- #1
Sergey S
- 11
- 0
I think this is a textbook-style question, if I am wrong, please redirect me to the forum section where I should have posted this. This is my first time here, so I am sorry if I am messing it up.
We have an n-dimensional vector [itex]\vec{r}[/itex] with a constant length [itex]\|\vec{r}\|=1[/itex]. The vector rotates with an angular speed [itex]\vec{ω}[/itex], which is also an n-dimensional vector and is a given function of time. I want to find first derivative of [itex]\vec{r}[/itex] with respect to time: [itex]\frac{d\vec{r}}{dt}-?[/itex]
2. Homework Equations
I don't really know how to tackle the problem. I think I've solved this for the 2-dimensional case, where [itex]\vec{r}\in\mathbb R^2[/itex] rotates with angular speed [itex]\dot{Θ}[/itex] (so [itex]Θ[/itex] would denote angular displacement from the original orientation). I doubt that it will be of any help with the general n-dimensional case, but here it is:
We can rewrite [itex]\vec{r}[/itex] as a product of a constant unit vector [itex]\vec{a}=const[/itex] and a rotational matrix [itex]T_Θ[/itex]:
[itex]\vec{r}=T_Θ\vec{a}[/itex], where [itex]T_Θ=\begin{pmatrix}
\cosΘ & -\sinΘ\\
\sinΘ & \cosΘ\\
\end{pmatrix}[/itex] (1)
So time derivative of [itex]\vec{r}[/itex] can be written as:
[itex]\frac{d\vec{r}}{dt}=\frac{d(T_Θ)}{dt}\vec{a}[/itex] (2)
Then we do math:
[itex]\frac{d(T_Θ)}{dt}=\dot{Θ}\begin{pmatrix}
-\sinΘ& -\cosΘ\\
\cosΘ& -\sinΘ\\
\end{pmatrix}=
(-1)\dot{Θ}\begin{pmatrix}
\cos(\frac{\pi}{2}-Θ)& \sin(\frac{\pi}{2}-Θ)\\
-\sin(\frac{\pi}{2}-Θ)& \cos(\frac{\pi}{2}-Θ)\\
\end{pmatrix}=
(-1)\dot{Θ}\begin{pmatrix}
\cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
\sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
\end{pmatrix}
[/itex]
Matrix [itex]\begin{pmatrix}
\cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
\sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
\end{pmatrix}
[/itex] is a rotation matrix that rotates a vector through the angle [itex]Θ-\frac{\pi}{2}[/itex]. You can do the same rotation with two matrices, first of which would rotate the vector through the angle Θ (it is [itex]T_Θ[/itex]), and second through the angle [itex]-\frac{\pi}{2}[/itex] (we'll name the later matrix[itex]T_0[/itex]).
[itex]T_0=\begin{pmatrix}
0& 1\\
-1& 0\\
\end{pmatrix}[/itex]
So we rewrite equation (2) as:
[itex]\frac{d\vec{r}}{dt}=(-1)\dot{Θ}T_ΘT_0\vec{a}[/itex]
Rotating a vector through the angle [itex](-\frac{\pi}{2})[/itex] and then multiplying it with (-1) is the same as rotating through the angle [itex]\frac{\pi}{2}[/itex], so:
[itex]\frac{d\vec{r}}{dt}=\dot{Θ}T_ΘT_1\vec{a}[/itex], where [itex]T_1=\begin{pmatrix}
0& -1\\
1& 0\\
\end{pmatrix}[/itex]
So using (1) we can write:
[itex]\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}
0& -1\\
1& 0\\
\end{pmatrix}\vec{r}[/itex]
Sorry for the long post. I guess what I've shown above might be the clumsiest way to deal with the problem. And it is hardly of any help with the n-dimensional case.
Homework Statement
We have an n-dimensional vector [itex]\vec{r}[/itex] with a constant length [itex]\|\vec{r}\|=1[/itex]. The vector rotates with an angular speed [itex]\vec{ω}[/itex], which is also an n-dimensional vector and is a given function of time. I want to find first derivative of [itex]\vec{r}[/itex] with respect to time: [itex]\frac{d\vec{r}}{dt}-?[/itex]
2. Homework Equations
The Attempt at a Solution
I don't really know how to tackle the problem. I think I've solved this for the 2-dimensional case, where [itex]\vec{r}\in\mathbb R^2[/itex] rotates with angular speed [itex]\dot{Θ}[/itex] (so [itex]Θ[/itex] would denote angular displacement from the original orientation). I doubt that it will be of any help with the general n-dimensional case, but here it is:
We can rewrite [itex]\vec{r}[/itex] as a product of a constant unit vector [itex]\vec{a}=const[/itex] and a rotational matrix [itex]T_Θ[/itex]:
[itex]\vec{r}=T_Θ\vec{a}[/itex], where [itex]T_Θ=\begin{pmatrix}
\cosΘ & -\sinΘ\\
\sinΘ & \cosΘ\\
\end{pmatrix}[/itex] (1)
So time derivative of [itex]\vec{r}[/itex] can be written as:
[itex]\frac{d\vec{r}}{dt}=\frac{d(T_Θ)}{dt}\vec{a}[/itex] (2)
Then we do math:
[itex]\frac{d(T_Θ)}{dt}=\dot{Θ}\begin{pmatrix}
-\sinΘ& -\cosΘ\\
\cosΘ& -\sinΘ\\
\end{pmatrix}=
(-1)\dot{Θ}\begin{pmatrix}
\cos(\frac{\pi}{2}-Θ)& \sin(\frac{\pi}{2}-Θ)\\
-\sin(\frac{\pi}{2}-Θ)& \cos(\frac{\pi}{2}-Θ)\\
\end{pmatrix}=
(-1)\dot{Θ}\begin{pmatrix}
\cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
\sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
\end{pmatrix}
[/itex]
Matrix [itex]\begin{pmatrix}
\cos(Θ-\frac{\pi}{2})& -\sin(Θ-\frac{\pi}{2})\\
\sin(Θ-\frac{\pi}{2})& \cos(Θ-\frac{\pi}{2})\\
\end{pmatrix}
[/itex] is a rotation matrix that rotates a vector through the angle [itex]Θ-\frac{\pi}{2}[/itex]. You can do the same rotation with two matrices, first of which would rotate the vector through the angle Θ (it is [itex]T_Θ[/itex]), and second through the angle [itex]-\frac{\pi}{2}[/itex] (we'll name the later matrix[itex]T_0[/itex]).
[itex]T_0=\begin{pmatrix}
0& 1\\
-1& 0\\
\end{pmatrix}[/itex]
So we rewrite equation (2) as:
[itex]\frac{d\vec{r}}{dt}=(-1)\dot{Θ}T_ΘT_0\vec{a}[/itex]
Rotating a vector through the angle [itex](-\frac{\pi}{2})[/itex] and then multiplying it with (-1) is the same as rotating through the angle [itex]\frac{\pi}{2}[/itex], so:
[itex]\frac{d\vec{r}}{dt}=\dot{Θ}T_ΘT_1\vec{a}[/itex], where [itex]T_1=\begin{pmatrix}
0& -1\\
1& 0\\
\end{pmatrix}[/itex]
So using (1) we can write:
[itex]\frac{d\vec{r}}{dt}=\dot{Θ}\begin{pmatrix}
0& -1\\
1& 0\\
\end{pmatrix}\vec{r}[/itex]
Sorry for the long post. I guess what I've shown above might be the clumsiest way to deal with the problem. And it is hardly of any help with the n-dimensional case.
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