# Two Signs for Rate of Change of Angle in Polar Coordinates

• whoareyou
In summary, the equation for the velocity vector is:$$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$

## Homework Statement

I didn't know if this was considered "advanced" physics, but it's an intermediate classical mechanics course so I'll just post my question here. Basically, if you have a cardioid ##r(\theta)=k(1+\cos(\theta))##, you can show that the ##\dot{\theta}=\frac{v}{\sqrt{2kr}}##. That means for a given ##r## with ##v## constant, the rate of change of the angle is both positive and negative. But what does this actually mean?

## Homework Equations

Description of (2D) motion in Polar Coordinates

## The Attempt at a Solution

I was thinking it could possibly have something to do with the cosine function being even (i.e. ##\cos(-\theta) = \cos(\theta)## but I don't understand the implications.

What is v? The velocity along your curve? A non-zero v does not allow r to be constant then.
How is the change in angle "both positive and negative"? The sign of the change in angle depends on the direction of v, of course.

mfb said:
What is v? The velocity along your curve? A non-zero v does not allow r to be constant then.
How is the change in angle "both positive and negative"? The sign of the change in angle depends on the direction of v, of course.

It's a constant speed ##v## along the curve. ##r## isn't constant. The graph of ##r## is a cardioid.

whoareyou said:
It's a constant speed ##v## along the curve. ##r## isn't constant. The graph of ##r## is a cardioid.
If ##\vec{r}## is the position vector drawn from the origin to the moving particle traveling at constant velocity v along the curve, then we can write:
$$\vec{r}(θ)=r(θ)\vec{i}_r(θ)$$
where ##\vec{i}_r(θ)## is the unit vector in the radial direction at angular position θ. The time derivative of this position vector is the the velocity vector of the particle along the curve:
$$\vec{v}=\frac{d\vec{r}(θ)}{dt}=\frac{d[r(θ)\vec{i}_r(θ)]}{dt}$$
Do you know how to take the time derivative of the right hand side of this equation to determine the velocity vector?

Chet

Chestermiller said:
If ##\vec{r}## is the position vector drawn from the origin to the moving particle traveling at constant velocity v along the curve, then we can write:
$$\vec{r}(θ)=r(θ)\vec{i}_r(θ)$$
where ##\vec{i}_r(θ)## is the unit vector in the radial direction at angular position θ. The time derivative of this position vector is the the velocity vector of the particle along the curve:
$$\vec{v}=\frac{d\vec{r}(θ)}{dt}=\frac{d[r(θ)\vec{i}_r(θ)]}{dt}$$
Do you know how to take the time derivative of the right hand side of this equation to determine the velocity vector?

Chet

$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$

whoareyou said:
$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$
Right. Are you also aware that
$$\frac{d\vec{i}_r(θ)}{dθ}=-\vec{i}_θ$$
So, substitute this and your equation for r(θ) into the equation for velocity, and also factor the dθ/dt. What do you get for the velocity vector?

Chet

Chestermiller said:
Right. Are you also aware that
$$\frac{d\vec{i}_r(θ)}{dθ}=-\vec{i}_θ$$
So, substitute this and your equation for r(θ) into the equation for velocity, and also factor the dθ/dt. What do you get for the velocity vector?

Chet

Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$

whoareyou said:
Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$
Oooops. You're right. Senior Moment. OK, Please continue.

Chet

Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$

whoareyou said:
Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$
OK. Nice. Now also factor out the k, and then take the dot product of ##\vec{v}## with itself to get the square of its magnitude v2. What do you get?

Chet

##v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r##. I did all of this before we did the problem (this is how I got the expression for ##\dot{\theta}## in the original post).

whoareyou said:
##v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r##. I did all of this before we did the problem (this is how I got the expression for ##\dot{\theta}## in the original post).
Oh. I was confused. I thought that is what you were trying to show.

Now I see that you were saying that dθ/dt has to be both positive and negative, and you were wondering how that can be. It's not positive and negative at the same time, right. If v is pointing clockwise, then it is one sign, and if v is pointing counter clockwise, then it's the other sign. Is that what the issue was? I guess mfb already said that in post #2.

Chet

But ##\dot{\theta}## is a magnitude. How can this magnitude be negative?

Suppose you associate the minus sign with the unit vector in the θ direction, rather than with the magnitude ##\dot{θ}##. So you have the vector ##\dot{θ}(-\vec{i}_θ)##. Does that work for you?EDIT: What I really meant to say is that, in the case where the object is moving in the negative θ direction, we express the θ component of velocity as ##\vert r\dot{θ}\vert (-\vec{i}_θ)##.

Chet

Last edited:

## 1. What is the definition of rate of change of angle in polar coordinates?

The rate of change of angle in polar coordinates is a measurement of how fast the angle of a point in a polar coordinate system is changing with respect to time or another variable.

## 2. How is the rate of change of angle in polar coordinates calculated?

The rate of change of angle in polar coordinates is calculated using the formula Δθ/Δt, where Δθ represents the change in angle and Δt represents the change in time or another variable.

## 3. What are the two signs for rate of change of angle in polar coordinates?

The two signs for rate of change of angle in polar coordinates are positive (+) and negative (-). A positive sign indicates that the angle is increasing, while a negative sign indicates that the angle is decreasing.

## 4. Why is it important to understand the rate of change of angle in polar coordinates?

Understanding the rate of change of angle in polar coordinates is important because it allows us to analyze the movement and behavior of objects in polar coordinate systems. It can also help us make predictions and solve problems involving polar coordinates.

## 5. How is the rate of change of angle in polar coordinates used in real-world applications?

The rate of change of angle in polar coordinates is used in many real-world applications, such as tracking the movement of objects in circular motion, analyzing changes in weather patterns, and predicting changes in stock market trends. It is also important in fields such as physics, engineering, and astronomy.

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