Basic question pertaining to Polar Coordinates & how to employ them

Click For Summary
SUMMARY

The discussion focuses on the integration of velocity and acceleration in angular motion using plane polar coordinates. The position vector is defined as \(\mathbf{r} = r\mathbf{e}_r(\theta)\), requiring both \(r(t)\) and \(\theta(t)\) for complete specification. The velocity vector is expressed as \(\dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \mathbf{e}_\theta\). Integration of the radial and angular components is performed as \(r = \int v_r\,dt\) and \(\theta = \int \frac{v_\theta}{r}\,dt\), highlighting the need to resolve components due to the non-fixed nature of polar unit vectors.

PREREQUISITES
  • Understanding of polar coordinates and their unit vectors
  • Knowledge of vector calculus and integration techniques
  • Familiarity with angular motion concepts
  • Basic proficiency in physics, particularly mechanics
NEXT STEPS
  • Study the derivation of polar coordinate transformations
  • Learn about the application of vector calculus in physics
  • Explore advanced topics in angular motion and dynamics
  • Investigate numerical methods for integrating motion equations
USEFUL FOR

Students and professionals in physics, particularly those focusing on mechanics, as well as engineers and mathematicians dealing with angular motion and polar coordinates.

warhammer
Messages
164
Reaction score
33
I have a question that might be considered vague or even downright idiotic but just wanted to know that once we find out the velocity & acceleration of a body in angular motion in plane polar coordinates, and are asked to integrate the expressions in order to find position at some specified time 't' , would we proceed to integrate the velocity expression (say) like we do usually with respect to time when the said expression involves r cap as well as theta cap; since the polar unit vectors are not "fixed" unlike the unit vectors of i,j,k in Cartesian system?
 
Science news on Phys.org
The position vector is \mathbf{r} = r\mathbf{e}_r(\theta). You need both r(t) and \theta(t) to fully specify the position. The velocity vector is \dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta so if \dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta then <br /> \begin{align*}<br /> r &amp;= \int v_r\,dt \\<br /> \theta &amp;= \int \frac{v_\theta}{r}\,dt<br /> \end{align*}

(You can always resolve components because \mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1 and \mathbf{e}_r \cdot \mathbf{e}_\theta = 0 are constant.)
 
  • Like
Likes   Reactions: warhammer and vanhees71
pasmith said:
The position vector is \mathbf{r} = r\mathbf{e}_r(\theta). You need both r(t) and \theta(t) to fully specify the position. The velocity vector is \dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta so if \dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta then <br /> \begin{align*}<br /> r &amp;= \int v_r\,dt \\<br /> \theta &amp;= \int \frac{v_\theta}{r}\,dt<br /> \end{align*}

(You can always resolve components because \mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1 and \mathbf{e}_r \cdot \mathbf{e}_\theta = 0 are constant.)
Thank you so much for your help sir
 
  • Like
Likes   Reactions: vanhees71 and berkeman

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K