Mixing
The acceleration of circulation motion
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
We know that in circular motion the position vector is $rhatr$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r dhatr/dt + fracdrdt .hatr.$$ now I can't understand why $dr/dt$ isn't zero. The $r$ or radii isn't changing throughout the circle. So should the velocity be only $rdhatr/dt$?
kinematics vectors differentiation
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
asked Aug 12 at 2:34
Nobody recognizeable
444216
add a comment |Â
up vote
3
down vote
favorite
We know that in circular motion the position vector is $rhatr$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r dhatr/dt + fracdrdt .hatr.$$ now I can't understand why $dr/dt$ isn't zero. The $r$ or radii isn't changing throughout the circle. So should the velocity be only $rdhatr/dt$?
kinematics vectors differentiation
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
asked Aug 12 at 2:34
Nobody recognizeable
444216
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We know that in circular motion the position vector is $rhatr$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r dhatr/dt + fracdrdt .hatr.$$ now I can't understand why $dr/dt$ isn't zero. The $r$ or radii isn't changing throughout the circle. So should the velocity be only $rdhatr/dt$?
kinematics vectors differentiation
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
asked Aug 12 at 2:34
Nobody recognizeable
444216
We know that in circular motion the position vector is $rhatr$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r dhatr/dt + fracdrdt .hatr.$$ now I can't understand why $dr/dt$ isn't zero. The $r$ or radii isn't changing throughout the circle. So should the velocity be only $rdhatr/dt$?
kinematics vectors differentiation
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
asked Aug 12 at 2:34
Nobody recognizeable
444216
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
edited Aug 12 at 9:14
Qmechanic♦
96.5k121631020
96.5k121631020
asked Aug 12 at 2:34
Nobody recognizeable
444216
asked Aug 12 at 2:34
Nobody recognizeable
444216
asked Aug 12 at 2:34
Nobody recognizeable
444216
444216
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Saying that the position is $rhat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $rhat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$rhat r$ is always true, so I can apply it to circular motion."
answered Aug 12 at 3:07
Aaron Stevens
2,572318
add a comment |Â
up vote
3
down vote
You are indeed right to say that the term $fracdrdt$ is zero. Indeed the velocity vector is given just by $rfracdhatrdt$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $rhatr=r(cos(omega t)i+sin(omega t)j$.
Now we know the derivative is $rfracdhatrdt$, which is equal to the derivative of the displacement vector with $r$ as constant, you get:
$$rfracdhatrdt=$r(-sin(omega t)i+cos(omega t)j$$
Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields :
$$(-sin(omega t)cos(omega t) + sin(omega t)cos(omega t))=0$$
As was to be shown.
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Saying that the position is $rhat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $rhat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$rhat r$ is always true, so I can apply it to circular motion."
answered Aug 12 at 3:07
Aaron Stevens
2,572318
add a comment |Â
up vote
5
down vote
accepted
Saying that the position is $rhat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $rhat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$rhat r$ is always true, so I can apply it to circular motion."
answered Aug 12 at 3:07
Aaron Stevens
2,572318
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Saying that the position is $rhat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $rhat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$rhat r$ is always true, so I can apply it to circular motion."
answered Aug 12 at 3:07
Aaron Stevens
2,572318
Saying that the position is $rhat r$ does not only describe circular motion. This is just a general position vector.
Therefore, your time derivative is just a general velocity vector. The first term represents velocity tangent to the position vector, the second parallel to it. Therefore, if you are undergoing circular motion, the second term is $0$.
Based on your comments, I think I see your issue. You are thinking "I see $rhat r$ is used in the case of circular motion, so it only describes circular motion". The correct way to see it is "$rhat r$ is always true, so I can apply it to circular motion."
answered Aug 12 at 3:07
Aaron Stevens
2,572318
edited Aug 12 at 3:13
answered Aug 12 at 3:07
Aaron Stevens
2,572318
answered Aug 12 at 3:07
Aaron Stevens
2,572318
answered Aug 12 at 3:07
Aaron Stevens
2,572318
2,572318
add a comment |Â
add a comment |Â
up vote
3
down vote
You are indeed right to say that the term $fracdrdt$ is zero. Indeed the velocity vector is given just by $rfracdhatrdt$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $rhatr=r(cos(omega t)i+sin(omega t)j$.
Now we know the derivative is $rfracdhatrdt$, which is equal to the derivative of the displacement vector with $r$ as constant, you get:
$$rfracdhatrdt=$r(-sin(omega t)i+cos(omega t)j$$
Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields :
$$(-sin(omega t)cos(omega t) + sin(omega t)cos(omega t))=0$$
As was to be shown.
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
add a comment |Â
up vote
3
down vote
You are indeed right to say that the term $fracdrdt$ is zero. Indeed the velocity vector is given just by $rfracdhatrdt$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $rhatr=r(cos(omega t)i+sin(omega t)j$.
Now we know the derivative is $rfracdhatrdt$, which is equal to the derivative of the displacement vector with $r$ as constant, you get:
$$rfracdhatrdt=$r(-sin(omega t)i+cos(omega t)j$$
Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields :
$$(-sin(omega t)cos(omega t) + sin(omega t)cos(omega t))=0$$
As was to be shown.
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You are indeed right to say that the term $fracdrdt$ is zero. Indeed the velocity vector is given just by $rfracdhatrdt$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $rhatr=r(cos(omega t)i+sin(omega t)j$.
Now we know the derivative is $rfracdhatrdt$, which is equal to the derivative of the displacement vector with $r$ as constant, you get:
$$rfracdhatrdt=$r(-sin(omega t)i+cos(omega t)j$$
Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields :
$$(-sin(omega t)cos(omega t) + sin(omega t)cos(omega t))=0$$
As was to be shown.
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
You are indeed right to say that the term $fracdrdt$ is zero. Indeed the velocity vector is given just by $rfracdhatrdt$.
You can see why this is true too if you consider the dot product of the velocity vector and the radius vector. Consider the displacement or position vector to be $rhatr=r(cos(omega t)i+sin(omega t)j$.
Now we know the derivative is $rfracdhatrdt$, which is equal to the derivative of the displacement vector with $r$ as constant, you get:
$$rfracdhatrdt=$r(-sin(omega t)i+cos(omega t)j$$
Now if you take the dot product of the two vectors you should get zero as is true in circular motion, that is the velocity vector is perpendicular to the radius vector. It is indeed true as the dot product yields :
$$(-sin(omega t)cos(omega t) + sin(omega t)cos(omega t))=0$$
As was to be shown.
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
answered Aug 12 at 3:15
Tausif Hossain
2,2671518
2,2671518
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f422310%2fthe-acceleration-of-circulation-motion%23new-answer', 'question_page');
);
Post as a guest
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
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
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
