Motion equation given the following function, is this a potential function?
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I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.
I'm considering the vector field
$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$
($k$ positive constant). I've checked the condition
$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.
Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.
The other question is I have the potential function
$$
U(x,y,z) = k sqrtx^2 + y^2
$$
and I clearly have
$$
nabla U = F
$$
How do I derive the motion equation? My intuition is that since
$$
F = ma = m fracd^2xdt^2
$$
I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?
calculus physics vector-fields
add a comment |Â
up vote
0
down vote
favorite
I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.
I'm considering the vector field
$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$
($k$ positive constant). I've checked the condition
$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.
Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.
The other question is I have the potential function
$$
U(x,y,z) = k sqrtx^2 + y^2
$$
and I clearly have
$$
nabla U = F
$$
How do I derive the motion equation? My intuition is that since
$$
F = ma = m fracd^2xdt^2
$$
I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?
calculus physics vector-fields
You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.
I'm considering the vector field
$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$
($k$ positive constant). I've checked the condition
$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.
Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.
The other question is I have the potential function
$$
U(x,y,z) = k sqrtx^2 + y^2
$$
and I clearly have
$$
nabla U = F
$$
How do I derive the motion equation? My intuition is that since
$$
F = ma = m fracd^2xdt^2
$$
I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?
calculus physics vector-fields
I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.
I'm considering the vector field
$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$
($k$ positive constant). I've checked the condition
$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.
Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.
The other question is I have the potential function
$$
U(x,y,z) = k sqrtx^2 + y^2
$$
and I clearly have
$$
nabla U = F
$$
How do I derive the motion equation? My intuition is that since
$$
F = ma = m fracd^2xdt^2
$$
I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?
calculus physics vector-fields
calculus physics vector-fields
asked Sep 3 at 10:01
user8469759
1,1801515
1,1801515
You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30
add a comment |Â
You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30
You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
add a comment |Â
up vote
2
down vote
accepted
You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.
You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.
edited Sep 3 at 13:00
answered Sep 3 at 12:17
denklo
2335
2335
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
add a comment |Â
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
But is my function $U$ a valid potential?
â user8469759
Sep 3 at 13:02
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Sure, why shouldn't it be?
â denklo
Sep 3 at 13:08
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
Thank you, If I post a more complicated question do you think you can give me some insight on that one?
â user8469759
Sep 3 at 13:54
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
â denklo
Sep 3 at 14:04
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
Posted math.stackexchange.com/questions/2903927/â¦. If you have time please have a look.
â user8469759
Sep 3 at 14:33
add a comment |Â
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You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
â Botond
Sep 3 at 10:42
Your vector field is undefined at the origin, so it shouldnâÂÂt be a surprise that none of the derivatives exist there, either.
â amd
Sep 3 at 21:17
It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
â user8469759
Sep 4 at 10:30