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Dynamics-Coordinate Motion

  1. Feb 1, 2007 #1
    Since Im taking dynamics right now, I thought I would write a short thread on coordinate motion that goes into more depth than most second year dynamics courses which tend to gloss things over.

    The only thing you need you follow along is basic dynamics, calculus III, and minor familiarity with matrix theory (Most of it is just stuff from the first few weeks).

    Before I get into coordinate motion, I will breifly review the Dot Product, Cross Product, and the Euclidean Norm, which will be used in the analysis of coordinate motion.

    Note: I might be a little bit slow in posting, but I'll try to do it as fast as I can. (And Im trying to relearn latex.

    Hopefully if your taking dynamics this will expose to a little more depth of the subject and make coordinate motion more clear and rigorous.
     
    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 1, 2007 #2
    The Dot Product *Review*

    [​IMG]
    Above is a graphical representation of the dot product.


    The dot product is a scalar product of two vectors and is denoted as:

    [tex]\vec{P} \cdot \vec{Q} = ||\vec{P}||_{2} ||\vec{Q}||_2 cos( \theta)[/tex]

    In component form:

    [tex] \vec{P} \cdot \vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z [/tex]

    In matrix form:

    [tex] \vec{P} \cdot \vec{Q} =[ P_x P_y P_z] \left[\begin{array}{c}Q_x\\Q_y\\Q_z\end{array}\right] [/tex]

    Important Points From this Review
    • Any vector dotted with itself is the magnitude of said vector
    • Any vector dotted with an orthogonal vector is zero
    • The Matlab command for the dot product is >>V=dot(P,Q)

    Side
    If you're woundering about the [tex]||_2 [/tex] notation, I will explain it shortly.
     
    Last edited: Feb 2, 2007
  4. Feb 1, 2007 #3
    The Cross Product *Review*

    [​IMG]
    Above is a graphical representation of the cross product.

    The cross product is a vector product of two vectors defined as:

    [tex] \vec{V}=\vec{P} \times \vec{Q} [/tex]

    or:

    [tex] ||\vec{V}||_2=\vec{P} \times \vec{Q} [/tex]


    or:

    [tex] \vec{V}=\vec{P} \times \vec{Q} = det \left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\P_x&P_y&P_z\\Q_x&Q_y&Q_z\end{array}\right| [/tex]

    In matrix form:

    [tex] \left\{\begin{array}{c}V_x\\V_y\\V_z\end{array}\right\} = [\tilde{P}]\{Q\} = \left[\begin{array}{ccc}0&-P_z&P_y\\P_z&0&-P_x\\-P_y&P_x&0\end{array}\right] \left\{\begin{array}{c}Q_x\\Q_y\\Q_z \end{array}\right\} [/tex]

    Important points from this review:
    • The vector produced is orthogonal to both vectors P & Q
    • The cross product is not commutative
    • The Matlab code for the cross product of two vectors is: >>V=cross(P,Q)
     
    Last edited by a moderator: Feb 3, 2007
  5. Feb 1, 2007 #4
    The Vector Magnitude *Review*

    Earlier, I used the notation [tex] ||_2 [/tex] when I wrote the dot and cross product. This notation denotes the Euclidean Norm of a vector, but what most books would call the magnitude. This is just another notation, but I included with the hopes of learning something new.

    The Euclidean Norm (or magnitude), is given by:

    [tex] ||a||_2 = \sqrt{a_1^2 + a_2^2 + a_3^2} [/tex]

    To be more general, if x is an n-dimensional vector:

    [tex] \vect{x}=(x_1, x_2, ... ,x_n) [/tex]

    Then the p-norm of x is given by:

    [tex] ||x||_p = ( \sum^{n}_{i=1} |X_i|^p)^{1/p} = (|x_1|^p + |x_2|^p+...+|x_n|^p)^{1/p} [/tex]

    When p=2, we get the Eulidean Norm (Magnitude).

    Important Points From this Review
    • Introduction to a more formal definition of the "Magnitude of a Vector"
    • The existance of a p=1 norm
    • The existance of a p=[tex]\infty[/tex] norm
    • The only norm used will be the p=2 norm, so don't worry about other norms: it was just to show that they exist.
    • The matlab code for the norm is: >>norm(a,p) [by defeault, p=2]

    side
    Write out the Norm using p=2, you will find that you get the equation of magnitude you are accustomed to!
     
    Last edited: Feb 2, 2007
  6. Feb 2, 2007 #5
    The Coordinate System *Background*

    Depending on the complexity of the problem, there are several different coordinate systems available. Picking the right coordinate system makes calculations easier; however, no matter which coordinate system you choose you should still arive at the same answer.

    There are 4 primary coordinate systems that are used in dynamics:

    • Cartesian
    • Curvilinear
    • Cylindrical
    • Spherical

    An important fact about any coordinate system is that all three base vectors, are orthonomral to eachother: they are orthogonal (mutually perpendicular), and have a Euclidean Norm (Magnitude) of 1. Each of the unit vectors lies in a principal direction. The three vectors together form a mutually orthogonal triad. (for now, think of the three base vectors as being in the x,y,z direction-a cartesian coordinate system).

