Trajectory of a particle in polar coordinates

[whatever itis]

Homework Statement

This is a problem from Matveev's Mechanics.
A particle moves in a plane with constant radial acceleration $a$, and a normal acceleration $2v\omega$, where $v$ is its velocity and $\omega$ a positive constant. Taking the direction of acceleration at the origin of coordinates as the polar axis, derive the equation for the particle's trajectory.

Relevant Equations

$$\ddot r-r\dot \theta^2=a$$
$$r\ddot \theta +2\dot r\dot\theta=2v \omega$$
Where $v$ would be
$$\sqrt{\dot r ^ 2 +(r\dot \theta)^2}$$

I tried using the formula for acceleration in polar coordinates, but I don't know how to solve the differential equations.

How do I solve them? Is there a simpler way to do the problem?

 

[erobz]

Homework Statement: This is a problem from Matveev's Mechanics.
A particle moves in a plane with constant radial acceleration $a$, and a normal acceleration $2v\omega$, where $v$ is its velocity and $\omega$ a positive constant. Taking the direction of acceleration at the origin of coordinates as the polar axis, derive the equation for the particle's trajectory.
Relevant Equations: $$\ddot r-r\dot \theta^2=a$$
$$r\ddot \theta +2\dot r\dot\theta=2v \omega$$
Where $v$ would be
$$\sqrt{\dot r ^ 2 +(r\dot \theta)^2}$$

I tried using the formula for acceleration in polar coordinates, but I don't know how to solve the differential equations.

How do I solve them? Is there a simpler way to do the problem?

Substitute $\dot \theta = \omega = \text{const.}$, what does that imply for its derivative $\ddot \theta = \dot \omega$?

 

[Charles Link]

I worked on it, and I think I got it about as far it will go in a differential equation for $r$. I first eliminated $v$ from the 3 equations, and wrote $\dot{\theta}=u$ to simplify things, and eliminated the $u$, but getting something very messy in the differential equation for $r$ by itself. It involves, among other things, the third derivative of $r$ (w.r.t. time) to the second power.

I don't agree with @erobz above. In this problem $\omega$ is not $\dot{\theta}$. (Edit: at least it's not obvious to me that $\dot{\theta}$ is constant.)

 

[erobz]

I worked on it, and I think I got it about as far it will go in a differential equation for $r$. I first eliminated $v$ from the 3 equations, and wrote $\dot{\theta}=u$ to simplify things, and eliminated the $u$, but getting something very messy in the differential equation for $r$ by itself. It involves, among other things, the third derivative of $r$ (w.r.t. time) to the second power.

I don't agree with @erobz above. In this problem $\omega$ is not $\dot{\theta}$. (Edit: at least it's not obvious to me that $\dot{\theta}$ is constant.)

They did a heck of a good job picking a parameter name if it isn’t. ( why! )

For what it’s worth…I just get a radial trajectory if it is.

 

[whatever itis]

@erobz Thanks,I guess I should have taken that hint. The solution $r=a(1- \cos \omega t)/ \omega^2$ works, which agrees with the answers.

 

[erobz]

@erobz Thanks,I guess I should have taken that hint. The solution $r=a(1- \cos \omega t)/ \omega^2$ works, which agrees with the answers.

It's interesting you get that by solving $\ddot r - r \omega^2 - a =0$?

But for the tangential equation I get $0 = ( r \omega )^2$ after I simplify which would imply $r = 0$ ($\omega$ constant). So I don't know how that all fits together...

Perhaps that is why @Charles Link suspects differently.

 

[Charles Link]

I think your differential equations could use a little work. The particular solution to $\ddot{r}-\omega^2 r=a$ is $r=-a/\omega^2$, and the homogeneous solution is $r=A \exp(\omega t )+ B \exp(- \omega t )$ with $A$ and $B$ to be determined. (Note that $a$ might be negative). (and the $\omega^2 r$ needs to have a plus sign in front of it to get $\cos(\omega t )$ and $\sin(\omega t)$ in the homogeneous solution).

@erobz has it correct that with $\dot{\theta}=\omega$ that $\ddot{\theta}=0$ with $\omega$ constant, so that $r=0$, unless $\omega=0$.

I am led to believe the book's answer may be in error.

 

[Charles Link]

It may interest you, for $\omega=0$, with $\ddot{r}=a$, the particular solution is $r=\frac{at^2}{2}$, and the homogeneous solution is $r=Ct+D$. This is the well-known solution of $r=\frac{at^2}{2}+v_o t + r_o$.

This turns out to be consistent with the above solution with finite $\omega$, if you take the limit as $\omega \rightarrow 0$ and expand the $\exp(\omega t)$ and $\exp(-\omega t)$ in a power series. Third order and higher power terms in $\omega$ wind up going to zero. You then can solve for $A$ and $B$ to give you $C$ and $D$ for the homogeneous terms that are basically $v_o$ and $r_o$.

 

[kuruman]

The solution $r=a(1- \cos \omega t)/ \omega^2$ works, which agrees with the answers.

It doesn't work. If you put it in the left-hand side of the equation $\ddot r-r\dot \theta^2=a$, you will get $-a=a$ because the double differentiation of the cosine brings out a negative sign up front. Using a hyperbolic cosine does work. You can also see that hyperbolic sines and cosines are involved from the homogeneous solution provided by @Charles Link in post #7 that has $r \sim \exp(\pm\omega t )$ and not $r \sim \exp(\pm i \omega t )$

 

[kuruman]

It may interest you, for $\omega=0$, with $\ddot{r}=a$, the particular solution is $r=\frac{at^2}{2}$, and the homogeneous solution is $r=Ct+D$.

