Mixing
Show that the motion is a circular path
Question:
The position $mathbf r$ of a particle satisfies
$$ ddot {mathbf r} = fdot {mathbf r} times mathbf k$$
where $f$ is a known constant and $mathbf k$ is the unit vector in the $z$ direction.
Given that $dot {mathbf r} cdot mathbf k equiv 0$, show that the path of the particle is a circle.
Attempt:
Firstly, there are a few quantities that I know are constant:
$$frac{d}{dt}big(|dot{mathbf r}|^2 big) = 2ddot {mathbf r} cdot dot {mathbf r} = 2f(dot{mathbf r} times mathbf k) cdot dot{mathbf r} = 0 implies |dot{mathbf r}|^2 equiv text{constant}$$
$$frac{d}{dt}big(mathbf r cdot mathbf kbig) = dot{mathbf r} cdot mathbf k =0 implies {mathbf r} cdot mathbf k equiv text{constant}$$
Moreover, I noticed that
$$frac{d^2dot{mathbf r}}{dt^2} = f^2dot{mathbf r}$$
and solving this gives
$$dot{mathbf r}(t) = mathbf c_1 cos(ft)+mathbf c_2 sin (ft)$$
but I am unsure how to deduce from this that $mathbf r$ traces out a circle.
differential-equations classical-mechanics
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
add a comment |
Question:
The position $mathbf r$ of a particle satisfies
$$ ddot {mathbf r} = fdot {mathbf r} times mathbf k$$
where $f$ is a known constant and $mathbf k$ is the unit vector in the $z$ direction.
Given that $dot {mathbf r} cdot mathbf k equiv 0$, show that the path of the particle is a circle.
Attempt:
Firstly, there are a few quantities that I know are constant:
$$frac{d}{dt}big(|dot{mathbf r}|^2 big) = 2ddot {mathbf r} cdot dot {mathbf r} = 2f(dot{mathbf r} times mathbf k) cdot dot{mathbf r} = 0 implies |dot{mathbf r}|^2 equiv text{constant}$$
$$frac{d}{dt}big(mathbf r cdot mathbf kbig) = dot{mathbf r} cdot mathbf k =0 implies {mathbf r} cdot mathbf k equiv text{constant}$$
Moreover, I noticed that
$$frac{d^2dot{mathbf r}}{dt^2} = f^2dot{mathbf r}$$
and solving this gives
$$dot{mathbf r}(t) = mathbf c_1 cos(ft)+mathbf c_2 sin (ft)$$
but I am unsure how to deduce from this that $mathbf r$ traces out a circle.
differential-equations classical-mechanics
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
add a comment |
Question:
The position $mathbf r$ of a particle satisfies
$$ ddot {mathbf r} = fdot {mathbf r} times mathbf k$$
where $f$ is a known constant and $mathbf k$ is the unit vector in the $z$ direction.
Given that $dot {mathbf r} cdot mathbf k equiv 0$, show that the path of the particle is a circle.
Attempt:
Firstly, there are a few quantities that I know are constant:
$$frac{d}{dt}big(|dot{mathbf r}|^2 big) = 2ddot {mathbf r} cdot dot {mathbf r} = 2f(dot{mathbf r} times mathbf k) cdot dot{mathbf r} = 0 implies |dot{mathbf r}|^2 equiv text{constant}$$
$$frac{d}{dt}big(mathbf r cdot mathbf kbig) = dot{mathbf r} cdot mathbf k =0 implies {mathbf r} cdot mathbf k equiv text{constant}$$
Moreover, I noticed that
$$frac{d^2dot{mathbf r}}{dt^2} = f^2dot{mathbf r}$$
and solving this gives
$$dot{mathbf r}(t) = mathbf c_1 cos(ft)+mathbf c_2 sin (ft)$$
but I am unsure how to deduce from this that $mathbf r$ traces out a circle.
differential-equations classical-mechanics
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
Question:
The position $mathbf r$ of a particle satisfies
$$ ddot {mathbf r} = fdot {mathbf r} times mathbf k$$
where $f$ is a known constant and $mathbf k$ is the unit vector in the $z$ direction.
Given that $dot {mathbf r} cdot mathbf k equiv 0$, show that the path of the particle is a circle.
