Off center circular motion (polar coordinates)

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SUMMARY

The discussion focuses on deriving the velocity vector of a particle moving with constant speed v around a circle of radius b, using polar coordinates with the origin located on the circle. The key equation utilized is \(\frac{d\vec{r}}{dt} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}\). Participants emphasize the importance of recognizing that the velocity vector must be derived from the circular motion principles rather than solely focusing on the radial component. The solution involves translating the origin and applying the time-dependent version of the circle's equation in polar coordinates.

PREREQUISITES
  • Understanding of polar coordinates and their applications in physics
  • Familiarity with circular motion concepts and velocity vectors
  • Knowledge of calculus, specifically derivatives and rates of change
  • Ability to translate between Cartesian and polar coordinate systems
NEXT STEPS
  • Study the derivation of velocity vectors in circular motion using polar coordinates
  • Learn how to convert equations from Cartesian to polar coordinates
  • Explore the implications of constant speed in circular motion
  • Investigate the geometric interpretation of velocity vectors at various points on a circle
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to polar coordinates and velocity vectors.

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Homework Statement


A particle moves with constant speed v around a circle of radius b. Find the velocity vector in polar coordinates using an origin lying on the circle.

https://www.desmos.com/calculator/maj7t9ple1
Imagine the r starts at (0,0).

Homework Equations



\frac{d\vec{r}}{dt} = \dot{r}\hat{r}+r\dot{\theta}\hat{\theta}

The Attempt at a Solution



We can make a triangle connecting the origin to the center of the circle, to a point where the particle is. the hypotenuse is r

I assume I need to find the rate of change of r, right? So, could I just do

r=b/cos(\theta) \frac{dr}{dt}=\frac{dr}{d \theta}\frac{d \theta}{dt}

\frac{dr}{dt}=b\frac{tan(\theta)}{cos(\theta)} \frac{d\theta}{dt}

My book doesn't do this, which leads me to believe I've made some horrible mistake.
 
Last edited:
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I assume I need to find the rate of change of r, right?
Nope. It asks you to find the velocity vector. That's the whole thing, not just the radial component.

You should already know the velocity vector if the origin in at the center of the circle - so write that down. Then just translate the origin. You don't have to start from scratch.

Not sure? Then look up the equation of a circle (offset origin) in polar coordinates... work out the time-dependent version of that (constant speed).

To understand the problem - sketch=ch the circle and put points on the circle for different positions (one at the origin, one opposite etc) and draw a little arrow on each point for the velocity there - write that velocity as a vector in cartesian and polar coordinates.
 
I think the easiest approach would be to find the motion using rectangular coordinates with the center of the circle at the origin, then translate the coordinates so the point on the circle is at the origin, then convert to polar coordinates.
 
Simon Bridge said:
Nope. It asks you to find the velocity vector. That's the whole thing, not just the radial component.

You should already know the velocity vector if the origin in at the center of the circle - so write that down. Then just translate the origin. You don't have to start from scratch.

Not sure? Then look up the equation of a circle (offset origin) in polar coordinates... work out the time-dependent version of that (constant speed).

To understand the problem - sketch=ch the circle and put points on the circle for different positions (one at the origin, one opposite etc) and draw a little arrow on each point for the velocity there - write that velocity as a vector in cartesian and polar coordinates.

Sorry, I need to head out for a while, I'll work on it some more when I get back!
 

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