This turns out to be consistent with the above solution with finite $\omega$, if you take the limit as $\omega \rightarrow 0$ and expand the $\exp(\omega t)$ and $\exp(-\omega t)$ in a power series. Third order and higher power terms in $\omega$ wind up going to zero. You then can solve for $A$ and $B$ to give you $C$ and $D$ for the homogeneous terms that are basically $v_o$ and $r_o$.

The homogeneous solution in post #7 is what you get for a bead of mass $m$ on a frictionless rod rotating with constant angular velocity $\omega$. If you place this in a uniform gravitational field $a$ that rotates with the rod, you get this problem. Turn off the rotation and you have the free-falling bead.

 

[Charles Link]

@kuruman If you look at my post 3, you may find it of interest. What I got for the differential equation for this problem for $r$ did not have $\dot{\theta}$ as a constant, and to solve it was well beyond anything that could be routinely solved, except for perhaps numerical evaluation in a very advanced class.

 

[kuruman]

@kuruman If you look at my post 3, you may find it of interest. What I got for the differential equation for this problem for r did not have θ˙ as a constant, and to solve it was well beyond anything that could be routinely solved, except for perhaps numerical evaluation in a very advanced class.

When I first looked at this problem I had your reaction and I echo @erobz's lament in post #4. There is ambiguity in the interpretation of the information given in the statement of the problem regarding the term $2\omega v$.

Is $\omega$ just the name of a constant that has nothing to do with $\dot{\theta}$? Is it coincidental that it has the same dimensions as $\dot{\theta}$? If it is just a constant unrelated to $\dot{\theta}$, why bother with the factor of $2$ and not just absorb it within this constant? Then there is $v$ which is referred to as the "velocity", a vector, but that's probably bad word choice when the problem was translated from Russian.

I tried to gauge the level of difficulty of this textbook and it looks like it's appropriate for an Intermediate Mechanics course for second or maybe third year undergraduate students who've had some diff eq experience. So I agree with you assessment that the hard core interpretation is not the case here and chalk it up as a poorly formulated problem statement.

 

[whatever itis]

@erobz @kuruman @Charles Link Thanks for help you all

 

[Charles Link]

So I agree with you assessment that the hard core interpretation is not the case here and chalk it up as a poorly formulated problem statement.

I think they might have miscalculated and thought one of the homogeneous solutions is $\cos(\omega t )$ when it is not. In addition, they seemed to not check for the inconsistencies that result if you assume $\dot{\theta}=\omega$ is a constant.

 

[vela]

You can look through the book at archive.org. It's problem 2.15.

https://archive.org/details/MatveevMechanicsAndTheoryOfRelativity/page/n75/mode/2up

The answer in the book isn't a function of time. It's actually $r=a(1-\cos\varphi)/\omega^2$, which described a cardioid. In the text, he defines normal acceleration as the component of acceleration perpendicular to the trajectory, so I think it's incorrect to identify it with $a_\theta$.

 

[Charles Link]

@vela Very helpful, but with your additions, I haven't yet been able to generate the new answer given in the text.

 

[vela]

Using Mathematica, I calculated the various components of the acceleration with the book's solution for $r$ and got
\begin{align*}
a_t &= \frac{a}{\omega^2}\left(\dot{\theta}^2\cos\frac \theta 2 + 2 \ddot{\theta}\sin\frac \theta 2\right) \\
a_n &= \frac{a}{\omega^2}\left(3\dot{\theta}^2 \sin\frac \theta 2\right) \\
a_r &= \frac{a}{\omega^2}\left[\dot{\theta}^2(2\cos \theta-1) + \ddot{\theta}\sin \theta\right] \\
a_\theta &= \frac{a}{\omega^2}\left[2\dot{\theta}^2 \sin \theta+ \ddot{\theta}(1-\cos\theta) \right]
\end{align*} where $a_t$ and $a_n$ are the components respectively tangential and normal to the trajectory and $a_r$ and $a_\theta$ are the components in the $\hat r$ and $\hat \theta$ directions.

The speed is $v = 2\left(\tfrac{a}{\omega^2}\sin\tfrac\theta 2\right)\dot{\theta}.$ If $a_n = 2v\omega$, it follows that $\dot{\theta} = \tfrac 43\omega > 0$. But then $a_r$ isn't constant.

 

[Charles Link]

See https://www.physicsforums.com/threa...ch-result-in-the-cornu-spiral-comment.902917/
This is a problem that I invented of my own and solved a number of years ago. I was thinking this one might have a similar type of solution, but so far I haven't found it. I tried the book's answer, but so far I have been unsuccessful at verifying it or disproving it.

 

[renormalize]

But then $a_r$ isn't constant.

Using Mathematica, I have also confirmed @vela 's result that the problem statement and the purported solution are incompatible. The plot of $r\left(\varphi\right)=a\left(1-\text{cos}\varphi\right)/\omega^2$ for $a=\omega=1$ shows that $r$ is clearly bounded:

But the condition on the radial acceleration is $\ddot{r}-r\dot{\varphi}^2=a\Rightarrow\ddot{r}=a+r\dot{\varphi}^{2}$ where $a$ is a positive constant. Integrating this twice w.r.t $t$ gives $r\left(t\right)=\frac{1}{2}at^{2}+v_{0}t+r_{0}+\intop^{t}dt_{2}\intop^{t_{2}}dt_{1}r\left(t_{1}\right)\dot{\varphi}^{2}\left(t_{1}\right)$. Note that $r\left(t\right)$ grows at least as fast as $t^2$ when $t\rightarrow\infty$, and possibly faster depending upon the behavior of the non-negative integrand $r\left(t_{1}\right)\dot{\varphi}^{2}\left(t_{1}\right)$. Thus, $r\left(t\right)$ grows without bound for large $t$, which shows that the problem statement is inconsistent with the stated solution.