Movement along the cardioid ##r=k(1+cos\theta)##

[davidbenari]

Homework Statement

A particle moves with velocity $v$ along the cardioid $r=k(1+cos\theta)$. Find $\vec{r''}\cdot \hat{r}$ and interpret the result. At what angles will this coincide with the centripetal acceleration?

Homework Equations

The Attempt at a Solution

To find $\vec{r''}\cdot \hat{r}$ I will simply use the formula for the radial acceleration in cylindrical coordinates. This is $\rho''-\rho\theta'^2$ which after lengthy manipulation gives: $-k(\theta'' sin\theta + \theta'^2cos\theta)-k\theta'^2(1+cos\theta)$. Which I have no idea how to "interpret", haha.

To solve the second part well I know the position vector is $\vec{r}=r\hat{\rho}$ and that its derivative with respect to $\theta$ will give a tangent vector to the curve. This is simply $\frac{d\vec{r}}{d\theta}=\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta}$

The unit tangent vector is simply: $\hat{t}=\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta})$. Which helps me to find the orthogonal normal vector and thus the centripetal acceleration as $\frac{v^2}{R}\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\theta}-r\hat{\rho})$. Note I've switched components on the and sign on one component as well.

This helps me notice that the centripetal acceleration coincides with my above result when the centripetal acceleration is purely in direction of $\rho$. This happens when $\frac{dr}{d\theta}=k(-\sin\theta)=0$ which is for Pi and 0.

Now, help me interpret the acceleration expression I derived above to which I'm expected to provide an interpretation. Also, is my procedure correct?

Thanks for reading this tedious thing.

 

[Chestermiller]

Let's take a step back. You kinda have the right idea.

What is the general equation for the velocity of a particle in cylindrical coordinates in terms of dr/dt and dθ/dt?
What is the magnitude of this vector?
For your problem, given the shape of the path, what is the magnitude of the vector (in terms of dθ/dt, but not dr/dt)? (This must be equal to the constant v).

Chet

 

[davidbenari]

So the velocity vector is $r' \hat{r} + r \theta ' \hat{\theta}$.

This would reduce to $k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}$ which, because of its constant magnitude we conclude $k^2\theta'^2(1+cos\theta)^2+k^2\theta'^2sin^2\theta=v^2$.

This looks similar to what I found as a term for the radial acceleration but its not exactly the same. Was I supposed to see something important here?

 

[Chestermiller]

So the velocity vector is $r' \hat{r} + r \theta ' \hat{\theta}$.

This would reduce to $k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}$ which, because of its constant magnitude we conclude $k^2\theta'^2(1+cos\theta)^2+k^2\theta'^2sin^2\theta=v^2$.

This looks similar to what I found but its not exactly the same. Was I supposed to see something important here?

Yes. You can solve this for dθ/dt.

 

[davidbenari]

$\theta'=\sqrt{\frac{v^2}{k^2[2+2cos\theta]}}$.

 

[davidbenari]

Am I supposed to solve this DE ?

 

[davidbenari]

Or should I plug it into my expression?

 

[davidbenari]

Since $\vec{r}''\cdot \hat{r}=\frac{d}{dt}(-k\theta'sin\theta)-k(1+cos\theta)\theta'^2=\frac{d}{dt}(\vec{v}\cdot\hat{r})-k(1+cos\theta)\theta'^2$ I'm starting to think the interpretation is simply that the first term is a radial acceleration by virtue of my change in radial velocity, and the other is a radial acceleration given by the curvature of my curve.

 

[davidbenari]

Where this isn't exactly curvature, but a type of curvature that is seen from the origin, if that even makes sense.

 

[Chestermiller]

Well, you are going to need to get r''. First get r' in terms of θ', then substitute your equation in post #5 for θ'. Then take the derivative of the resulting equation with respect to t to get r''. The objective it to express r''-r(θ')² exclusively in terms of functions of θ.

Chet

 

[davidbenari]

I've got $\frac{v^2}{k}(\frac{sin^2\theta}{2+2cos\theta}-\frac{cos\theta}{2+2cos\theta}-1/2]$

 

[davidbenari]

Out of curiosity, what made you think making everything as a function of $\theta$ would be helpful? I'm not seeing it clearly right now. Thanks.

 

[Chestermiller]

I've got $\frac{v^2}{k}(\frac{sin^2\theta}{2+2cos\theta}-\frac{cos\theta}{2+2cos\theta}-1/2]$

I'm not going to check your "arithmetic", but I do recommend reducing the final equation to a common denominator and also factoring out that 2.

Chet

 

[Chestermiller]

Out of curiosity, what made you think making everything as a function of $\theta$ would be helpful? I'm not seeing it clearly right now. Thanks.

Lots of experience.

Chet

 

[davidbenari]

$\frac{v^2}{2k}(\frac{sin^2\theta-2cos\theta-1}{1+cos\theta})$

 

[davidbenari]

I have a feeling this is wrong since it diverges at $\theta=pi$

 

[davidbenari]

Hey Chet I've done it all over again and got $\frac{-3v^2}{4k}$ looks interesting... Any ideas?

 

[davidbenari]

Could this be it? haha

 

[Chestermiller]

Chet

Could this be it? haha

Sure, if you did the algebra correctly. Try doing it a third time and see what you get.

