your question is no. You always need a radial force (technically this specific radial force. Not any radial force will do) for circular motion (uniform or not). However, you can have the tangential force as zero (uniform circular motion) or non zero (non-uniform circular motion).

---

$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let's look at the general acceleration vector in polar coordinates$^*$:

$$mathbf a=left(ddot r-rdottheta^2right)hat r+left(rddottheta+2dot rdotthetaright)hattheta$$

If we want our object to remain on the same circle, we must have $dot r=0$ and $ddot r=0$. This means our acceleration must have the form:
$$mathbf a=-rdottheta^2hat r+rddotthetahattheta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force of
$$F_r=-mrdottheta^2$$
and a tangential force of
$$F_theta=mrddottheta$$

The radial force is responsible for changing the direction of the particle. The tangential force is responsible for changing the speed of the particle as it moves around the circle.

If we additionally require uniform circular motion, then $ddottheta=0$, and we can only have a radial force without a tangential force. i.e.
$$mathbf F=-mrdottheta^2hat r$$

Therefore, the answer to your question is no. You always need a radial force (technically this specific radial force. Not any radial force will do) for circular motion (uniform or not). However, you can have the tangential force as zero (uniform circular motion) or non zero (non-uniform circular motion).

---

$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

$endgroup$

share|cite|improve this answer

edited 11 hours ago

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

answered 11 hours ago

Aaron StevensAaron Stevens

16.9k44 gold badges2828 silver badges6464 bronze badges

2

$begingroup$

In circular motion, $$vec{a}_{text{radial}}neq0,$$ but $vec{a}_{text{tangential}}$ may or may not be equal to zero.

share|cite|improve this answer

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

In circular motion, $$vec{a}_{text{radial}}neq0,$$ but $vec{a}_{text{tangential}}$ may or may not be equal to zero.

share|cite|improve this answer

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

In circular motion, $$vec{a}_{text{radial}}neq0,$$ but $vec{a}_{text{tangential}}$ may or may not be equal to zero.

share|cite|improve this answer

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

edited 3 hours ago

Sebastiano

37922 silver badges2020 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

answered 10 hours ago

UniqueUnique

1,06511 gold badge55 silver badges1818 bronze badges

0

$begingroup$

I think the premise of your question is false, because the accelerated tangential motion will lead to a change in the centrifugal force on the particle, which in turn will cause radial acceleration. You can't have one kind of acceleration without causing the other in the first place.

But of course, when the acceleration and the resulting adjustment of the radius are finished, you can end up with a new circular orbit at a the new radius. Perhaps this is what you meant?

share|cite|improve this answer

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

I think the premise of your question is false, because the accelerated tangential motion will lead to a change in the centrifugal force on the particle, which in turn will cause radial acceleration. You can't have one kind of acceleration without causing the other in the first place.

But of course, when the acceleration and the resulting adjustment of the radius are finished, you can end up with a new circular orbit at a the new radius. Perhaps this is what you meant?

share|cite|improve this answer

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

$endgroup$

add a comment | 

$begingroup$

I think the premise of your question is false, because the accelerated tangential motion will lead to a change in the centrifugal force on the particle, which in turn will cause radial acceleration. You can't have one kind of acceleration without causing the other in the first place.

But of course, when the acceleration and the resulting adjustment of the radius are finished, you can end up with a new circular orbit at a the new radius. Perhaps this is what you meant?

share|cite|improve this answer

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

answered 1 hour ago

Thomas BlankenhornThomas Blankenhorn

1,09622 gold badges22 silver badges77 bronze badges

-2

$begingroup$

For circular motion, we should have
$$v_r=0,qquad v_{phi}=const.$$
This means that both the radial as well as tangential acceleration should be zero. If the tangential acceleration is non-zero, this means that the tangential velocity changes which would make the orbit spiral depending on the sign of the acceleration.

share|cite|improve this answer

answered 12 hours ago

RichardRichard

16499 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is for uniform circular motion. Non-uniform circular motion is also possible. Your speed can change while still remaining on the same circle.
  $endgroup$
  – Aaron Stevens
  12 hours ago
  

  
  
  
  

add a comment | 

-2

$begingroup$

For circular motion, we should have
$$v_r=0,qquad v_{phi}=const.$$
This means that both the radial as well as tangential acceleration should be zero. If the tangential acceleration is non-zero, this means that the tangential velocity changes which would make the orbit spiral depending on the sign of the acceleration.

share|cite|improve this answer

answered 12 hours ago

RichardRichard

16499 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is for uniform circular motion. Non-uniform circular motion is also possible. Your speed can change while still remaining on the same circle.
  $endgroup$
  – Aaron Stevens
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

For circular motion, we should have
$$v_r=0,qquad v_{phi}=const.$$
This means that both the radial as well as tangential acceleration should be zero. If the tangential acceleration is non-zero, this means that the tangential velocity changes which would make the orbit spiral depending on the sign of the acceleration.

share|cite|improve this answer

answered 12 hours ago

RichardRichard

16499 bronze badges

$endgroup$

share|cite|improve this answer

answered 12 hours ago

RichardRichard

16499 bronze badges

answered 12 hours ago

RichardRichard

16499 bronze badges

answered 12 hours ago

RichardRichard

16499 bronze badges