1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radial and Circumferential Components in terms of t

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex] \vec{R} = (t + \sin t) \hat{i} + (t + \cos t) \hat{j} [/itex] denotes at time t the position of a moving particle. Determine the radial and circumferential components of acceleration as a function of t.


    2. Relevant equations

    [tex] v_r = \dot{r} [/tex]
    [tex] v_{\theta} = r\dot{\theta} [/tex]


    3. The attempt at a solution

    I tried to write r in terms of t using [itex]r = \sqrt{x^2 + y^2} [/itex], but the derivative was complicated. I do not know how to write [itex]\theta[/itex] in terms of t. We were asked to determine the tangential and normal components of acceleration in the previous part of the question, but I do not see how they will avail me. Could anybody point me in the right direction?

    Thanks,
    Trapezoid
     
  2. jcsd
  3. Apr 16, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Trapezoid! :smile:
    The tangential component (i assume that's what they mean by circumferential) is just d|v|/dt :wink:

    (and you know the magnitude and direction of the total acceleration)
     
  4. Apr 16, 2012 #3
    Hi tiny-tim,

    It is my understanding that the tangential and circumferential components are different. When I say the circumferential component of acceleration, I refer to the acceleration in the direction of [itex]\theta[/itex], ie: the change in the rate of change of [itex]\theta[/itex]. I'm having trouble finding [itex]\theta[/itex] as a function of t..

    Does that make sense? Have I misunderstood?

    Thanks,
    Trapezoid
     
  5. Apr 17, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ah, not a terminology i've come across before :redface:

    ok, then r is the position, r'' is the acceleration, r''.r/|r| is the radial component, and what's left is the circumferential component :smile:
     
  6. Apr 17, 2012 #5
    Thanks tiny-tim,

    Let me make sure that I understand correctly. Is [itex]\frac{r}{|r|}[/itex] the unit vector for motion in the radial direction?
     
  7. Apr 17, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    yes, the unit vector in the radial direction is r/|r|
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook