A particle moving in a parabolic path in the ##x-y## plane

[brotherbobby]

Homework Statement

A particle moves in the $x-y$ plane with constant acceleration $a_0$ directed along the negative y–axis. The equation of motion of the particle has the form $y=px-qx^2$ where $p$ and $q$ are positive constants. Find the velocity of the particle at the origin of coordinates.

Relevant Equations

1. If $\dot x = v_x=\dfrac{dx}{dt}$ is the $x$ component of velocity and $\dot y = v_y=\dfrac{dy}{dt}$ the $y$ component of velocity, then the (scalar) velocity or speed at a point is given by $v = \left( \dot x^2+\dot y^2\right)^{\frac{1}{2}}$.
2. Acceleration components $a_x=\ddot x$ and $a_y=\ddot y$.
3. Acceleration can also be expressed as a function of position, for instance, $a_x=v_x\dfrac{dv_x}{dx}$.

Problem statement : I copy and paste the problem as it appears in the text down below. I have only changed the symbol of the given acceleration from $a\rightarrow a_0$, owing to its constancy.

Attempt : I must admit that I could proceed very little.
Given $a_0=\dfrac{d^2y}{dt^2}\Rightarrow \dot y = a_0t+b$. But the constant $b$ is unknown.
Likewise, it is also given that $\dfrac{d^2x}{dt^2}=0\Rightarrow \dot x = c$, but the constant $c$ is unknown.

Differentiating the path of the particle $y = px-qx^2$ on both sides with respect to time, $\dot y = p\dot x-2qx\dot x\Rightarrow a_0t+b = pc-2qcx$.
This equation can yield the position $x$ as a function of the unknowns $b,c$.
Substituting for $x$ in the equation of the curve can yield $y$ again as a function of the same unknowns.

I can't see how to progress from here.

Request : A suggestion as to how to proceed to solve the problem will be welcome.

 

[PeroK]

Isn't that just projectile motion?

 

[brotherbobby]

Yes. But how to go about finding the velocity at the origin? The problem is set in a twisted way.

 

[PeroK]

You can write down $x$ and $y$ as functions of $t$ to get you started.

 

[Orodruin]

You already know that $x = v_x t$. This means that any derivatives with respect to $t$ may be rewritten as derivatives with respect to $x$ as $df/dt = v_x (df/dx)$. For the second derivative you would have $d^2f/dt^2 = v_x^2 (d^2f/dx^2)$. Use these relations for $f = y$ and evaluate at the origin. Poff, two unknowns ($v_x$ and $v_y$) and two equations.

 

[PeroK]

Hint. Take $t =0$ when the particle is at the origin.

 

[brotherbobby]

Hint. Take $t =0$ when the particle is at the origin.

How do we know that?

 

[brotherbobby]

You already know that $x = v_x t$. This means that any derivatives with respect to $t$ may be rewritten as derivatives with respect to $x$ as $df/dt = v_x (df/dx)$. For the second derivative you would have $d^2f/dt^2 = v_x^2 (d^2f/dx^2)$. Use these relations for $f = y$ and evaluate at the origin. Poff, two unknowns ($v_x$ and $v_y$) and two equations.

Yes, let me try.

 

[Orodruin]

How do we know that?

It is a freedom of choice. You can call whatever time you wish $t=0$.

 

[PeroK]

How do we know that?

We get to choose when to set our stopwatch to zero!

 

[Orodruin]

Hint. Take $t =0$ when the particle is at the origin.

However, I suggest not working with $t$ at all as mentioned above.

 

[erobz]

However, I suggest not working with $t$ at all as mentioned above.

The velocity of the particle at $t=0$ can be fully represented in terms of just $p$ and $q$? I'm having a mind block getting rid of $\dot x = c$ in the final result. (not trying to intrude, just looking for a "yes" or "no")

 

[PeroK]

The velocity of the particle at $t=0$ can be fully represented in terms of just $p$ and $q$? I'm having a mind block getting rid of $\dot x = c$ in the final result. (not trying to intrude, just looking for a "yes" or "no")

The answer should drop out from either method, although it's not a particularly enlightening problem, IMO.

 

[Orodruin]

The velocity of the particle at $t=0$ can be fully represented in terms of just $p$ and $q$? I'm having a mind block getting rid of $\dot x = c$ in the final result. (not trying to intrude, just looking for a "yes" or "no")

Yes, as long as acceleration is known.

