Off center circular motion (polar coordinates)

[Astrum]

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.

 

[Simon Bridge]

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.

 

[tms]

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.

 

[Astrum]

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!

 

[Nickie]

Your approach is partially correct, but there are a few mistakes. Let's break it down step by step:

1. The velocity vector in polar coordinates is given by:

\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}

2. We can find \dot{r} and \dot{\theta} using the chain rule:

\dot{r} = \frac{dr}{dt} = \frac{dr}{d\theta} \frac{d\theta}{dt}

\dot{\theta} = \frac{d\theta}{dt}

Note that \dot{\theta} is simply the angular velocity of the particle, which is constant since the particle moves at a constant speed around the circle.

3. Now, we need to find \frac{dr}{d\theta}. This can be done using the Pythagorean theorem:

r^2 = x^2 + y^2

But, since the origin is on the circle, we know that x^2 + y^2 = b^2. Therefore, we can rewrite the equation as:

r^2 = b^2

Differentiating both sides with respect to \theta, we get:

2r\frac{dr}{d\theta} = 0

Therefore, \frac{dr}{d\theta} = 0. This means that \dot{r} = 0 and the velocity vector simplifies to:

\vec{v} = r\dot{\theta}\hat{\theta}

4. Putting it all together, we get:

\vec{v} = (b\dot{\theta})\hat{\theta}

Note that the magnitude of the velocity vector is just b\dot{\theta}, which is the tangential speed of the particle. This makes sense since the particle is moving at a constant speed around the circle.

I hope this helps clarify the process for finding the velocity vector in polar coordinates. Keep in mind that there may be other approaches to solving this problem, but this is one way to approach it.