Curvilinear Motion of a particle

[Saladsamurai]

Homework Statement

A particle moves along the curve y(x)=x^2-4 with constant speed of 5 m/s. Determine the point on the curve where the max magnitude of acceleration occurs and compute its value.

Homework Equations

radius of curvature
|a|=\sqrt{a_n^2+a_t^2}

The Attempt at a Solution

I have assumed that 'constant speed' means v_t=5 \Rightarrow a_t=0

So now I have to find where a_n is maximum. Any hints to get me going here?

Thanks,
Casey

 

[Saladsamurai]

So I got the right answer. But I don't know how to spell out a proper procedure for future problems.

I found y'(x)=0; at x=0.

So x=0 is a critical point. Then I just plugged that into a_n=\frac{[1+\frac{dy}{dx}^2]^{3/2}}{\frac{d^2y}{dx^2}} and that worked.

But do I know that x=0 is a MAXIMUM, if I do the second derivative test, I get positive 2, so I would have thought it was a minimum.

 

[Saladsamurai]

Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?

 

[Saladsamurai]

Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?

But really. Do it. Do it. (Dodgeball reference)

 

[Hootenanny]

You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x'''(t)=0 and not when y''(x)=0.

 

[Saladsamurai]

You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x''(t)=0 and not when y''(x)=0.

I see. So how would I determine, if and where a maximum would be if I am given y(x)?

 

[Saladsamurai]

You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x'''(t)=0 and not when y''(x)=0.

I see. So how would I determine, if and where a maximum would be if I am given y(x)?

Any takers? 'cause I don't know...

 

[belliott4488]

Have you studied parametric equations? I'm guessing that's exactly what you're doing right now, from the looks of this problem.

If you can restate the problem by writing independent eqations for y(t) and x(t) (i.e. parametrically), then you'll be set. The key is that you know how x and y are related, and you know that the magnitude of the velocity (i.e. the speed) is constant. Do you know how to express the velocity vector for parametric equations?

 

[Saladsamurai]

I have actually never studied them. This is for a dynamics course. I have probably studied them, but never heard them worded as "parametric". Give me an example, if you would.

 

[belliott4488]

Oh, then maybe that's not the way to go. It's where you have separate equations for x(t) and y(t) and then you use the chain rule a lot to express dy/dx in terms of dx/dt and dy/dt.

If you haven't seen it, then you probably have to stick with expressing the normal acceleration as a function of the curvature. Do you know how to do that? I'm not sure I remember, myself ...

 

[Saladsamurai]

No...you're right, it does have to do with the chain rule. I am just confused by it...that, is I do not have a nice procedure as to where to start or how to set it up.

I have y as a function of x. Now what?

 

[Rainbow Child]

Can I jump in?
First of all constant velocity means u^2=u_x^2+u_y^2=C. Thus first take the derivative of y(t)=x(t)^2-4 to find the relation bewteen u_x,\,u_y and plug it in u^2

 

[Saladsamurai]

Can I jump in?
First of all constant velocity means u^2=u_x^2+u_y^2=C. Thus first take the derivative of y(t)=x(t)^2-4 to find the relation bewteen u_x,\,u_y and plug it in u^2

You lost me. I am given y as a function of x... not time. i'll look it over some more, though.

 

[Rainbow Child]

The y,\,x are actually y(t),\,x(t)[/itex]. Can you take the derivative of y(t)=x(t)^2-4 with respect to t?<br /> <br /> Hint. u_x=\dot{x}(t),u_y=\dot{y}(t)

 

[Saladsamurai]

I am sorry, I cannot follow any of your notations. I have y(x) not y(t). . . where are you getting t? What is u? Where am I?

 

[Rainbow Child]

Ok!

The partile is moving in the plane x-y. Thus x,y varies with time, i.e. x(t),y(t) and the velocity \vec{u} has two compontents, i.e. u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t} and the magnitude of the velocity is

u^2=u_x^2+u_y^2

Do you follow up to now?

 

[Saladsamurai]

Ok!

The partile is moving in the plane x-y. Thus x,y varies with time
Do you follow up to now?

Not this part. I get all the maths after that, but why do i automatically assume that it varies with time? It says nothing about time.

 

[Rainbow Child]

The partile moves on the parabola y=x^2-4 doesn't it?
That means that it's coordinates (x,y) are not constant thus they vary with time. Correct?

