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Path of a Projectile in Polar Coordinates

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched from a mountain at a given angle and velocity (which is large). Using polar coordinates find the position of the particle at time t. I'm ignoring drag (for now).

    2. Relevant equations
    I tried using the polar kinematic equations: http://mathenthusiast.com/physics/mechanics/kinematic-equations-in-polar-coordinates/

    3. The attempt at a solution
    I have a python program that uses a numerical Euler method to find the path of the projectile. I want to use the polar kinematic equations to ensure I did things correctly and to see the discrepancy between numerical and kinematic answers. I first tried using cartesian coordinates, but ran into issues with splitting the gravitational force into x and y components. The velocity is high enough that I can't treat gravity as acting only in the y direction.

    Unfortunately, the answers I get are quite different. When using the polar kinematic equations, my velocity in the theta unit vector direction is increasing over time (enough so that my total velocity increases as well). Can someone show me what the equations of motion should look like in polar coordinates?
     
    Last edited: Mar 23, 2015
  2. jcsd
  3. Mar 23, 2015 #2

    BvU

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    Hello Ng, welcome back to PF :smile: !

    Could you elaborate a little ? Ignoring drag makes the trajectory a parabola and you can transform to polar and back with ease.
    It's unclear to me why you would want an x-component in gravity. Earth isn't flat enough ?
    If you show your work and some drawings, perhaps we can comment.
     
  4. Mar 23, 2015 #3
    Thanks!

    It seems to me there should be a way to do things in polar. I've found lots of google help with cartesian mechanics though and very little in polar coordinates.

    The projectile is moving very fast ie multiple km/s, so the gravitational acceleration vector can change direction from straight down.

    Here is a quick write up of what I've been trying (which seems to be wrong) and a drawing. I think i'm misunderstanding something in my acceleration terms and/or unit vectors.

    Edit: I know that the acceleration component for my theta is wrong...I'm just how to find the correct term...seems like I need a 1/(r initial) otherwise my theta values go crazy.
     

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    Last edited: Mar 23, 2015
  5. Mar 24, 2015 #4

    BvU

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    So is your coordinate system wrt the firing muzzle or wrt the center of the earth ?
    Anyhow, the things you claim constant are not constant if earth isn't flat enough!
     
  6. Mar 24, 2015 #5
    Center of the earth. What I want are equations that describe the projectile's motion giving me theta, r, omega, and radial velocity using my starting values for v, theta, angle of projectile launch.

    Which "constants"? The idea with using the center of the earth as the origin is that there should be no real forces acting in the theta direction. So my only acceleration in that direction would be due to the movement of the coordinate unit vectors. Whereas gravity should always act only in the r direction. I should have mentioned that I'm ignoring the change in the gravitational field strength of the earth as I move farther away. I'd be happy to include that, but don't know if it's possible.
     
  7. Mar 24, 2015 #6

    BvU

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    Ok, I see. (I saw the ##\hat r ## and the ##\hat \theta## at the firing position, so I drew the wrong conclusion).

    But ar = ##\vec g## confused me no end. I understand you mean ##\vec a = - \vec g\ ## (not constant) and ## a_r = -|\vec g|\ ##, a number (constant).
    Perhaps you can do something nice with $$
    \vec a = (\ddot r− r\dot \theta^2) {\bf \hat r} + (r\ddot\theta + 2\dot r \dot \theta) {\bf \hat θ}
    $$which splits up in $$
    \ddot r− r\dot \theta^2 = -|\vec g|\quad {\mathsf {and} }\quad r\ddot\theta + 2\dot r \dot \theta = 0
    $$but I don't directly see what.

    Why is ##d\theta\over dt## a constant ?
    I understand ##\vec F## is central, so the angular momentum ##\vec L = \vec r\times\vec p## is constant.
    This follows from the second equation: ##{d\over dt}\left (r^2\dot\theta\right ) = r^2\ddot\theta + 2r\dot r\dot\theta = r\left(r\ddot\theta+2\dot r\dot \theta\right) = 0##

    You can't just copy the SUVAT equations and replace x, v and a by ##\theta##, ##\dot\theta## and ##a_\theta##, you'll have to solve the equation of motion.

    Including that means you are looking for Kepler problem solutions and ICBM trajectories. Pretty hefty.
     
    Last edited: Mar 24, 2015
  8. Mar 24, 2015 #7
    Sorry about that. They should probably be from the origin. I've seen it done that way a lot online. ## a_r = -|\vec g|\ ## is what I meant. I was confusing ##\ddot r ## with##\ a_r ##. This makes much more sense:
    $$\ddot r− r\dot \theta^2 = -|\vec g|\quad {\mathsf {and} }\quad r\ddot\theta + 2\dot r \dot \theta = 0 $$
    I haven't done physics in a few years and am in the process of reviewing, so please bear with me if and when my syntax/terminology is bad or if I miss something obvious.

    ##d\theta\over dt## I thought should be constant because the acceleration in the theta direction should be zero. I'm confused which "second equation" you are referring to. I agree with your time derivative, but I'm not sure where ##r^2\dot\theta ## comes from. How would I go about solving my equations of motion. If you provide a link, I'll be happy to do some reading.

    Lets leave out distance dependent gravity acceleration for now. I can test against static gravitational acceleration in my numerical solution and add that in later (probably numerically).
     
  9. Mar 24, 2015 #8

    BvU

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    ( Currently I don't see processed LaTeX, only source TeX input. Also on mathenthusiast. Must be some add-in that got switched off - But it's on IE11 as well as on GoogleChrome. Someone knows what to do to fix that ?)

    Second equation is $$r\ddot\theta + 2\dot r \dot \theta = 0 $$the ##\hat\theta## component of ##\vec a##. That's the one that is zero, not ##\ddot \theta## !

    ##r^2\dot\theta ## comes from ##\vec L = \vec r\times\vec p = m\; \vec r\times\vec v= m\; \vec r\times\left (\vec\omega\times\vec r \right )##

    The icbm guys do a nice job in solving.
     
    Last edited: Mar 25, 2015
  10. Mar 25, 2015 #9

    BvU

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    You have two equations$$
    \ddot r− r\dot \theta^2 = -|\vec g|\quad {\rm[1]} \quad {\mathsf {and} }\quad r\ddot\theta + 2\dot r \dot \theta = 0 \quad {\rm[2]} \quad
    $$and you have the constant L, so ##
    r^2\dot\theta## is constant. That means ##r^4\dot\theta^2## is constant too. Call it C

    Eq [1] gives you ##\ddot r= r\dot \theta^2 -|\vec g| = C/r^3 -|\vec g|## so for your integration step you use this to get ##\dot r(t+dt) = \dot r(t)+\ddot r(t) * dt ## and r with ## r(t+dt) = \dot r +\dot r(t) * dt ## .

    And for theta you use ##\dot\theta = \sqrt C /r^2##

    Since all this is numerical anyway, using g(r) is just as easy as using g(rfiring)
     
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