Polar Coordinates for Velocity and Acceleration of a ParticleDate: 03/11/97 at 16:59:12 From: Nick Manville Subject: Use of polar coordinates for velocity and acceleration of a particle. In England this is a new topic on our A-level syllabus (our final-year exams) and our teacher really doesn't have a clue. Do you know of any internet resources that might be able to enlighten me on the subject? I know that velocity components are dr/dt and rdo/dt and that components of acceleration are d^2r/dt^2 - r(do/dt)^2, etc. I would appreciate any help at all in understanding how these are generated and/or how to use them. Many thanks, Nick Manville Date: 03/12/97 at 19:38:13 From: Doctor Charles Subject: Re: Use of polar coordinates for velocity and acceleration of a particle. I am one of the few brit doctors here and consider polar coordinates to be one of my more familiar areas. I had a look at a couple of web pages we have links to here and couldn't find much that was enlightening as such - basically just the definitions and a few formulae. I presume that you are looking for some understanding of the problem and here is just a small start which might help. Unfortunately - having studied some tensor calculus I know more about them than was certainly in my A-level syllabus so I have a little difficulty in reconciling the slightly easier A-level version with the more complete tensor version - but anyway... The thing which makes Cartesian co-ordinates so easy is that the basis vectors always point in the same direction (the i vector - in the direction of the x axis always points straight 'right'. This means that when you differentiate a vector you just have to differentiate its components. i.e. d/dt([x(t),y(t)]) = [ dx/dt, dy/dt ] or using i,j notation which is more helpful for understanding in this case: d/dt ( xi + yj ) = (dx/dt)i + (dy/dt)j because i, j are constant However with polar co-ordinates the basis vector in the 'r' direction always points away from the origin so at x=0, y=1 it points up but at x=1, y=0 it points right. Normally we use 'e sub r' for the basis vector in the radial direction (away from the origin and 'e sub theta' for the basis vector an the tangential (anticlockwise) direction. I'll denote these e_r, e_0 for ease of typing. So if we have a vector: r(t) e_r + th(t) e_0 and we want to differentiate it (using r' for dr/dt for speedy typing again) we get: r'(t) e_r + th'(r) e_0 + r(t) d/dt(e_r) + th(t) d/dt(e_0) So we need to know how e_r and e_0 differentiate. Well if we move in the r direction both vectors stay in the same direction, so d/dr(e_r) = 0 = d/dr(e_0) But if we move in the theta direction - around the origin - the basis vectors rotate. Because the vectors don't change side this is just like adding a very small displacement vector perpendicular to the original basis vector for each very small displacement of theta. (It's best to draw a diagram but I can't think of how best to do this right know.) Because the basis vectors stay as unit vectors the size of this displacement is d0 for a small displacement d0. Again my reasoning isn't sound but if you draw a diagram showing arrows for typical r basis vector at two points (r,th) and (r,th+dth) and then look at a vector parallel to the second at the same point of the first and look at the triangle the form you should be able to draw the vector d(th) * e_0 joining the tips of these two vectors. This is difficult to explain but perhaps you get my gist? Then we have d/dr(e_r) = 0, d/dth(e_r) = e_0, d/dr(e_0) = 0, d/dth(e_0) = -e_r (Can you see why the minus sign?) Any now we have done the hard part, the rest is easy(ish!). Using the chain rule (and partial derivatives) we have: d/dt(e_r) = dr/dt * d/dr(e_r) + dth/dt * d/dth(e_r) = r' * 0 + th' * e_0 = th' * e_0 and similarly d/dt(e_0) = -th' * e_r So if we have a particle at position (r,th) then its position vector is r(t) * e_r so we can find its velocity by differentiating: v = r'(t) * e_r + r(t) * d/dt(e_r) = r'(t) * e_r + r(t) * th'(t) * e_0 and acceleration: a = r''(t) * e_r + r'(t) * th'(t) * e_0 + r'(t)*th'(t)*e_0 + r(t)*th''(t)*e_0 - r(t)*th'(t)*-th'(t)*e_r = [r''(t) - r(t)*th'(t)^2] * e_r + [2*r'(t)*th'(t) + r(t)*th''(t)] * e_0 And this works with any vector - not just position vectors. -Doctor Charles, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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