 

[Saladsamurai]

Although I cannot think of a situation in which position is a function independent of time, it is still not clear to me why it is so obvious that x and y DO vary with time.

Are you trying to say that since \frac{d}{dx}(y)=v(x) and

v=\frac{dx}{dt}

I can do this: \frac{dy}{dt}=\frac{dx}{dt}*\frac{dy}{dx}?

Or am getting close?

 

[Rainbow Child]

First of all

Although I cannot think of a situation in which position is a function independent of time, it is still not clear to me why it is so obvious that x and y DO vary with time.

if you throw a stone from height H with horizontial velocity u_o then it will moves both horizontal and vertical, i.e. x(t),y(t) and the orbit of the stone is
y=H-\frac{g}{2\,u_o^2}\,x^2
Your "stone" is moving on the orbit
y=x^2-4

Is it clear now?

 

[Saladsamurai]

No. You merely gave me a specific example of when a particle moves in the x and y direction and its position varies with time.

In general why should always assume that just because a particle moves in the "x,y plane" that its position must vary with time?

 

[Rainbow Child]

Let's take two points of the parabola, e.g. (x_1,y_1)=(0,-4),\,(x_2,y_2)=(1,-3). Let the particle is at the 1st point at some time t_1, isn't it obvious that it would be at the 2nd point at a different time t_2?

 

[Saladsamurai]

Yes it is obvious, but because something is obvious does not necessarily make it true. That is my main point.

If I am given a particle's y(x) position, why should even consider time? Maybe you can't answer my question. . . it is quite possible that my question is absurd, and so am I.

But I just need proof as to why that assumption is correct.

 

[Saladsamurai]

Anyway, I am off to bed. thanks for your help thus far Rainbow! I do appreciate it!

Casey

 

[Rainbow Child]

Ok! I will post some hints for your original post, and I hope they will help

 

[Rainbow Child]

Now let's take the derivate of your equation, remembering that
u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t} and u=C, i.e.

\frac{d\,y(t)}{d\,t}=2\,x(t)\,\frac{d\,x(t)}{d\,t}\Rightarrow u_y=2\,x\,u_x

dropping the t dependence.
Similary taking the derivative once more, in order to find the accerelation

\frac{d\,u_y(t)}{d\,t}=2\frac{d\,x(t)}{d\,t}\,u_x+2\,x\,\frac{d\,u_x(t)}{d\,t}\Rightarrow a_y=2\,u_x^2+2\,x\,a_x

Thus the particle's velocity would be

u^2=u_x^2+u_y^2 \Rightarrow u^2=u_x^2\,(4\,x^2+1) \Rightarrow u_x=\frac{C}{\sqrt(4\,x^2+1)}
and the accerelation

a^2=a_x^2+a_y^2\Rightarrow a^2=\dots

From the last equation you will find a with respect to x thus you are ready to find the maximum value.

 

[belliott4488]

Thanks for taking over here, Rainbow Child ... I got called away to watch a movie (it's nice to have left behind the days of working on homework on Saturday evenings!). I'll check back in tomorrow to see how Saladsamurai is doing, although I think your last post covers things pretty well.

Saladsamurai - were you ever comfortable with the time dependence of x and y? It seemed that you were missing a vital link somewhere in your reasoning.

Basically, you were given an object moving at some speed along a path - that implies dependence on time right from the start. Motion is change in position over time, so it's just a question of how you describe it. You were given the path explicitly (y as a function of x) but weren't given the explicit time dependence of the object's position components (x(t) and y(t)). Nonetheless, you were given the speed, i.e. the magnitude of the velocity vector, and there are generic relations between that quantity and the velocity components. You can relate those to the derivative of y with respect to x by the chain rule, as Rainbow Child outlined.

Anyway, maybe it's all clear by now ... if not, try again.

 

[Integral]

Form the position vector:
R= x(t) \hat{i} + y(t) \hat{j}
then get the velocity vector:
\dot{R} = \frac {dx(t)} {dt} \hat{i} + \frac {dy(t)} {dt} \hat{J}

you are given that |\dot{R}| is constant (5\frac m s)

You need to continue by finding the acceleration and its derivative.