 

[brotherbobby]

Hint. Take $t =0$ when the particle is at the origin.

Had a small question before I can attempt.

If indeed we are free to choose when the particle is at the origin (say at $t=0$), do we have a similar choice for a component of the velocity? I mean, can we instead choose $v_y(t=0)=0$?
Can we do so simultaneously $\rightarrow v_y=y=0\quad \text{at}\; t=0$?
We remember that $a_y\ne0$, so the above requirement will not force the particle to "sit" at the origin.

However, $a_x=0$ for the problem above. Hence we cannot choose $v_x(t=0)=x(t=0)=0$, for that will mean that the particle has no $x-$ displacement.
But that means we can make one of them zero, either $v_x$ or $x$ at time $t=0$.

Am I correct in my assertion above?

 

[PeroK]

Had a small question before I can attempt.

If indeed we are free to choose when the particle is at the origin (say at $t=0$), do we have a similar choice for a component of the velocity? I mean, can we instead choose $v_y(t=0)=0$?

Yes, but not both at the same time. Also, you have to know that $v_y = 0$ at some time. In general. it might never be zero.

Can we do so simultaneously $\rightarrow v_y=y=0\quad \text{at}\; t=0$?

Definitely not!

We remember that $a_y\ne0$, so the above requirement will not force the particle to "sit" at the origin.

However, $a_x=0$ for the problem above. Hence we cannot choose $v_x(t=0)=x(t=0)=0$, for that will mean that the particle has no $x-$ displacement.

$v_x = 0$ is a degenerate case, which you may wish to consider.

But that means we can make one of them zero, either $v_x$ or $x$ at time $t=0$.

$v_x$ depends in general on $p$ and $q$.

 

[brotherbobby]

Yes, but not both at the same time. Also, you have to know that vy=0 at some time. In general. it might never be zero.

Making $y=0$ at some time $t=0$ involves a translation of coordinates. Which is valid.
Why can't we do the same for $v_y$? Give our frame a boost equal to $v_y$ at time $t=0$ so that the particle has $v_y=0$ at time $t=0$.
Let me draw a projectile motion to help you see my point, mistaken though I might well be.

The path of the projectile is shown in red. At some point P, the projectile clearly as its position $y\ne 0$ and y-velocity $v_y\ne 0$. So I propose making a translation of the origin to some $O'$ such that $y'=0$. Likewise I propose giving this frame a boost equal to the particle's y-velocity $v_y$ such that the particle has velocity $v_y'=0$.(We'd discuss the situation with the $x-$components a bit later, let's focus on the $y-$ones for now)

 

[PeroK]

Making $y=0$ at some time $t=0$ involves a translation of coordinates. Which is valid.
Why can't we do the same for $v_y$? Give our frame a boost equal to $v_y$ at time $t=0$ so that the particle has $v_y=0$ at time $t=0$.

Yes, but the particle is no longer at the origin at $t = 0$. You can't transalte the coordinates without affecting the equation you were given: $y = px - qx^2$. That's a coordinate dependent equation.

 

[brotherbobby]

I think this is a very interesting point, and I am prepared to wait before solving the problem.

Yes, but the particle is no longer at the origin at $t = 0$. You can't transalte the coordinates without affecting the equation you were given: $y = px - qx^2$. That's a coordinate dependent equation.

We are given the equation of the curve along which the particle moves : $y = px-qx^2$. Nowhere is time mentioned. But the frame (of reference) is given - the particle clearly moves through the origin as $(0,0)$ is a solution to the curve.

Now comes the choice.

I choose $t=0$ when the particle is at $(0,0)$. I likewise choose $v_y=0$ at the time $t=0$. You might say that this gives the frame of reference a boost - but why am I not free to do that?

 

[Steve4Physics]

How about this...

Treat the system as a simple projectile launched from the origin $(0, 0)$ at $t=0$ ($a_0$ corresponds to $g$).

The apex of the trajectory (point P) is $(d, H)$ and the projectile reaches it at time T.

1) $y=px+qx^2$: find the value of $x$ which maximises $y$ (simple calculus). This is $d$. The corresponding maximum value of $y$ is $H$. Note that $d$ and $H$ are expressed in terms of $p$ and $q$..

2)The vertical component of velocity changes from its initial value $u_{0,y}$ to zero when $y$ changes from $0$ to $H$. That means $u_{0,y}$ can be found using a standard kinematics (‘suvat’) equation; then $T$ can be found. Note that $u_{0,y}$ and $T$ are expressed in terms of $p, q$ and $a_0$.
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3) It’s now simple to find $u_{x,0}$ and hence $u$ (in terms of $p, q$ and $a_0$).

 

[PeroK]

I choose $t=0$ when the particle is at $(0,0)$. I likewise choose $v_y=0$ at the time $t=0$. You might say that this gives the frame of reference a boost - but why am I not free to do that?

Because now you've set $v_y = 0$ at the origin. You can check that this is invalid by looking for a counterexample. Any projectile motion with an initial velocity at the origin would do that. You can check that this satifies an equation of the form $y = px - qx^2$. $v_y$ would be zero at the apex: $(x_1, y_1)$. If we make this the origin, with $x' = x - x_1, \ y' = y - y_1$ we get the new equation:
$$y' + y_1 = p(x' + x_1) - q(x' + x_1)^2$$And this equation is not of the form $y' = px' - qx'^2$.

Whereas, a time translation $t' = t - t_0$ does not change the equation of the trajectory in the $x-y$ plane. So, you can take $t = 0$ at the origin without changing the equation you were given to work with.

 

[Orodruin]

Meanwhile I still would suggest not bothering to use the time parameter at all. It is a red herring.

 

[brotherbobby]

Thank you at @PeroK . I get your point.
But then it implies that the equation of the curve, viz. $y=px-qx^2$ is true only for a given (spatial) origin. We are really only free to choose time as we wish. A given $v_0$ at $(0,0)$ is non-negotiable.
Imagine I wanted to make the translation to some point $P (\ne O)$ along the trajectory $x\rightarrow x'=x+a,\quad y\rightarrow y'=y+b$. Here $(a,b)$ are the coordinates of $P$.
The new equation will not be of the form $y'=px'-qx'^2$, as you said. However, doesn't it seem counter-intuitive?
Surely, the particle has some velocity $v(P)$ at $P$. Could I not argue that the particle was actually launched from the point $P$ with the initial velocity $v(P)$? It will be the same curve, surely : $y=px-qx^2$. And yet, when we transform coordinates as I just did to the new origin $P$, it has a different equation!

 

[PeroK]

Thank you at @PeroK . I get your point.
But then it implies that the equation of the curve, viz. $y=px-qx^2$ is true only for a given (spatial) origin.

Precisely!

We are really only free to choose time as we wish. A given $v_0$ at $(0,0)$ is non-negotiable.
Imagine I wanted to make the translation to some point $P (\ne O)$ along the trajectory $x\rightarrow x'=x+a,\quad y\rightarrow y'=y+b$. Here $(a,b)$ are the coordinates of $P$.
The new equation will not be of the form $y'=px'-qx'^2$, as you said. However, doesn't it seem counter-intuitive?
Surely, the particle has some velocity $v(P)$ at $P$. Could I not argue that the particle was actually launched from the point $P$ with the initial velocity $v(P)$? It will be the same curve, surely : $y=px-qx^2$. And yet, when we transform coordinates as I just did to the new origin $P$, it has a different equation!

It can't be the same equation for a translation of your $x, y$ coordinates. You can see that in general for any graph in the $x-y$ plane.

 

[brotherbobby]

Precisely!

It can't be the same equation for a translation of your $x, y$ coordinates.

I get that, But "see" the curve for a minute. It is identical with $P$ as the origin as it is with $O$ as the origin.
If the curve is the same, why does it have a different equation in $x'$ and $y'$ when both the coordinate axes are parallel to the original $x-y$ axes?
Same curve has a different equation upon change of coordinate origin? So it seems to be.

 

[PeroK]

Same curve has a different equation upon change of coordinate origin? So it seems to be.

It must. Even the equation of a straight line changes if you translate the $x$ and/or $y$ coordinates.

 

[nasu]

Yes. But how to go about finding the velocity at the origin? The problem is set in a twisted way.

Do you know how the equation of trajectory looks like for projectile motion? What are the p and q constants in it?

 

[brotherbobby]

It must. Even the equation of a straight line changes if you translate the $x$ and/or $y$ coordinates.

