Two Signs for Rate of Change of Angle in Polar Coordinates

[whoareyou]

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.

 

[mfb]

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.

 

[whoareyou]

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]

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

 

[whoareyou]

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]

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

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

 

[whoareyou]

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]

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

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

Chet

 

[whoareyou]

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

 

[Chestermiller]

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

 

[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).

 

[Chestermiller]

$v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r$. 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

 

[whoareyou]

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

 

[Chestermiller]

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