Note that the chain rule states:
\frac {dy} {dt} = \frac {dy} {dx} * \frac{dx} {dt}

rainbows child has been pointing you in the right direction, I am concerned that you cannot follow his notation, you may need to dig out your old calculus book and review the concepts of a position vector and the velocity in multi-variables.

 

[Saladsamurai]

Form the position vector:
R= x(t) \hat{i} + y(t) \hat{j}
then get the velocity vector:
\dot{R} = \frac {dx(t)} {dt} \hat{i} + \frac {dy(t)} {dt} \hat{J}

you are given that |\dot{R}| is constant (5\frac m s)

You need to continue by finding the acceleration and its derivative.

Note that the chain rule states:
\frac {dy} {dt} = \frac {dy} {dx} * \frac{dx} {dt}

rainbows child has been pointing you in the right direction, I am concerned that you cannot follow his notation, you may need to dig out your old calculus book and review the concepts of a position vector and the velocity in multi-variables.

Well, my problem was not so much the calculus that is confusing me (though changing the dummy variable every other post does not help), what was confusing was that the problem never said anything explicitly about time. It wa given as a function of x.

I just wanted to know how time was brought into the picture.

But, I'll just say I get it now and move on and maybe it will make more sense later.

Thanks for your help guys,
Casey

 

[Integral]

Dummy variable? What dummy variable, I did not introduce any dummy variables, I expressed the problem in terms of, x,y and t.

 

[belliott4488]

Casey - the problem involves a moving object, gives its speed as an input variable, and asks about something having to do with acceleration. Time has to enter into it. The tricky part, which we've been trying to help with, is how to express the time dependence (i.e. derivatives) when you're given only y as a function of x.

 

[Hootenanny]

Well, my problem was not so much the calculus that is confusing me (though changing the dummy variable every other post does not help), what was confusing was that the problem never said anything explicitly about time. It wa given as a function of x.

I just wanted to know how time was brought into the picture.

But, I'll just say I get it now and move on and maybe it will make more sense later.

Thanks for your help guys,
Casey

Time is inherent in any kinematics, since by definition acceleration is the time derivative of velocity, which itself is the time derivative of position. The question defines the path which the particle follows, which as rainbow has said is in the x-y plane. However, the path in itself tells you nothing of velocity nor of acceleration, all the path equation tells you is how one spatial variable (y) is related to another spatial variable (x). For example, in your question when x=2, then y=0 or when x=4, y=12. That's all great, but you don't need to know how the spatial variables behave with respect to each other, you need to know how they behave with respect to time; the equation of path itself cannot describe how the particle moves with respect to time, for this you need velocity and acceleration.

In kinematics x, y, z are all functions of t, so your function should actually be written;

y\left(x(t)\right) = x^2(t)-4

If x, y, z were not functions of t, then you wouldn't have a path, because the particle's position would not evolve with time, it would just stay where it was.

As Integral has said, no dummy variable was introduced, the path was simply expressed in terms of two spatial and one temporal variable. I hope this helps clear things up.

 

[Saladsamurai]

Time is inherent in any kinematics, since by definition acceleration is the time derivative of velocity, which itself is the time derivative of position. The question defines the path which the particle follows, which as rainbow has said is in the x-y plane. However, the path in itself tells you nothing of velocity nor of acceleration, all the path equation tells you is how one spatial variable (y) is related to another spatial variable (x). For example, in your question when x=2, then y=0 or when x=4, y=12. That's all great, but you don't need to know how the spatial variables behave with respect to each other, you need to know how they behave with respect to time; the equation of path itself cannot describe how the particle moves with respect to time, for this you need velocity and acceleration.

In kinematics x, y, z are all functions of t, so your function should actually be written;

y\left(x(t)\right) = x^2(t)-4

If x, y, z were not functions of t, then you wouldn't have a path, because the particle's position would not evolve with time, it would just stay where it was.

As Integral has said, no dummy variable was introduced, the path was simply expressed in terms of two spatial and one temporal variable. I hope this helps clear things up.

Okay. I am with you now. So in order to deal with the original question, at what point is acceleration max, I need to take the second derivative wrt time of

y(x)=x^2-4 which as you say, since x must be a function of time, may

be written as y(x(t))=x^2(t)-4

Am I correct so far?

