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Two Signs for Rate of Change of Angle in Polar Coordinates

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  • #1
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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

  • #2
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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
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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.
 
  • #4
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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
 
  • #5
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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
$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$
 
  • #6
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$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$
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
 
  • #7
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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? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$
 
  • #8
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Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$
Oooops. You're right. Senior Moment. OK, Please continue.

Chet
 
  • #9
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Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$
 
  • #10
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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
 
  • #11
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##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
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##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
 
  • #13
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But ##\dot{\theta}## is a magnitude. How can this magnitude be negative?
 
  • #14
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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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