MHB Stuck on Polar Coordinate Components?

[Carla1985]

I'm stuck on the second part of this question.

Suppose a particle moves in a plane with its trajectory given by the polar equation $r=2b\sin(\theta)$ for some constant $b>0$.

(i) Show that this can be written in Cartesian coordinates as $x^2+(y-b)^2=b^2$.

This is the equation for a circle of centre $(0,b)$ and radius $b$.

[Hint: recall that $r^2=x^2+y^2$ and $y=r\sin(\theta)$]

(ii) Suppose that the transverse component of the acceleration is zero.

(a) Prove that $r^2\dot{\theta}=h$ is constant.

(b) Assuming that $r\ne0$, show that $\dot{r}=2bhr^{-2}\cos(\theta)$ and hence find $\ddot{r}$.

(c) Use your answers to (b) to show that the radial component of the acceleration is $-8b^2h^2r^{-5}$.

So far, I've got:

$r=2b\sin(\theta)$

$\dot{r}=2b\cos(\theta)\dot{\theta}$

So, the transverse coordinate is $4b\cos(\theta)\dot{\theta}^2+2b\sin(\theta)\dot{ \theta}$.

 

[chisigma]

I'm stuck on the second part of this question.

Suppose a particle moves in a plane with its trajectory given by the polar equation $r=2b\sin(\theta)$ for some constant $b>0$...

Of course it must be r>0 so that i wonder if the equation of the trajectory is, may be, $ r= 2\ b\ |\sin \theta|$...

Kind regards

$\chi$ $\sigma$

 

[Carla1985]

I don't understand, sorry. I'm on part ii) a. I've found a hint in some of our notes to differentiate

r2θ˙

​

but that just gets really messy and I don't see how it helped :/

 

[I like Serena]

Welcome to MHB, Carla1985! :)

As you can see here or here, the angular acceleration is $a_\theta = r\ddot\theta + 2 \dot r \dot\theta$.
Is that perchance in your notes?

Since $a_\theta = 0$, it follows that:

$r\ddot\theta + 2 \dot r \dot\theta = 0$​

You already saw in (i) that it is useful to multiply by r:

$r^2\ddot\theta + 2r \dot r \dot\theta = 0$

$\frac{d}{dt}(r^2\dot\theta) = 0$

$r^2\dot\theta = constant$ $\qquad \blacksquare$​

 

[Carla1985]

Ah that makes sense. Thank you :)

 

[I like Serena]

Of course it must be r>0 so that i wonder if the equation of the trajectory is, may be, $ r= 2\ b\ |\sin \theta|$...

When $\pi < \theta < 2\pi$ there is no positive value for r... but with a negative value it still fits the orbit, which is a circle around (0,b) with radius b.

 

[Carla1985]

Ah yes, I think that's what we proved in part 1, that its trajectory is a circle :)

 

[Carla1985]

I've made a mistake somewhere in the last 2 parts but can't find where.
r^.=2bhr^-2cos
r^..=-2bhr^-2sinθ-8b²h²r^-5cos²θ

then for the radial component
(-2bhr^-2sinθ-8b²h²r^-5cos²θ-2bh²r^-4sinθ)

which isn't what I want lol

 

[I like Serena]

I've made a mistake somewhere in the last 2 parts but can't find where.
r^.=2bhr^-2cos
r^..=-2bhr^-2sinθ-8b²h²r^-5cos²θ

then for the radial component
(-2bhr^-2sinθ-8b²h²r^-5cos²θ-2bh²r^-4sinθ)

which isn't what I want lol

Your $\dot r$ looks good! (Except for the θ that didn't make it.)

And in your $\ddot r$, it seems that you forgot the $\dot \theta$.
Makes sense because you lost a $\theta$ before that. ;)

I'm assuming you continued with:

$a_r = \ddot r - r \dot \theta^2$​

Did you consider that you can eliminate $\sin \theta$ by using $r = 2b\sin\theta$?

 

[Carla1985]

Your $\dot r$ looks good! (Except for the θ that didn't make it.)

