# Polar acceleration problem

## Homework Statement:

A particle moves on a plane with rθ = constant. It's transverse acceleration is zero. Force acting on it is related to r as:
(a)1/r
(b)1/r^2
(c) 1/r^3
(d) 1/r^4

## Relevant Equations:

## \dot {\hat {r} }=\dot θ \hat {\theta}##

## \dot {\hat \theta }=-\dot {\theta} \hat r##
##\vec r=r \hat r##
##\vec v=\dot r \hat r + r \dot \theta \hat \theta##
##\vec a = (\ddot r - r \dot \theta^2)\hat r + (2 \dot r \dot \theta + r \ddot \theta)\hat \theta##
Given that,
##2 \dot r \dot \theta + r \ddot \theta =0##

Also,
##r \theta=constant##
##\Rightarrow \dot r \theta + r \dot \theta=0##
##\Rightarrow \dot r = -\frac{r \dot \theta} {\theta}##
##\Rightarrow \frac{-2r \dot \theta^2}{\theta} + r \ddot \theta=0##
##\Rightarrow \frac{2 \dot \theta^2}{\theta} = \ddot \theta##
##\Rightarrow 2 \dot \theta^2=\theta \ddot \theta##
##\Rightarrow \vec a=(\ddot r-r \dot\theta^2) \hat r##
##\Rightarrow \vec a=(\ddot r-\frac{r \theta\ddot \theta}{2}) \hat r##
##\Rightarrow \vec a=(\ddot r-\frac{(const.)\ddot \theta}{2}) \hat r##

Which is independent of r.
What is wrong with my work?

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PeroK
Homework Helper
Gold Member
Homework Statement:: A particle moves on a plane with rθ = constant. It's transverse acceleration is zero. Force acting on it is related to r as:
(a)1/r
(b)1/r^2
(c) 1/r^3
(d) 1/r^4
Relevant Equations:: ## \dot {\hat {r} }=\dot θ \hat {\theta}##

## \dot {\hat \theta }=-\dot {\theta} \hat r##

##\vec r=r \hat r##
##\vec v=\dot r \hat r + r \dot \theta \hat \theta##
##\vec a = (\ddot r - r \dot \theta^2)\hat r + (2 \dot r \dot \theta + r \ddot \theta)\hat \theta##
Given that,
##2 \dot r \dot \theta + r \ddot \theta =0##

Also,
##r \theta=constant##
##\Rightarrow \dot r \theta + r \dot \theta=0##
##\Rightarrow \dot r = -\frac{r \dot \theta} {\theta}##
##\Rightarrow \frac{-2r \dot \theta^2}{\theta} + r \ddot \theta=0##
##\Rightarrow \frac{2 \dot \theta^2}{\theta} = \ddot \theta##
##\Rightarrow 2 \dot \theta^2=\theta \ddot \theta##
##\Rightarrow \vec a=(\ddot r-r \dot\theta^2) \hat r##
##\Rightarrow \vec a=(\ddot r-\frac{r \theta\ddot \theta}{2}) \hat r##
##\Rightarrow \vec a=(\ddot r-\frac{(const.)\ddot \theta}{2}) \hat r##

Which is independent of r.
What is wrong with my work?
Why is this independent of ##r## when you haven't solved for ##\ddot r## or ##\ddot \theta##?

Wow! It's so difficult to know when to stop manipulating the equations...
##\vec a=(\ddot r - r \theta \ddot \theta/2) \hat r##
Now, ##r \theta=## const.
##r\dot \theta + \theta \dot r=0##
##r\ddot \theta +2\dot r \dot \theta + \theta \ddot r=0##
So, ##\ddot r = \frac{-2 \dot r \dot \theta - r \ddot \theta}{\theta}##

Also, since tangential part of acceleration is zero,
##\ddot \theta=\frac{-2\dot r \dot \theta}{r}##
So,
##\ddot r=\frac{-2\dot r \dot \theta+ 2 \dot r \dot \theta}{\theta}=0##
##\vec a=-const \ddot \theta \hat r = const \frac{ \dot r \dot \theta \hat r} {r}##

So, (a) is correct

Writing Latex is hard...
Thanks for help!

