Polar acceleration problem

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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?
 

Answers and Replies

  • #2
PeroK
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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##?
 
  • #3
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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:biggrin:

Writing Latex is hard...
Thanks for help!
 
  • #4
PeroK
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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:biggrin:

Writing Latex is hard...
Thanks for help!
You still have ##\dot \theta## in that answer!
 
  • #5
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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?
 
  • #6
PeroK
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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.
 
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Wow...
You're great!:bow:
 
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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})$$
 
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  • #9
PeroK
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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##.
 
  • #10
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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.
 
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  • #11
PeroK
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That looks worse!
 
  • #12
pasmith
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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 [tex]
r^2 \ddot \theta + 2r \dot r \dot \theta = 0[/tex] you obtain [tex]
\ddot \theta = - 2\frac{ \dot r \dot \theta}{r}.[/tex] Substituting [itex]\dot \theta = -C\dot r/ r^2[/itex] you should have obtained [tex]
\ddot \theta = \frac{2C \dot r^2}{r^3}.[/tex] You, however, have an additional minus sign here.

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

$$\begin{align*}
4\frac{c\dot{r}^2}{r^3}=\frac{c\ddot{r}}{r^2}\\
\frac{4}{r}=\frac{\ddot{r}}{\dot{r}^2}\\
\ddot{r}=\dot{r}\frac{d\dot{r}}{dr}\\
4\int_{r_0}^{r_f}\frac{dr}{r}=\int_{\dot{r}_0}^{\dot{r}_f}\frac{d\dot{r}}{\dot{r}}\\
\dot{r}= br^4\\
b=\frac{r_0^4}{\dot{r}_0}
\end {align*}$$
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.
 
  • #13
haruspex
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since tangential part of acceleration is zero,
Be careful with the terminology.
You are given that the transverse component is zero, i.e. the ##\hat\theta## component, also known as the circumferential component. It is normal to the radial component.
The tangential component is in the direction of the velocity and is normal to the centripetal component.
 

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