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

[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