    Typically, the unit vector is denoted with a lower case e and a subscript indicating the name of the direction. *For example, [tex]\vec{e}_x[/tex] is a unit vector in the x-direction.

    A formal way of writing this is to say:

    [tex]\vec{e_i} \times \vec{e_j} = \varepsilon_{ijk} \vec{e_k} [/tex]

    where:

    [tex]\varepsilon_{ijk} =1[/tex] if i,j,k are in order (i=1,j=2,k=3) or (i=2,j=3,k=1)...etc.

    [tex]\varepsilon_{ijk} =-1[/tex] if i,j,k are not in order (i=1,j=3,k=2) or (i=2,j=1,k=3)...etc.

    and

    [tex]\varepsilon_{ijk} =0[/tex] if any two indices are repeated.

    This formal definition is for a right handed coordinate system, and is simply a consequence of the cross product mentioned earlier. -The cross product of a vector to itself is zero, and the cross product of two orthogonal vectors is a third vector orthogonal to both; however, it is not commutative and the sign will change depending on the order of i,j,k. (This is why you see the 1 or -1 depending on the order.)

    side
    *The unit vector in the x-direction is not given the notation [tex]\vec{e}_x[/tex], but for now its just to drive the point across of what the notation means.
     
    Last edited: Feb 2, 2007
  7. Feb 2, 2007 #6
    Cartesian Coordinates

    Cartesian, or Rectilinear Coordinates, is the one most people are accustomed, shown here schematically.

    The three directions are the x,y,z axis. The unit vectors in these directions are given by the vectors [tex]\vec{i}[/tex],[tex]\vec{j}[/tex],[tex]\vec{k}[/tex], respectively. *Notice in the cartesian coordinate system the e-subscript for unit vectors is not used, i.e. you dont see [tex]\vec{e_x}[/tex],[tex]\vec{e_y}[/tex],[tex]\vec{e_z}[/tex].

    This coordinate system is fixed and is inertial, meaning the coordinate system does not accelerate. It remains stationary or moves with constant velocity. For this reason, the unit vectors that describe the orthonomal directions are linearly time invariant (LTI) - they dont change with time.

    The position of a particle P, can be described by a position vector from the origin of the coordinate system to the particle. We denote this vector as:

    [tex]\vec{r}(t) = x \vec{i}+y\vec{j}+ z \vec{k} [/tex]

    where x,y,z are functions of time

    Recalling the definition of the Euclidean Norm earlier, the distance from r to the origin O becomes:

    [tex] ||\vec{r}||_2 = \sqrt{ \vec{r}(t) \cdot \vec{r}(t)} = \sqrt{x^2+y^2+z^2}[/tex]

    During a small incremental time step, [tex]t+ \Delta t[/tex], the position changes by an amount [tex]\vec{r}(t+ \Delta t) [/tex], as seen here. From this, the velocity of the particle is given as:

    [tex] \vec{v}(t) = \frac{d \vec{r}(t)}{dt} = \dot{ \vec{r}} (t) = lim_{\Delta t \rightarrow 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t} [/tex]

    Using the chain rule, this gives:

    [tex]\vec{v}(t) = (\dot{x}\vec{i} + {\color{red}{x\dot{\vec{i}}}}) +(\dot{y}\vec{j}+ {\color{red}y\dot{\vec{j}}} )+(\dot{z}\vec{k} +{\color{red}z\dot{\vec{k}}}) = \dot{x}\vec{i} + \dot{y}\vec{j}+\dot{z}\vec{k} = v_x\vec{i} +v_y\vec{j}+v_z\vec{k}[/tex]

    Notice above that the derivatives of the unit vectors vanish because they are time invariant - this is a consequence of being a fixed frame of reference.

    The acceleration can be found in a simliar fashion:

    [tex]\vec{a}(t) = \frac{d^2 \vec{r}} {dt^2} =\ddot{x}\vec{i} + \ddot{y}\vec{j}+\ddot{z}\vec{k} = a_x\vec{i} +a_y\vec{j}+a_z\vec{k}[/tex]

    side
    The formal forumla for the velocity and acceleration vectors were given; however, no mention was made to their directions. I will explain their directions in the next section on curvilinear coordinates.
     
    Last edited: Feb 3, 2007
  8. Feb 3, 2007 #7
    The Cartesian Coordinate System *Addendum*

    At this point, it is better to talk about the directions of the velocity and acceleration vectors before moving on to curvilinear motion.

    ___Velocity____________
    Here is a good graphic that describes the direction of the velocity vector.

    Consider the vector [tex]\vec{r_i}[/tex]. Denote [tex]i=t[/tex], the initial time and [tex]\vec{r_f}[/tex] the vector at some later time [tex]f=t+6\Delta t [/tex] (it will become clear why I picked [tex]6\Delta t [/tex] shortly).