 

[Hootenanny]

Okay. I am with you now. So in order to deal with the original question, at what point is acceleration max, I need to take the second derivative wrt time of

y(x)=x^2-4 which as you say, since x must be a function of time, may

be written as y(x(t))=x^2(t)-4

Am I correct so far?

You are indeed correct. Hence, we can express the position vector;

r(t) = x(t)\hat{i}+y(t)\hat{j} = x(t)\hat{i} + \left(x^2(t) - 4\right)\hat{j}

Do you follow, and can you now make the next step?

 

[Saladsamurai]

You are indeed correct. Hence, we can express the position vector;

r(t) = x(t)\hat{i}+y(t)\hat{j} = x(t)\hat{i} + \left(x^2(t) - 4\right)\hat{j}

Do you follow, and can you now make the next step?

Oh man...

Is it \vec{v}(t)=\frac{d\vec{r}}{dt}=[\frac{dx}{dt}\hat i+2x\frac{dx}{dt}\hat j]\frac{m}{s}?

Thus, \vec{a}(t)=\frac{d\vec{v}}{dt}= Oh God. . something is amiss. . . how do I differentiate dx/dt WRT t ?

 

[Hootenanny]

Oh man...

Is it v(t)=\frac{d\vec{r}}{dt}=[\frac{dx}{dt}\hat i+2x\frac{dx}{dt}\hat j]\frac{m}{s}?

Correct! And can you now find the acceleration...?

Aside:

It is common practice in physics and applied mathematics to use Newton's notation to make things a little easier to read;

\frac{dx}{dt} = \dot{x}\hspace{1cm};\hspace{1cm}\frac{d^2x}{dt^2} = \ddot{x}(t)

So your velocity would become;

\underline{v}(t) = \underline{\dot{r}}(t)= \dot{x}(t)\hat{i}+ 2x\cdot\dot{x}(t)\hat{j}

But either notation is equally correct, so just use whichever you feel most comfortable with.
​

 

[Saladsamurai]

I am thinking product rule is needed for the j component?\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2x}{dt}\hat{i}+[2x*\frac{d^2x}{dt}+2*\frac{dx}{dt}]\hat{j}

Or am I way off the mark now?

 

[Hootenanny]

I am thinking product rule is needed for the j component?\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2x}{dt}\hat{i}+[2x*\frac{d^2x}{dt}+2*\frac{dx}{dt}]\hat{j}

Or am I way off the mark now?

Nope, your spot on again ! So, now you need to find the maximum of this function...

 

[Saladsamurai]

Nope, your spot on again ! So, now you need to find the maximum of this function...

Please tell me that requires further differentiation. . . Wouldn't I now have to find the 1st derivative of this function, set it to zero and solve; then use the 2nd derivative to find concave up/down?

I am going fruit loops here! Thanks Hootenanny from trudging through this with me

 

[Hootenanny]

Please tell me that requires further differentiation. . . Wouldn't I now have to find the 1st derivative of this function, set it to zero and solve; then use the 2nd derivative to find concave up/down?

Indeed you would, but let's just start with the first derivative for now (and don't forget you have a condition on v)

I am going fruit loops here! Thanks Hootenanny from trudging through this with me

No problem, it's a pleasure

 

[Saladsamurai]

Before I go any further here, I have a question. I still do not have x(t). How am I going to cope with that? I have only differentiated the general function of time. I still don't have an explicit function x(t)=something. So where is this all going?

 

[Hootenanny]

As promised I'm back again and after some addition thought (on the train this morning), I offer a few more hints. You have correctly found the acceleration vector;

\underline{a}(t) = \ddot{x}(t)\underline{i} + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]\underline{j}

And you are asked to maximise the magnitude of the acceleration vector;

a(t) = \sqrt{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2}

However, you are asked to find the point on the curve at which maximum acceleration occurs. Therefore, you must maximise a(t) with respect to x(t). Hence, you should solve;

\frac{d}{dx}a(t) = \frac{d}{dx}\sqrt{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2} := 0

I would also be useful to note that the point at which a(t) is maximal coincides with the point at which the function under the root is maximal (although the value of the maximum may be different). Do you see why? Hence, you can find the point on the curve at which the magnitude of the acceleration is maximal by solving;

\frac{d}{dx}a^2(t) = \frac{d}{dx}\left\{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2\right\} := 0