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

## 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.

## Answers and Replies

mfb
Mentor
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.

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.

Chestermiller
Mentor
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

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

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

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

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? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$

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

Chet

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

Chestermiller
Mentor
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).

Chestermiller
Mentor
##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?

Chestermiller
Mentor
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

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