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

1. Jan 24, 2015

### whoareyou

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Description of (2D) motion in Polar Coordinates

3. 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.

2. Jan 24, 2015

### Staff: 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.

3. Jan 24, 2015

### whoareyou

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

4. Jan 24, 2015

### Staff: Mentor

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

5. Jan 24, 2015

### whoareyou

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

6. Jan 24, 2015

### Staff: Mentor

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

7. Jan 24, 2015

### whoareyou

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

8. Jan 24, 2015

### Staff: Mentor

Oooops. You're right. Senior Moment. OK, Please continue.

Chet

9. Jan 24, 2015

### whoareyou

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

10. Jan 24, 2015

### Staff: Mentor

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

11. Jan 24, 2015

### whoareyou

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

12. Jan 24, 2015

### Staff: Mentor

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

13. Jan 26, 2015

### whoareyou

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

14. Jan 26, 2015

### Staff: 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

Last edited: Jan 27, 2015