By the way, the centripetal acceleration is equal to the component of the acceleration vector perpendicular to the velocity vector. Do you know how to determine that?

Chet

 

[davidbenari]

I think I do. By the way, I think I successfully solved that part in my original post, don't you think? Namely I just found the normal vector in the frenet-serret sense and solved when that would become purely radial and thus the centripetal acceleration would then be what I found as an expression previously. That works doesn't it?

 

[Chestermiller]

I think I do. By the way, I think I successfully solved that part in my original post, don't you think? Namely I just found the normal vector in the frenet-serret sense and solved when that would become purely radial and thus the centripetal acceleration would then be what I found as an expression previously. That works doesn't it?

Yes. I didn't notice that.

Chet

 

[Chestermiller]

David,

To understand even better what is happening here, plot the graph. It will then make sense visually to you.

Chet

 

[davidbenari]

The thing about my result being equal to the centripetal acceleration and $0,\pi$ makes sense to me, however what doesn't make sense to me is that the radial acceleration should be a constant of this motion. I've already graphed it. Any ideas?

 

[Chestermiller]

The thing about my result being equal to the centripetal acceleration and $0,\pi$ makes sense to me, however what doesn't make sense to me is that the radial acceleration should be a constant of this motion. I've already graphed it. Any ideas?

Well, from the graph, you can probably see by eye that the radii of curvature are the same at 0 and pi.

Chet

 

[davidbenari]

But how does that prove that the radial acceleration should be a constant of this motion? I think that would only prove the centripetal acceleration is equal at 0 and pi. Plus, I think the radius of curvature at the kink (probably theta=pi in your graph) is ill-defined.

 

[Chestermiller]

But how does that prove that the radial acceleration should be a constant of this motion? I think that would only prove the centripetal acceleration is equal at 0 and pi. Plus, I think the radius of curvature at the kink (probably theta=pi in your graph) is ill-defined.

I haven't plotted it myself yet, so I was just guessing.

 

[davidbenari]

its basically just this

 

[Chestermiller]

I doesn't look like you are going to get a satisfactory understanding of what you want from the figure. If it were me, and I were motivated to resolve this issue (which I'm not), I would calculate the radial and circumferential components of the acceleration at a selection of angles from 0 to 360, and plot these on the figure. That way I could see by eyeball how it was playing out. If that still didn't satisfy me, I would also make another similar plot, but this time with the normal and tangential components of acceleration. Hopefully that would all do the trick.

Chet

 

[davidbenari]

I'll check it out. Thanks for everything though.

 

[Macykc2]

I was going to create a whole new thread with reference to this post, but I guess I should check if this will bump it.
My question lies in posts 2 and 3, I understand where we get the velocity function in terms of cylindrical cords, but I'm missing the bridge between the v² part, why is it "concluded" that we use that equation?

 

[Chestermiller]

I was going to create a whole new thread with reference to this post, but I guess I should check if this will bump it.
My question lies in posts 2 and 3, I understand where we get the velocity function in terms of cylindrical cords, but I'm missing the bridge between the v² part, why is it "concluded" that we use that equation?

If I understand your question correctly, you are asking why the dot product of the velocity vector with itself is equal to the square of its magnitude. Is that correct?

 

[Macykc2]

So the velocity vector is $r' \hat{r} + r \theta ' \hat{\theta}$.

This would reduce to $k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}$ which, because of its constant magnitude we conclude $k^2\theta'^2(1+cos\theta)^2+k^2\theta'^2sin^2\theta=v^2$.

My confusion is with the last sentence, I was able to get to, $k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}$, but I'm not sure of the reasoning behind the next part where he mentions it's a constant magnitude and concludes the second equation.

 

[Chestermiller]

My confusion is with the last sentence, I was able to get to, $k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}$, but I'm not sure of the reasoning behind the next part where he mentions it's a constant magnitude and concludes the second equation.

The problem statement implies that the tangential velocity v is constant.

 

[Macykc2]

Alright I've figured out that part, and I've also solved for θ', then I put that equation into r' and ended up with $r'=\frac{v[1+cos\theta - sin\theta]}{\sqrt{1+2cos\theta}}$
Now you say I'm supposed to differentiate with respect to t, dv/dt would make acceleration, but differentiating the cos and sin functions, does that just end up with θ' or is it now θ"?
Also I'm sorry if you read this before I fix the math code stuff, I've never used it!

 

[Chestermiller]

Alright I've figured out that part, and I've also solved for θ', then I put that equation into r' and ended up with $r'=\frac{v[1+cos\theta - sin\theta]}{\sqrt{1+2cos\theta}}$
Now you say I'm supposed to differentiate with respect to t, dv/dt would make acceleration, but differentiating the cos and sin functions, does that just end up with θ' or is it now θ"?
Also I'm sorry if you read this before I fix the math code stuff, I've never used it!

You are aware that, when using cylindrical coordinates, to get the acceleration of a particle, you need to take into consideration the changes in the unit vectors in the radial and circumferential directions with respect circumferential angle, correct? What this problem is asking for is the component of the particle acceleration in the radial polar coordinate direction. @davidbenari correctly expressed this component of the acceleration in his first post. I think it would be appropriate if david handled your questions with regard to this problem. How about it, David?

Chet