MIT16_07F09_Lec05

MIT16_07F09_Lec05 - S Widnall J Peraire 16.07 Dynamics Fall...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
S. Widnall, J. Peraire 16.07 Dynamics Fall 2008 Version 2.0 Lecture L5 - Other Coordinate Systems In this lecture, we will look at some other common systems of coordinates. We will present polar coordinates in two dimensions and cylindrical and spherical coordinates in three dimensions. We shall see that these systems are particularly useful for certain classes of problems. Polar Coordinates ( r θ ) In polar coordinates, the position of a particle A , is determined by the value of the radial distance to the origin, r , and the angle that the radial line makes with an arbitrary fixed line, such as the x axis. Thus, the trajectory of a particle will be determined if we know r and θ as a function of t , i.e. r ( t ) , θ ( t ). The directions of increasing r and θ are defined by the orthogonal unit vectors e r and e θ . The position vector of a particle has a magnitude equal to the radial distance, and a direction determined by e r . Thus, r = r e r . (1) Since the vectors e r and e θ are clearly different from point to point, their variation will have to be considered when calculating the velocity and acceleration. Over an infinitesimal interval of time dt , the coordinates of point A will change from ( r, θ ), to ( r + dr , θ + ) as shown in the diagram. 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
We note that the vectors e r and e θ do not change when the coordinate r changes. Thus, d e r /dr = 0 and d e θ /dr = 0 . On the other hand, when θ changes to θ + , the vectors e r and e θ are rotated by an angle . From the diagram, we see that d e r = e θ , and that d e θ = e r . This is because their magnitudes in the limit are equal to the unit vector as radius times in radians. Dividing through by , we have, d e r d e θ = e θ , and = e r . Multiplying these expressions by dθ/dt θ ˙ , we obtain, d e r d e r d e θ dt dt = θ ˙ e θ , and dt = θ ˙ e r . (2) Note Alternative calculation of the unit vector derivatives An alternative, more mathematical, approach to obtaining the derivatives of the unit vectors is to express e r and e θ in terms of their cartesian components along i and j . We have that e r = cos θ i + sin θ j e θ = sin θ i + cos θ j . Therefore, when we differentiate we obtain, d e r d e r = 0 , = sin θ i + cos θ j e θ dr d e θ d e θ = 0 , = cos θ i sin θ j ≡ − e r . dr Velocity vector We can now derive expression (1) with respect to time and write v = r ˙ = r ˙ e r + r e ˙ r , or, using expression (2), we have v = r ˙ e r + ˙ e θ . (3) Here, v r = r ˙ is the radial velocity component, and v θ = ˙ is the circumferential velocity component. We also have that v = v r 2 + v θ 2 . The radial component is the rate at which r changes magnitude, or stretches, and the circumferential component, is the rate at which r changes direction, or swings.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern