Motion in 2 dimensions

[aronclark1017]

TL;DR Summary: Will motion in y direction eventually stop

If an object moves with initial velocity in both x and y direction and acceleration in only the x direction. If the angle of movement with the x axis will eventually become zero does that mean the object will eventually not move in the y direction.

 

[PeroK]

In that scenario, the component of velocity in the y direction remains constant. It would not reduce to zero.

 

[kuruman]

To illustrate @PeroK's response.

Think of a rock projected at an angle from the edge of a cliff. In the absence of air resistance, it will maintain its horizontal velocity because there is no acceleration in that direction. As time goes by, the velocity will come closer to the vertical but never exactly along the vertical. In other other words, the path, which is a parabola, will never become a straight vertical line.

 

[aronclark1017]

TL;DR Summary: Will motion in y direction eventually stop

If an object moves with initial velocity in both x and y direction and acceleration in only the x direction. If the angle of movement with the x axis will eventually become zero does that mean the object will eventually not move in the y direction.

View attachment 364746

Hmm yes it seems to me that as the x velocity becomes large that one must imagine this as zooming out on the diagram in such case the y velocity vector would be become invisible but still must exist.

 

[TSny]

The vertical side of each triangle is the y-component of the velocity, which remains constant in time.
The horizontal side is the x-component of velocity, which increases with time.

The constant y-component of velocity represents the distance the object moves in the y-direction during one unit of time. So, even after a long time, the object will move the same distance in the y-direction per unit of time as it did at the beginning of the motion.

 

[aronclark1017]

To illustrate @PeroK's response.

Think of a rock projected at an angle from the edge of a cliff. In the absence of air resistance, it will maintain its horizontal velocity because there is no acceleration in that direction. As time goes by, the velocity will come closer to the vertical but never exactly along the vertical. In other other words, the path, which is a parabola, will never become a straight vertical line.

Is moving in x direction with acceleration, y direction no acceleration

 

[aronclark1017]

Seems to me that there could be trouble here taking the limit as the x velocity becomes large such that the direction of motion in the y direction is neglected. The ratio of Y/X becomes so small as X becomes large that arctan Y//X could assumed to be 0 in some applications but here in this case of real physics it must exist else is big trouble could be

 

[jbriggs444]

Seems to me that there could be trouble here taking the limit as the x velocity becomes large such that the direction of motion in the y direction is neglected. The ratio of Y/X becomes so small as X becomes large that arctan Y//X could assumed to be 0 in some applications but here in this case of real physics it must exist else is big trouble could be

If one takes the limits

$\lim_{t \to \infty} x(t) = \infty$
$\lim_{t \to \infty} y(t) = 1$
$\lim_{t \to \infty} \frac{x(t)}{y(t)} = 0$
$\lim_{t \to \infty} \text{arctan} \frac{x(t)}{y(t)} = 0$

If you then try to play naive arithmetic games with infinities and limits, you get:

$y(\infty) = x(\infty) \tan \theta(\infty) = \infty \times 0$ which is indeterminate.

 

[PeroK]

Seems to me that there could be trouble here taking the limit as the x velocity becomes large such that the direction of motion in the y direction is neglected. The ratio of Y/X becomes so small as X becomes large that arctan Y//X could assumed to be 0 in some applications but here in this case of real physics it must exist else is big trouble could be

Look at it this way. If you consider a graph of you against the other eight billion on Earth, then it would be impossible to see you on the graph. But, that doesn't mean you don't exist.

 

[kuruman]

Is moving in x direction with acceleration, y direction no acceleration

So just swap x and y. The point of this is that the velocity changes only in the one direction but stays constant and non-zero in the other direction. Look at the figure on the right. It shows the direction of the velocity vector in equal time intervals starting with v₀ at t = 0.

Seems to me that there could be trouble here taking the limit as the x velocity becomes large such that the direction of motion in the y direction is neglected.

There is no trouble. Do you really think that as time increases, the velocity vector could be along the vertical x- axis only? If that were the case, this would mean that the velocity component in the y-direction will, at some point, have to go to zero which cannot happen because the acceleration is zero along that direction. If the velocity in the y-direction is, say, 2 m/s one cannot argue that 2 is equal to zero for small values of 2!

(Edited for clarity)

 

[gmax137]

I posted a few minutes ago, and just deleted that post. I haven't had my coffee yet. Maybe I will come back here later. Sorry for any confusion!

 

[haruspex]

such that the direction of motion in the y direction is neglected

Whether a feature can be neglected depends on the purpose of the analysis. If you want to know the direction in the limit then the y velocity can be neglected. If you want to know whether some y value will ever be reached it cannot.

 

[wrobel]

Spoiler: offtop by association

Looking through this thread I remember a problem about circular motion. Let a point move along a circle of radius r such that the magnitude $a>0$ of its acceleration is a constant. Find a time and arc length in which the speed of the point becomes a constant. Initial velocity equals zero.

 

[haruspex]

Spoiler: offtop by association

Looking through this thread I remember a problem about circular motion. Let a point move along a circle of radius r such that the magnitude $a>0$ of its acceleration is a constant. Find a time and arc length in which the speed of the point becomes a constant. Initial velocity equals zero.

Involves $\Gamma(1/4)$, right?

 

[TSny]

Involves $\Gamma(1/4)$, right?

Yes, I agree. And the arc length comes out nice.

 

[JimWhoKnew]

Spoiler: offtop by association

Looking through this thread I remember a problem about circular motion. Let a point move along a circle of radius r such that the magnitude $a>0$ of its acceleration is a constant. Find a time and arc length in which the speed of the point becomes a constant. Initial velocity equals zero.

If my calculations are correct, it is possible to construct infinitely many answers to that problem (as stated). Including "never".

 

[haruspex]

If my calculations are correct

I suggest using the subjunctive here.

 

[Gavran]

If an object moves with initial velocity in both x and y direction and acceleration in only the x direction. If the angle of movement with the x axis will eventually become zero does that mean the object will eventually not move in the y direction.

$$ \lim_{x\to\infty}\frac1x=0 $$ does not mean that $1$ approaches $0$ as $x$ approaches $\infty$.

 

[wrobel]

If my calculations are correct, it is possible to construct infinitely many answers to that problem (as stated). Including "never".

Bring the proofs please.

 

[JimWhoKnew]

Bring the proofs please.

Since this problem came about as an off topic, I feel uncomfortable to hijack the thread. Even more so when the OP places a picture of a young boy. I apologize.

I wanted to hide the calculations under a "spoiler" button, but can't because of the bug.

With the usual$$x=r\cos\theta(t)\quad,\quad y=r\sin\theta(t)$$the condition $$\ddot{x}^2+\ddot{y}^2=a^2$$becomes$$\ddot{\theta}^2+\dot{\theta}^4=\frac{a^2}{r^2}:= b^2 \quad .$$I don't know all the solutions to this equation, but one solution that satisfies the initial condition, can be given in terms of the lemniscate sine:$$\dot{\theta}(t)=\sqrt{b}\cdot\mathrm{sl}(t~\sqrt{b})\quad.$$$\mathrm{sl}(x)~$ is a periodic function on the real axis, so with this solution the speed $\left|\dot{\theta}\right|$ never becomes constant.
At the extrema of $\dot{\theta}$ , $~\ddot{\theta}=0$ and I get $~\vec{a}=-a\hat{r}~$ . This is the familiar result for constant circular motion. So for every $n$ in $\mathbb{N}$ , we can construct a new solution where the speed becomes constant, by$$\dot{\theta}(t)=\begin{cases}
\sqrt{b}\cdot\mathrm{sl}(t~\sqrt{b}) \; , & 0\le t \le (n-1/2)\varpi/\sqrt{b}\\
\sqrt{b}\cdot\mathrm{sl}([n-1/2]\varpi) \; , & (n-1/2)\varpi/\sqrt{b}<t\\
\end{cases} \qquad.$$Note that the second line is just $~\pm\sqrt{b}$ . In simple words: when $~\dot{\theta}~$ reaches our chosen extremum, we fix the direction of $~\vec{a}~$ to satisfy $~\vec{a}=-a\hat{r}~$ at all following time.

Acceptable?

(What is your solution?)

 

[haruspex]

Since this problem came about as an off topic, I feel uncomfortable to hijack the thread. Even more so when the OP places a picture of a young boy. I apologize.

I wanted to hide the calculations under a "spoiler" button, but can't because of the bug.

With the usual$$x=r\cos\theta(t)\quad,\quad y=r\sin\theta(t)$$the condition $$\ddot{x}^2+\ddot{y}^2=a^2$$becomes$$\ddot{\theta}^2+\dot{\theta}^4=\frac{a^2}{r^2}:= b^2 \quad .$$I don't know all the solutions to this equation, but one solution that satisfies the initial condition, can be given in terms of the lemniscate sine:$$\dot{\theta}(t)=\sqrt{b}\cdot\mathrm{sl}(t~\sqrt{b})\quad.$$$\mathrm{sl}(x)~$ is a periodic function on the real axis, so with this solution the speed $\left|\dot{\theta}\right|$ never becomes constant.
At the extrema of $\dot{\theta}$ , $~\ddot{\theta}=0$ and I get $~\vec{a}=-a\hat{r}~$ . This is the familiar result for constant circular motion. So for every $n$ in $\mathbb{N}$ , we can construct a new solution where the speed becomes constant, by$$\dot{\theta}(t)=\begin{cases}
\sqrt{b}\cdot\mathrm{sl}(t~\sqrt{b}) \; , & 0\le t \le (n-1/2)\varpi/\sqrt{b}\\
\sqrt{b}\cdot\mathrm{sl}([n-1/2]\varpi) \; , & (n-1/2)\varpi/\sqrt{b}<t\\
\end{cases} \qquad.$$Note that the second line is just $~\pm\sqrt{b}$ . In simple words: when $~\dot{\theta}~$ reaches our chosen extremum, we fix the direction of $~\vec{a}~$ to satisfy $~\vec{a}=-a\hat{r}~$ at all following time.

Acceptable?

(What is your solution?)

Ok, I do see now how multiple solutions arise.
At any instant, we have $\dot\omega^2=b^2-\omega^4$, giving two solutions for $\dot\omega$. So we could arbitrarily choose the positive or negative solution at any time. If we ever let $b^2=\omega^4$ then the system behaviour is locked.

So that permits extra solutions in which $\dot\omega$ is discontinuous, but it would seem that if we require it to be continuous there is a unique solution, once the initial sign of the tangential acceleration is chosen. How does that square with the lemniscate sine solution? What am I missing?

Pedantic note: the problem statement is “Find a time and arc length in which the speed of the point becomes a constant.”, so a solution to the equation which does not reach a locked state is not a solution to the problem.

 

[JimWhoKnew]

but it would seem that if we require it to be continuous there is a unique solution

Please post your solution, and how you derived it.

 

[wrobel]

The point of JimWhoKnew can be illustrated by the following simple analogy.
Consider an initial value problem
$\dot x^2+x^2=1,\quad x(0)=0,\quad t\ge 0.$
It has a solution $x(t)=\sin t$ for all $t\ge 0$;
and another solution
$x(t)=\sin t$ for $t\in[0,\pi/2]$ and $x(t)=1$ for $t>\pi/2$
etc
All the solutions are of $C^1[0,\infty)$

 

[haruspex]

Please post your solution, and how you derived it.

It's not based on a solution, merely the fact that an explicit equation like $\dot\omega=\sqrt{b^2-\omega^4}$ (positive square root implied), together with the initial condition $\omega=0$, should guarantee unique development.
Moreover,
1) $\omega$ must be non-decreasing and bounded above by $\sqrt b$.
2) writing $\omega=\sqrt b-x$ yields $\dot x=-\sqrt{4xb^{\frac 32}+O(x^2)}$, $\Delta t=\int_0^{\sqrt b}\frac{dx}{\sqrt{4xb^{\frac 32}+O(x^2)}}<\frac 1{\sqrt b}$

 

[haruspex]

The point of JimWhoKnew can be illustrated by the following simple analogy.
Consider an initial value problem
$\dot x^2+x^2=1,\quad x(0)=0,\quad t\ge 0.$
It has a solution $x(t)=\sin t$ for all $t\ge 0$;
and another solution
$x(t)=\sin t$ for $t\in[0,\pi/2]$ and $x(t)=1$ for $t>\pi/2$
etc
All the solutions are of $C^1[0,\infty)$

Ok, so it is the difference between a solution that satisfies a differential equation and one that corresponds to a physical process which the differential equation represents. If $\dot x=f(x)$ is cause and effect then once $x$ satisfies $f(x)=0$ then it is stuck there.
This never struck me before.

So do we conclude that the solutions @JimWhoKnew found are the same as alluded to in posts #14 and #15, but then retrace by effectively reversing the flow of time?

 

[JimWhoKnew]

and one that corresponds to a physical process which the differential equation represents.

I think that as a physical process, the problem is under-determined. The constancy of $~\left|\vec{a}\right|~$ constraint, necessitates the application of a time-varying external force, compatible with the circularity constraint. But there is still some residual freedom left, which may be related to the direction of $\vec{a}$ and to its higher order derivatives.
Consider, for example, the $~n=1~$ case of #21 when $~\dot{\theta}~$ reaches $~\sqrt{b}~$ (is it the same as your and @TSny's solution, hinted in #14 and #15 ?). If I'm in control of the external force, I can choose whether to shut it down at that instant, or to go for some more rounds on the roller-coaster. These are distinct (valid) physical processes, which are compatible with the presentation of the problem.

The point of @JimWhoKnew can be illustrated by the following simple analogy...

That was the guiding idea, but in this specific case one also has to verify the continuity of $\vec{a}$ , which does not vanish when $\ddot{\theta}$ does.

Edit: rephrased some sentences

 

[TSny]

I think that as a physical process, the problem is under-determined. The constancy of $~\left|\vec{a}\right|~$ constraint, necessitates the application of a time-varying external force, compatible with the circularity constraint. But there is still some residual freedom left, which may be related to the direction of $\vec{a}$ and to its higher order derivatives. Consider, for example, the $~n=1~$ case of #21 when $~\dot{\theta}~$ reaches $~\sqrt{b}~$ (is it the same as your and @TSny's solution, hinted in #14 and #15 ?). If I'm in control of the external force, I can choose whether to shut it down at that instant, or to go for some more rounds on the roller-coaster. These are distinct (valid) physical processes, which are compatible with the presentation of the problem.

Very nice explanation! I think it gets to the heart of the matter.

By choosing the tangential force appropriately as a function of time, we can switch back and forth between oscillatory motion and uniform circular motion while maintaining constant acceleration magnitude.

This problem reminds me of Norton's Dome . [edit: Also here] A discussion of this interesting problem, which emphasizes different choices for the force as a function of time, is presented in this video.

 

[JimWhoKnew]

If $\dot x=f(x)$ is cause and effect then once $x$ satisfies $f(x)=0$ then it is stuck there.
This never struck me before.

I don't know what you mean by "cause and effect", but as in #23
$\dot x^2+x^2=1,\quad x(0)=0,\quad t\ge 0.$
has a solution $x(t)=\sin t$ for all $t\ge 0$, and also
$\dot{x}=\pm\sqrt{1-x^2}$
at $~t=\pi/2\mp\varepsilon$ . Obviously $x(t)$ doesn't get stuck at $x=1$ .

The resolution comes from higher order derivatives. Suppose $~\dot x=f(x)~$ and $f(x(t_0))=0$ . Then \begin{align}
x(t_0+\varepsilon) & =x(t_0)+\varepsilon f(x(t_0))+\varepsilon^2 \ddot{x}(t_0)/2+\dots \nonumber\\
& =x(t_0)+\varepsilon^2 \ddot{x}(t_0)/2+\dots \nonumber
\end{align}and$$\dot{x}(t_0+\varepsilon) =f(x(t_0+\varepsilon))=f(x(t_0)+\varepsilon^2 \ddot{x}(t_0)/2+\dots) \quad.$$So although the contribution from the higher order derivatives is extremely tiny, it can get $\dot{x}$ out of the zero trap.

Late edit: corrected equation

 

[DaveC426913]

the vertical but never exactly along the vertical. In other other words, the path, which is a parabola, will never become a straight vertical line.

*cough* ellipse segment! *cough*

 

[JimWhoKnew]

But there is still some residual freedom left, which may be related to the direction of $\vec{a}$ and to its higher order derivatives.

I realize now that if $~\dot{\theta}(0)=0$ , then $~\vec{a}(0)~$ must be tangential. So the residual freedom seems to be very limited.

 

[haruspex]

I realize now that if $~\dot{\theta}(0)=0$ , then $~\vec{a}(0)~$ must be tangential. So the residual freedom seems to be very limited.

Yes, as I wrote in post #21, you can (only) elect to switch the sign of the tangential acceleration arbitrarily, but if you choose not to you must arrive at $\dot{\omega}=0$.

I don't know what you mean by "cause and effect",

As in, the given differential equation accurately represents the dynamics, so applying it in a simulation should produce the same result.
But here we do not know the basis of the equation, and indeed it cannot represent dynamics because the quadratic in the differential equation has two solutions.

but as in #23
$\dot x^2+x^2=1,\quad x(0)=0,\quad t\ge 0.$
has a solution $x(t)=\sin t$ for all $t\ge 0$, and also
$\dot{x}=\sqrt{1-x^2}$
near the maximum. But obviously $x(t)$ doesn't get stuck at $x=1$ .

The SHM analogy is interesting. In the first order ODE form it has the same issue: two solutions, and simulation would not discover the cyclic nature. But the usual second order ODE has neither of those problems. The algebraic process of deriving the first order from the second order involves multiplying by $\dot x$, thus creating the "stuck" solution.

The resolution comes from higher order derivatives.

Yes, that dawned on me overnight. E.g. if we require $\ddot\omega$ to be continuous then we only get the cyclic solution.