And in your $\ddot r$, it seems that you forgot the $\dot \theta$.
Makes sense because you lost a $\theta$ before that. ;)

I'm assuming you continued with:

$a_r = \ddot r - r \dot \theta^2$​

Did you consider that you can eliminate $\sin \theta$ by using $r = 2b\sin\theta$?

I substituted $\dot \theta$ with hr^-2 from in the last step? And no I didnt think of that but I'l give it a try

 

[Carla1985]

Tried eliminating sinθ and its close but not quite

Im left with -h²r^-3 and -h²r^-1 so still don't cancel out and I still have a cos²θ in the middle term that shouldn't be there

 

[I like Serena]

Did you know that $\sin^2\theta + \cos^2\theta = 1$?

 

[Carla1985]

Did you know that $\sin^2\theta + \cos^2\theta = 1$?

Yes, that's what I thought I was working towards but I have a - between them not a x

 

[I like Serena]

Yes, that's what I thought I was working towards but I have a - between them not a x

You can still replace $\cos^2\theta$ by $1 - \sin^2 \theta$.

 

[Carla1985]

Still doesn't work. I get:

sin - (the term i want to be left with) + sin² -sin
The 3 terms with sin have different coefficients so still don't cancel out

 

[I like Serena]

Sorry, but I'm way overdue to sleep.
See you tomorrow! (Sleepy)

 

[Carla1985]

nps, me too. thanks for all the help, I am sure il get it in the morning after some sleep :)

 

[I like Serena]

nps, me too. thanks for all the help, I am sure il get it in the morning after some sleep :)

Hey Carla1985! ;)

Did you get it by now?
To be honest, I did not.
I haven't properly checked my calculations yet, but I'm not getting the result stated in your problem, just like you.
I can get an expression in just $r$ and $\theta$ which seems to be the purpose of the exercise.
but it's just not what is suggested.
Perhaps they were wrong? Or perhaps I made a mistake. :(

 

[Carla1985]

I made progress. I had the differentiation slightly wrong to start with. So I have:

\[

r=2bsin\theta \\
\dot{r}=2bhr^{-2}cos\theta \\
\ddot{r}=-8b^2h^2r^{-5}cos^2\theta-2bh^2r^{-4}sin\theta \\
\\
which\ gives: \\
\ddot{r}-r\dot{\theta}^2=-8b^2h^2r^{-5}cos^2\theta-2bh^2r^{-4}sin\theta-2bh^2r^{-4}sin\theta \\
so\ =-8b^2h^2r^{-5}cos^2\theta-4bh^2r^{-4}sin\theta \\

\]

Im not sure how to get rid of the cos onwards, apparently we divide through by something :/

 

[I like Serena]

I made progress. I had the differentiation slightly wrong to start with. So I have:

\[

r=2bsin\theta \\
\dot{r}=2bhr^{-2}cos\theta \\
\ddot{r}=-8b^2h^2r^{-5}cos^2\theta-2bh^2r^{-4}sin\theta \\
\\
which\ gives: \\
\ddot{r}-r\dot{\theta}^2=-8b^2h^2r^{-5}cos^2\theta-2bh^2r^{-4}sin\theta-2bh^2r^{-4}sin\theta \\
so\ =-8b^2h^2r^{-5}cos^2\theta-4bh^2r^{-4}sin\theta \\

\]

Im not sure how to get rid of the cos onwards, apparently we divide through by something :/

Looks good! :)

So let's continue:

$a_r = -8b^2h^2r^{-5}\cos^2\theta-4bh^2r^{-4}\sin\theta$

$a_r = -8b^2h^2r^{-5}(1 - \sin^2\theta)-4bh^2r^{-4}\sin\theta$

$a_r = -8b^2h^2r^{-5} + 8b^2h^2r^{-5}\sin^2\theta-4bh^2r^{-4}\sin\theta$​

We can substitute $r = 2b \sin \theta$ (in reverse), getting:

$a_r = -8b^2h^2r^{-5} + 2h^2r^{-5}\cdot r^2-2h^2r^{-4} \cdot r$

$a_r = -8b^2h^2r^{-5}$ $\qquad \blacksquare$​

There you go! ;)