PeroK
Homework Helper
Gold Member
Wow! It's so difficult to know when to stop manipulating the equations...
##\vec a=(\ddot r - r \theta \ddot \theta/2) \hat r##
Now, ##r \theta=## const.
##r\dot \theta + \theta \dot r=0##
##r\ddot \theta +2\dot r \dot \theta + \theta \ddot r=0##
So, ##\ddot r = \frac{-2 \dot r \dot \theta - r \ddot \theta}{\theta}##

Also, since tangential part of acceleration is zero,
##\ddot \theta=\frac{-2\dot r \dot \theta}{r}##
So,
##\ddot r=\frac{-2\dot r \dot \theta+ 2 \dot r \dot \theta}{\theta}=0##
##\vec a=-const \ddot \theta \hat r = const \frac{ \dot r \dot \theta \hat r} {r}##

So, (a) is correct

Writing Latex is hard...
Thanks for help!
You still have ##\dot \theta## in that answer!

You still have ##\dot \theta## in that answer!
Oh no...

So I used ##\dot \theta r +\theta \dot r=0##
To get
##\vec a = \frac{-\dot r^2 \theta}{r^2}\hat r##
And so
##\vec a = \frac{-\dot r^2 const}{r^3} \hat r##

So option C is correct?

And also, why leave answer in ##\dot r##?
I could convert it in terms of ##\dot \theta## and have only r in numerator.

How do you know which one to choose?

PeroK
Homework Helper
Gold Member
Oh no...

So I used ##\dot \theta r +\theta \dot r=0##
To get
##\vec a = \frac{-\dot r^2 \theta}{r^2}\hat r##
And so
##\vec a = \frac{-\dot r^2 const}{r^3} \hat r##

So option C is correct?

And also, why leave answer in ##\dot r##?
I could convert it in terms of ##\dot \theta## and have only r in numerator.

How do you know which one to choose?
You already showed that ##\ddot r = 0##, so ##\dot r## is constant.

Kaguro

Wow...
You're great!

PeroK
I take issue with the selection of answers provided in the problem statement. Allow me to offer my analysis of this problem using Lagrangian dynamics. The Lagrangian for the system is\begin{align*} \mathcal{L}=\frac{1}{2}mv^2-V(r,\theta)\\ v^2=\dot{x}^2+\dot{y}^2 \end{align*}
where m is the mass of the particle, ##v## is the velocity and ##V(r,\theta)## is the potential energy. In our two dimensional Cartesian coordinate system,\begin{align*} x=r\cos(\theta)\\ y=r\sin(\theta)\\ \dot{x}=\dot{r}\cos(\theta)-r\dot{\theta}sin(\theta)\\ \dot{y}=\dot{r}\sin(\theta)+r\dot{\theta}sin(\theta)\\ v^2=\dot{r}^2+r^2\dot{\theta}^2 \end{align*}
and the Lagrangian becomes,$$\mathcal{L}=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r,\theta)$$with the prescription,\begin{align*} \frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial{\dot{r}}}=\frac{\partial{\mathcal{L}}}{\partial{r}}\\ \frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}}=\frac{\partial{\mathcal{L}}}{\partial{\theta}} \end{align*}
we get the following equations,\begin{align*} m(\ddot{r}-r\dot{\theta}^2)=-\frac{\partial{V(r,\theta)}}{\partial{r}}=F_r\\ mr^2\ddot{\theta}^2=-\frac{\partial{V(r,\theta)}}{\partial{\theta}}=F_{\theta}=0 \end{align*}From the problem statement,$$\theta=\frac{c}{r}$$ where ##c## is a constant and thus,\begin{align*} \dot{\theta}=\frac{-c\dot{r}}{r^2}\\ \ddot{\theta}=2c\frac{\dot{r}^2}{r^3}-\frac{c\ddot{r}}{r^2}=0\\ \ddot{r}=2\frac{\dot{r}^2}{r}=\dot{r}\frac{d\dot{r}}{dr}\\ \int_{\dot{r}_0}^{\dot{r}_f}\frac{d\dot{r}}{\dot{r}}=2\int_{r_0}^{r_f}\frac{dr}{r} \end{align*}and thus,
\begin{align*} \dot{r}=br\\ b=2\frac{\dot{r}_0}{r_0}\\ \ddot{r}=b\dot{r}=b^2r \end{align*}
Plugging the result into the formula for ##F_r## we get,$$F_r=mb^2r(1-\frac{2c^2}{r^2})$$

Last edited:
PeroK
PeroK
Homework Helper
Gold Member
we get the following equations,\begin{align*} m(\ddot{r}-r\dot{\theta}^2)=-\frac{\partial{V(r,\theta)}}{\partial{r}}=F_r\\ mr^2\ddot{\theta}^2=-\frac{\partial{V(r,\theta)}}{\partial{\theta}}=F_{\theta}=0 \end{align*}
You should check that second equation. It's missing the term ##2m \dot r \dot \theta##.