I agree. In $y=mx+c$, $c$ is well-known to be the $y-$intercept. "Move" the coordiates a distance of $c$ vertically and the equation of the same line is $y'=mx'$. Rotate the axes and even $m'$ will change.

However, I think the present conundrum requires a mathematical working out, not least because of its non-intuitive nature. I draw below for you what I have in mind - but briefly, it is this :

"If I transform the equation of the trajectory for a projectile, viz. $y = x\tan\theta-\dfrac{gx^2}{2v_0^2\cos^2\theta}$ where the quantities have their usual meanings, from the present origin $O$ to a new origin $O'(a,b)$ along the trajectory, and suitably use the new angle to the horizontal $\theta' (<\theta_0)$ and new initial velocity $v'(<v_0)$, will it alter the form of the equation from $y'=x'\tan\theta'-\dfrac{gx'^2}{2v'^2\cos^2\theta'}$?"Of course my intuition tells me it shouldn't, wrong though I know it is. The drawing below shows my confusion:

We have for the new velocity $v'=(v'^2_x+v'^2_y)^{\frac{1}{2}}$ where $v'_x=v_{0x}=v_0\cos\theta_0$ and $v'_y=(v^2_{0y}-2gb)^{\frac{1}{2}}$ and $v_{0y}=v_0\sin\theta_0$.

Likewise for the new angle $\tan\theta'=\dfrac{v'_y}{v'_x}$.

Already I can see your point, in that there is a $b$ of the new coordinate origin appearing in the equation but nowhere the $a$ of the new origin. Is it possible that the $b's$ will cancel out with one another, or with the $a's$ where they are related by $b = a\tan\theta_0-\dfrac{ga^2}{2v^2_0\cos^2\theta_0}$? I have to calculate to find out.

 

[brotherbobby]

Do you know how the equation of trajectory looks like for projectile motion? What are the p and q constants in it?

Sure. $p=\tan\theta_0$ and $q = \dfrac{g}{2v_0^2\cos^2\theta_0}$

 

[nasu]

Then what else do you need to find velocity at time t=0? What is the $v_0$ in the expression for q?

 

[brotherbobby]

Then what else do you need to find velocity at time t=0? What is the $v_0$ in the expression for q?

I copy and paste my working below.

This is the answer from the book, but it makes reference to the details of projectile motion, so perhaps less suitable. I suspect the way to the answer is @Orodruin 's method in post#5, which I will do in a moment.
As you see in the posts above, I am stuck with a conceptual problem - namely, should the equation of the trajectory of a projectile change if one were to effect a change in its coordinates to some point along the trajectory?

 

[PeroK]

As you see in the posts above, I am stuck with a conceptual problem - namely, should the equation of the trajectory of a projectile change if one were to effect a change in its coordinates to some point along the trajectory?

Imagine you fire a projectile straight up and want the speed at the origin, which was the original starting point. First, you move the origin to the highest point, where the speed is zero and that's your answer?

 

[brotherbobby]

Imagine you fire a projectile straight up and want the speed at the origin, which was the original starting point. First, you move the origin to the highest point, where the speed is zero and that's your answer?

I am sure you phrased the question incorrectly. Can you rephrase it for me?

 

[PeroK]

I am sure you phrased the question incorrectly. Can you rephrase it for me?

I think you should focus on the problem you started with. As a separate exercise you might want to study some graphs and look at how their equations change when you translate the $x$ and $y$ coordinates.

If you don't like taking $t = 0$ at the origin, then take $t = t_0$ at the origin. Or, use the alternative method using derivatives, where you don't need to introduce the time parameter.

In any case, you don't need the parameter $\theta$. That unnecessarily complicates the working. You are already given $p$ and $q$. That's what you should be working with.

 

[brotherbobby]

You already know that $x = v_x t$. This means that any derivatives with respect to $t$ may be rewritten as derivatives with respect to $x$ as $df/dt = v_x (df/dx)$. For the second derivative you would have $d^2f/dt^2 = v_x^2 (d^2f/dx^2)$. Use these relations for $f = y$ and evaluate at the origin. Poff, two unknowns ($v_x$ and $v_y$) and two equations.

Can you check the last line of your statement @Orodruin ?

This is what I am getting.

I have an added term for equation $2$. I wonder if I am mistaken.