Attempt:
Firstly, there are a few quantities that I know are constant:
$$frac{d}{dt}big(|dot{mathbf r}|^2 big) = 2ddot {mathbf r} cdot dot {mathbf r} = 2f(dot{mathbf r} times mathbf k) cdot dot{mathbf r} = 0 implies |dot{mathbf r}|^2 equiv text{constant}$$
$$frac{d}{dt}big(mathbf r cdot mathbf kbig) = dot{mathbf r} cdot mathbf k =0 implies {mathbf r} cdot mathbf k equiv text{constant}$$
Moreover, I noticed that
$$frac{d^2dot{mathbf r}}{dt^2} = f^2dot{mathbf r}$$
and solving this gives
$$dot{mathbf r}(t) = mathbf c_1 cos(ft)+mathbf c_2 sin (ft)$$
but I am unsure how to deduce from this that $mathbf r$ traces out a circle.
differential-equations classical-mechanics
differential-equations classical-mechanics
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
asked Nov 20 '18 at 1:54
glowstonetrees
2,285318
2,285318
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Let $mathbf u = mathbf r + frac1f (dot{mathbf r} times mathbf k)$, we have
$$dot{mathbf u} = dot{mathbf r} + frac1f(ddot{mathbf r} times mathbf k)
= dot{mathbf r} + ((dot{mathbf r} times mathbf k) times mathbf k)
= dot{mathbf r} + ((dot{mathbf r}cdot mathbf k) mathbf k - dot{mathbf r}) = mathbf 0$$
This means $mathbf u$ is a constant. Notice
$$frac{d}{dt}|mathbf r - mathbf u|^2
= 2(mathbf r-mathbf u)cdot dot{mathbf r} = -frac{2}{f} (dot{mathbf r} times mathbf k)cdot dot{mathbf r} = mathbf 0$$
The expression $|mathbf r-mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $mathbf u$ with radius $R$. Together with what you know $mathbf r cdot mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
add a comment |
Insert the solution you found of the derived equations into the original equations.
$$
-fvec c_1sin(ft)+fvec c_2cos(ft)=f(vec c_1×vec k)cos(ft)+f(vec c_2×vec k)sin(ft).
$$
Comparing coefficients,
$$
vec c_2=~~vec c_1×vec k\
vec c_1=-vec c_2×vec k\
$$
is only possible if both vectors are orthogonal to $vec k$, of the same length and orthogonal to each other. With $vec k=(0,0,1)$ you get $vec c_1=(a,b,0)$ and $vec c_2=(b,-a,0)$. This means that $dot{vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $vec r$ circulates.
answered Nov 20 '18 at 10:11
LutzL
56k42054
add a comment |
One way is to show that the curvature is constant. Recall that
$$kappa = frac{||mathbf{v}(t) times mathbf{a}(t)||}{||mathbf{v}(t)||^3} = frac{||dot{mathbf{r}}(t) times ddot{mathbf{r}}(t)||}{||dot{mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $dot{mathbf{r}}(t)$.
It might help to write:
begin{align}
dot{mathbf{r}}(t) &= mathbf{c_1}cos(ft) + mathbf{c_2}sin(ft)\
&= left[begin{array}{c}c_{11}\c_{12}\c_{13}end{array}right]cos(ft) + left[begin{array}{c}c_{21}\c_{22}\c_{23}end{array}right]sin(ft)\
&=(c_1cos(ft) + c_2sin(ft))hat{mathbf{i}} + (c_3cos(ft) + c_4sin(ft))hat{mathbf{j}} + (c_5cos(ft) + c_6sin(ft))hat{mathbf{k}}
end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $ddot{mathbf{r}}(t) = fdot{mathbf{r}}(t) times mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $mathbf u = mathbf r + frac1f (dot{mathbf r} times mathbf k)$, we have
$$dot{mathbf u} = dot{mathbf r} + frac1f(ddot{mathbf r} times mathbf k)
= dot{mathbf r} + ((dot{mathbf r} times mathbf k) times mathbf k)
= dot{mathbf r} + ((dot{mathbf r}cdot mathbf k) mathbf k - dot{mathbf r}) = mathbf 0$$
This means $mathbf u$ is a constant. Notice
$$frac{d}{dt}|mathbf r - mathbf u|^2
= 2(mathbf r-mathbf u)cdot dot{mathbf r} = -frac{2}{f} (dot{mathbf r} times mathbf k)cdot dot{mathbf r} = mathbf 0$$
The expression $|mathbf r-mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $mathbf u$ with radius $R$. Together with what you know $mathbf r cdot mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
add a comment |
Let $mathbf u = mathbf r + frac1f (dot{mathbf r} times mathbf k)$, we have
$$dot{mathbf u} = dot{mathbf r} + frac1f(ddot{mathbf r} times mathbf k)
= dot{mathbf r} + ((dot{mathbf r} times mathbf k) times mathbf k)
= dot{mathbf r} + ((dot{mathbf r}cdot mathbf k) mathbf k - dot{mathbf r}) = mathbf 0$$
This means $mathbf u$ is a constant. Notice
$$frac{d}{dt}|mathbf r - mathbf u|^2
= 2(mathbf r-mathbf u)cdot dot{mathbf r} = -frac{2}{f} (dot{mathbf r} times mathbf k)cdot dot{mathbf r} = mathbf 0$$
The expression $|mathbf r-mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $mathbf u$ with radius $R$. Together with what you know $mathbf r cdot mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