You should check that second equation. It's missing the term ##2m \dot r \dot \theta##.
Your right PeroK! It should read,$$mr^2\ddot{\theta}+ 2mr\dot{r}\dot{\theta}=-\frac{\partial{V(r,\theta)}}{\partial{\theta}}=F_{\theta}=0$$which gives,$$\ddot{\theta}=-2\frac{\dot{r}\dot{\theta}}{r}$$We also have,\begin {align*} \theta=\frac{c}{r}\\ \dot{\theta}=\frac{-c\dot{r}}{r^2}\\ \ddot{\theta}=\frac{2c\dot{r}^2}{r^3}-\frac{c\ddot{r}}{r^2} \end {align*}and thus\begin {align*} \ddot{\theta}=-2\frac{\dot{r}\dot{\theta}}{r}=-2\frac{\dot{r}}{r}\frac{c\dot{r}}{r^2}=\frac{2c\dot{r}^2}{r^3}-\frac{c\ddot{r}}{r^2}\\ 4\frac{c\dot{r}^2}{r^3}=\frac{c\ddot{r}}{r^2}\\ \frac{4}{r}=\frac{\ddot{r}}{\dot{r}^2}\\ \ddot{r}=\dot{r}\frac{d\dot{r}}{dr}\\ 4\int_{r_0}^{r_f}\frac{dr}{r}=\int_{\dot{r}_0}^{\dot{r}_f}\frac{d\dot{r}}{\dot{r}}\\ \dot{r}= br^4\\ b=\frac{r_0^4}{\dot{r}_0} \end {align*}
Plugging into ##F_r##$$F_r=m(\ddot{r}-r\dot{\theta}^2)=4mb^2r^7(1-\frac{c^2}{r^2})$$
I think I got it right this time...whatever...I found this problem interesting.

PeroK
PeroK
Homework Helper
Gold Member
That looks worse!

pasmith
Homework Helper
Your right PeroK! It should read,$$mr^2\ddot{\theta}+ 2mr\dot{r}\dot{\theta}=-\frac{\partial{V(r,\theta)}}{\partial{\theta}}=F_{\theta}=0$$which gives,$$\ddot{\theta}=-2\frac{\dot{r}\dot{\theta}}{r}$$We also have,\begin {align*} \theta=\frac{c}{r}\\ \dot{\theta}=\frac{-c\dot{r}}{r^2}\\ \ddot{\theta}=\frac{2c\dot{r}^2}{r^3}-\frac{c\ddot{r}}{r^2} \end {align*}and thus\begin {align*} \ddot{\theta}=-2\frac{\dot{r}\dot{\theta}}{r}=-2\frac{\dot{r}}{r}\frac{c\dot{r}}{r^2}=\frac{2c\dot{r}^2}{r^3}-\frac{c\ddot{r}}{r^2} \end{align*}
You have made a sign error: from $$r^2 \ddot \theta + 2r \dot r \dot \theta = 0$$ you obtain $$\ddot \theta = - 2\frac{ \dot r \dot \theta}{r}.$$ Substituting $\dot \theta = -C\dot r/ r^2$ you should have obtained $$\ddot \theta = \frac{2C \dot r^2}{r^3}.$$ You, however, have an additional minus sign here.

With the correct sign, comparing with $\ddot \theta = \frac{d}{dt}(-C\dot r /r^2)$ you should conclude $$\frac{2C\dot r ^2}{r^3} = -\frac{C\ddot r}{r^2} + \frac{2C \dot r ^2}{r^3}$$ and thus $$0 = \frac{C\ddot r}{r^2}.$$

\begin{align*} 4\frac{c\dot{r}^2}{r^3}=\frac{c\ddot{r}}{r^2}\\ \frac{4}{r}=\frac{\ddot{r}}{\dot{r}^2}\\ \ddot{r}=\dot{r}\frac{d\dot{r}}{dr}\\ 4\int_{r_0}^{r_f}\frac{dr}{r}=\int_{\dot{r}_0}^{\dot{r}_f}\frac{d\dot{r}}{\dot{r}}\\ \dot{r}= br^4\\ b=\frac{r_0^4}{\dot{r}_0} \end {align*}
Plugging into ##F_r##$$F_r=m(\ddot{r}-r\dot{\theta}^2)=4mb^2r^7(1-\frac{c^2}{r^2})$$
I think I got it right this time...whatever...I found this problem interesting.

haruspex