    Now, lets decrease the time step, one step at a time, until we come infinitely close to [tex]\vec{r_t}[/tex]. This is the result. When this occurs, the vectors [tex]\vec{r_i}[/tex] and [tex]\vec{r_f}[/tex] become almost indistinguishable. (Notice in this picture that as [tex]\Delta t[/tex] decreases [6-times total] the vector [tex]\vec{r_f}[/tex] moves to the left and becomes darker in color.)

    Additionally, the vector [tex]\Delta \vec{r} [/tex] becomes tangent to the red curve. The red curve is the path that the particle follows; therefore, the velocity vector lies tangent to the objects direction of motion. This is simply a graphical representation of what it means to take the limit of:

    [tex] \vec{v}(t) = \frac{d \vec{r}(t)}{dt} = \dot{ \vec{r}} (t) = lim_{\Delta t \rightarrow 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t} [/tex]

    ___Acceleration____________
    As mentioned, the acceleration is the limit of the time rate of change of the velocity, or the second derivative of the time rate of change of the position vector:

    [tex]\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2 \vec{r}}{dt^2} [/tex]

    To 'see' what's going on with the acceleration vector, we create whats called a hodograph. A hodograph describes the locus of points for the velocity vector centered about a common origin.

    At each instant in time, the particle will have a velocity vector that is tangent to the path. This is shown by the top curve, or "trajectory" in this diagram.

    If we were to take each of these velocity vectors along the trajectory and give them a common origin, O, we would get a graph similar to the one on the bottom of the figure. There would be a path traced out through the collection of the tip of each velocity vector.

    Consider the velocity vector on the bottom figure of the hodograph labeled v. This can be any velocity vector at any time t. The velocity vector to its right can be another velocity vector at a later time [tex]t+\Delta t[/tex]. Just as in the case of the position vector, if we take the limit as [tex]\Delta t[/tex] goes to zero, we find that the acceleration vector becomes tangent to the velocity vector on the hodograph.

    Caution
    The acceleration vector is tangent to the hodograph and is generally not tangent to the path of motion. Why? The acceleration vector is directly affected by the change in the both the magnitude and direction of the velocity vector. If only the magnitude of the velocity vector were changing, then the acceleration would be tangent to the path of motion as well as to the hodograph; however, whenever the velocity vector changes direction it tends to swing the velocity vector towards the inside, or concave side of the path. As a result, the acceleration cannot remain tangent to the path.


    *Note*
    When we take the limit as [tex]\Delta t \rightarrow 0[/tex] we call this the instantaneous velocity or the instantaneous acceleration. When we have a discretization of the data such that the velocity or accleration is found by using a finite difference, we call that the average velocity or the average acceleration over the time interval [tex]\Delta t[/tex]. Here is an example:

    -Instantaneous velocity:
    [tex]\vec{v}(t) = lim_{\Delta t \rightarrow 0} \frac{\vec{r}(t+ \Delta t) -\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt} [/tex]

    -Discritized average velocity:
    [tex]\vec{v}(t)_{avg} = \frac{\vec{r}(t+ \Delta t) -\vec{r}(t)}{\Delta t}=\frac{\Delta \vec{r}}{\Delta t}[/tex]

    Although the average velocity and acceleration are not as accurate as the instantaneous velocity and acceleration, dont underestimate its imporantance. In real world systems, the velocity and acceleration are measured by analog devices and converted into a digitial signal which is sent to the computer. An analog device outputs a continuous signal; however, mircoprocessors work in binary (1's and 0's). Therefore, the analog signal must first be converted into a digital signal the processor can use. This is done by an analog-to-digital converter (A/D). During this process, the continuous analog signal becomes discritized (or quantized) to finite time step on the x-axis and a finite amplitude zone on the y-axis. How small the time step is depends on the quality of the sampling rate of the A/D. The result is that the computer has to calculate the average velocity and acceleration of real world systems and use it as an approximation of the instantaneous velocity or acceleration.

    Here is an example of how an analog signal (the gray curve) becomes converted into a digital signal (the red curve). The dashed lines represent the discretization of time on the x-axis, and the voltage on the y-axis. During any time interval [tex]\Delta t[/tex], the red curve approximates the gray curve as being constant over said time span.
     
    Last edited: Feb 4, 2007
  9. Feb 9, 2008 #8
    It's a pity I didn't have all these nice tips at my first year uni.
     
  10. Apr 20, 2008 #9
  11. Apr 21, 2008 #10
    Nice, thanks - just saved me from posting a question.
    The 'search' function works
     
  12. Oct 7, 2008 #11

    Pyrrhus

    User Avatar
    Homework Helper

    Cyrus, maybe you should move the Tutorial with the other tutorials?
     
  13. Feb 17, 2009 #12
    does such a thread concern what i'm asking about or is it just that u want me to see what ur doing , ur not helping man , just say u don't have a clue about the case i asked u about and i'll be fine with it , thanks anyway
     
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