# Polar coordinate components

#### Carla1985

##### Member
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 $4b\cos(\theta)\dot{\theta}^2+2b\sin(\theta)\dot{ \theta}$.

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#### chisigma

##### Well-known member
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

##### Member
I dont 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 dont see how it helped :/

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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:

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

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

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

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#### Carla1985

##### Member
Ah that makes sense. Thank you #### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### Member
Ah yes, I think thats what we proved in part 1, that its trajectory is a circle #### Carla1985

##### Member
I've made a mistake somewhere in the last 2 parts but cant find where.
r.=2bhr-2cos
r..=-2bhr-2sinθ-8b2h2r-5cos2θ

(-2bhr-2sin
θ-8b2h2r-5cos2θ-2bh2r-4sinθ)

which isnt what I want lol

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I've made a mistake somewhere in the last 2 parts but cant find where.
r.=2bhr-2cos
r..=-2bhr-2sinθ-8b2h2r-5cos2θ

(-2bhr-2sin
θ-8b2h2r-5cos2θ-2bh2r-4sinθ)

which isnt 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

##### Member
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

##### Member
Tried eliminating sinθ and its close but not quite

Im left with -h2r-3 and -h2r-1 so still dont cancel out and I still have a cos2θ in the middle term that shouldnt be there

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Did you know that $\sin^2\theta + \cos^2\theta = 1$?

#### Carla1985

##### Member
Did you know that $\sin^2\theta + \cos^2\theta = 1$?
Yes, thats what I thought I was working towards but I have a - between them not a x

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, thats 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

##### Member
Still doesnt work. I get:

sin - (the term i want to be left with) + sin2 -sin
The 3 terms with sin have different coefficients so still dont cancel out

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Sorry, but I'm way overdue to sleep.
See you tomorrow! #### Carla1985

##### Member
nps, me too. thanks for all the help, im sure il get it in the morning after some sleep #### Klaas van Aarsen

##### MHB Seeker
Staff member
nps, me too. thanks for all the help, im 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

##### Member

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

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member

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

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! 