# Off center circular motion (polar coordinates)

• Astrum
In summary, the particle moves around the circle with a constant speed. The velocity vector is found using an origin located on the circle.
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.

Last edited:
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!

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.

## 1. What is off center circular motion in polar coordinates?

Off center circular motion in polar coordinates refers to the movement of an object in a circular path that is not centered at the origin. Instead, the center of the circle is located at a point with coordinates (r,θ), where r represents the distance from the origin and θ represents the angle from the positive x-axis.

## 2. How is off center circular motion different from regular circular motion?

In regular circular motion, the center of the circle is located at the origin. This means that the radius and angle remain constant throughout the motion. In off center circular motion, the center of the circle is shifted and the radius and angle vary as the object moves along the circular path.

## 3. What causes an object to move in off center circular motion?

An object will move in off center circular motion if there is a force acting on it that has both a radial and tangential component. The radial component causes the object to move towards or away from the center of the circle, while the tangential component causes the object to move along the circular path.

## 4. How is off center circular motion described mathematically?

In polar coordinates, the position of an object in off center circular motion can be described by the equations r = r0 and θ = ωt + θ0, where r0 and θ0 represent the initial position of the object, ω represents the angular velocity, and t represents time.

## 5. What are some applications of off center circular motion?

Off center circular motion has many applications in physics and engineering, such as in planetary orbits, satellite trajectories, and the motion of objects in a centrifuge. It is also used in the design of machinery and vehicles that require circular motion, such as gears, pulleys, and wheels.

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