Radial and Circumferential Components in terms of t

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 5K views
Trapezoid
Messages
9
Reaction score
0

Homework Statement



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.


Homework Equations



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


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
 
Physics news on Phys.org
Hi Trapezoid! :smile:
Trapezoid said:
Determine the radial and circumferential components of acceleration as a function of t.

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)
 
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
 
Trapezoid said:
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..

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:
 
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?
 
yes, the unit vector in the radial direction is r/|r|