add a comment |
Let $mathbf u = mathbf r + frac1f (dot{mathbf r} times mathbf k)$, we have
$$dot{mathbf u} = dot{mathbf r} + frac1f(ddot{mathbf r} times mathbf k)
= dot{mathbf r} + ((dot{mathbf r} times mathbf k) times mathbf k)
= dot{mathbf r} + ((dot{mathbf r}cdot mathbf k) mathbf k - dot{mathbf r}) = mathbf 0$$
This means $mathbf u$ is a constant. Notice
$$frac{d}{dt}|mathbf r - mathbf u|^2
= 2(mathbf r-mathbf u)cdot dot{mathbf r} = -frac{2}{f} (dot{mathbf r} times mathbf k)cdot dot{mathbf r} = mathbf 0$$
The expression $|mathbf r-mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $mathbf u$ with radius $R$. Together with what you know $mathbf r cdot mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
Let $mathbf u = mathbf r + frac1f (dot{mathbf r} times mathbf k)$, we have
$$dot{mathbf u} = dot{mathbf r} + frac1f(ddot{mathbf r} times mathbf k)
= dot{mathbf r} + ((dot{mathbf r} times mathbf k) times mathbf k)
= dot{mathbf r} + ((dot{mathbf r}cdot mathbf k) mathbf k - dot{mathbf r}) = mathbf 0$$
This means $mathbf u$ is a constant. Notice
$$frac{d}{dt}|mathbf r - mathbf u|^2
= 2(mathbf r-mathbf u)cdot dot{mathbf r} = -frac{2}{f} (dot{mathbf r} times mathbf k)cdot dot{mathbf r} = mathbf 0$$
The expression $|mathbf r-mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $mathbf u$ with radius $R$. Together with what you know $mathbf r cdot mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
edited Nov 20 '18 at 13:19
glowstonetrees
2,285318
2,285318
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
answered Nov 20 '18 at 2:44
achille hui
95.4k5130256
95.4k5130256
add a comment |
add a comment |
Insert the solution you found of the derived equations into the original equations.
$$
-fvec c_1sin(ft)+fvec c_2cos(ft)=f(vec c_1×vec k)cos(ft)+f(vec c_2×vec k)sin(ft).
$$
Comparing coefficients,
$$
vec c_2=~~vec c_1×vec k\
vec c_1=-vec c_2×vec k\
$$
is only possible if both vectors are orthogonal to $vec k$, of the same length and orthogonal to each other. With $vec k=(0,0,1)$ you get $vec c_1=(a,b,0)$ and $vec c_2=(b,-a,0)$. This means that $dot{vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $vec r$ circulates.
answered Nov 20 '18 at 10:11
LutzL
56k42054
add a comment |
Insert the solution you found of the derived equations into the original equations.
$$
-fvec c_1sin(ft)+fvec c_2cos(ft)=f(vec c_1×vec k)cos(ft)+f(vec c_2×vec k)sin(ft).
$$
Comparing coefficients,
$$
vec c_2=~~vec c_1×vec k\
vec c_1=-vec c_2×vec k\
$$
is only possible if both vectors are orthogonal to $vec k$, of the same length and orthogonal to each other. With $vec k=(0,0,1)$ you get $vec c_1=(a,b,0)$ and $vec c_2=(b,-a,0)$. This means that $dot{vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $vec r$ circulates.
answered Nov 20 '18 at 10:11
LutzL
56k42054
add a comment |
Insert the solution you found of the derived equations into the original equations.
$$
-fvec c_1sin(ft)+fvec c_2cos(ft)=f(vec c_1×vec k)cos(ft)+f(vec c_2×vec k)sin(ft).
$$
Comparing coefficients,
$$
vec c_2=~~vec c_1×vec k\
vec c_1=-vec c_2×vec k\
$$
is only possible if both vectors are orthogonal to $vec k$, of the same length and orthogonal to each other. With $vec k=(0,0,1)$ you get $vec c_1=(a,b,0)$ and $vec c_2=(b,-a,0)$. This means that $dot{vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $vec r$ circulates.
answered Nov 20 '18 at 10:11
LutzL
56k42054
Insert the solution you found of the derived equations into the original equations.
$$
-fvec c_1sin(ft)+fvec c_2cos(ft)=f(vec c_1×vec k)cos(ft)+f(vec c_2×vec k)sin(ft).
$$
Comparing coefficients,
$$
vec c_2=~~vec c_1×vec k\
vec c_1=-vec c_2×vec k\
$$
is only possible if both vectors are orthogonal to $vec k$, of the same length and orthogonal to each other. With $vec k=(0,0,1)$ you get $vec c_1=(a,b,0)$ and $vec c_2=(b,-a,0)$. This means that $dot{vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $vec r$ circulates.
answered Nov 20 '18 at 10:11
LutzL
56k42054
answered Nov 20 '18 at 10:11
LutzL
56k42054
answered Nov 20 '18 at 10:11
LutzL
56k42054
answered Nov 20 '18 at 10:11
LutzL
56k42054
56k42054
add a comment |
add a comment |
One way is to show that the curvature is constant. Recall that
$$kappa = frac{||mathbf{v}(t) times mathbf{a}(t)||}{||mathbf{v}(t)||^3} = frac{||dot{mathbf{r}}(t) times ddot{mathbf{r}}(t)||}{||dot{mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $dot{mathbf{r}}(t)$.
It might help to write:
begin{align}
dot{mathbf{r}}(t) &= mathbf{c_1}cos(ft) + mathbf{c_2}sin(ft)\
&= left[begin{array}{c}c_{11}\c_{12}\c_{13}end{array}right]cos(ft) + left[begin{array}{c}c_{21}\c_{22}\c_{23}end{array}right]sin(ft)\
&=(c_1cos(ft) + c_2sin(ft))hat{mathbf{i}} + (c_3cos(ft) + c_4sin(ft))hat{mathbf{j}} + (c_5cos(ft) + c_6sin(ft))hat{mathbf{k}}
end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $ddot{mathbf{r}}(t) = fdot{mathbf{r}}(t) times mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
add a comment |
One way is to show that the curvature is constant. Recall that
$$kappa = frac{||mathbf{v}(t) times mathbf{a}(t)||}{||mathbf{v}(t)||^3} = frac{||dot{mathbf{r}}(t) times ddot{mathbf{r}}(t)||}{||dot{mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $dot{mathbf{r}}(t)$.
It might help to write:
begin{align}
dot{mathbf{r}}(t) &= mathbf{c_1}cos(ft) + mathbf{c_2}sin(ft)\
&= left[begin{array}{c}c_{11}\c_{12}\c_{13}end{array}right]cos(ft) + left[begin{array}{c}c_{21}\c_{22}\c_{23}end{array}right]sin(ft)\
&=(c_1cos(ft) + c_2sin(ft))hat{mathbf{i}} + (c_3cos(ft) + c_4sin(ft))hat{mathbf{j}} + (c_5cos(ft) + c_6sin(ft))hat{mathbf{k}}
end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $ddot{mathbf{r}}(t) = fdot{mathbf{r}}(t) times mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
add a comment |
One way is to show that the curvature is constant. Recall that
$$kappa = frac{||mathbf{v}(t) times mathbf{a}(t)||}{||mathbf{v}(t)||^3} = frac{||dot{mathbf{r}}(t) times ddot{mathbf{r}}(t)||}{||dot{mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $dot{mathbf{r}}(t)$.
It might help to write:
begin{align}
dot{mathbf{r}}(t) &= mathbf{c_1}cos(ft) + mathbf{c_2}sin(ft)\
&= left[begin{array}{c}c_{11}\c_{12}\c_{13}end{array}right]cos(ft) + left[begin{array}{c}c_{21}\c_{22}\c_{23}end{array}right]sin(ft)\
&=(c_1cos(ft) + c_2sin(ft))hat{mathbf{i}} + (c_3cos(ft) + c_4sin(ft))hat{mathbf{j}} + (c_5cos(ft) + c_6sin(ft))hat{mathbf{k}}
end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $ddot{mathbf{r}}(t) = fdot{mathbf{r}}(t) times mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
One way is to show that the curvature is constant. Recall that
$$kappa = frac{||mathbf{v}(t) times mathbf{a}(t)||}{||mathbf{v}(t)||^3} = frac{||dot{mathbf{r}}(t) times ddot{mathbf{r}}(t)||}{||dot{mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $dot{mathbf{r}}(t)$.
It might help to write:
begin{align}
dot{mathbf{r}}(t) &= mathbf{c_1}cos(ft) + mathbf{c_2}sin(ft)\
&= left[begin{array}{c}c_{11}\c_{12}\c_{13}end{array}right]cos(ft) + left[begin{array}{c}c_{21}\c_{22}\c_{23}end{array}right]sin(ft)\
&=(c_1cos(ft) + c_2sin(ft))hat{mathbf{i}} + (c_3cos(ft) + c_4sin(ft))hat{mathbf{j}} + (c_5cos(ft) + c_6sin(ft))hat{mathbf{k}}
end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $ddot{mathbf{r}}(t) = fdot{mathbf{r}}(t) times mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
edited Nov 20 '18 at 2:16
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
answered Nov 20 '18 at 2:10
AlkaKadri
1,459411
1,459411
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
add a comment |
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $mathbf r$ traces out a circle?
– glowstonetrees
Nov 20 '18 at 2:14
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
@glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/…
– AlkaKadri
Nov 20 '18 at 2:28
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