Curvilinear Motion of a particle

  • #26
Now let's take the derivate of your equation, remembering that
[tex]u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t}[/tex] and [tex]u=C[/tex], i.e.

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

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

[tex]\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[/tex]

Thus the particle's velocity would be

[tex]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)}[/tex]
and the accerelation

[tex]a^2=a_x^2+a_y^2\Rightarrow a^2=\dots[/tex]

From the last equation you will find [itex]a[/itex] with respect to [itex]x[/itex] thus you are ready to find the maximum value.
 
  • #27
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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.
 
  • #28
Integral
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Form the position vector:
[tex] R= x(t) \hat{i} + y(t) \hat{j} [/tex]
then get the velocity vector:
[tex] \dot{R} = \frac {dx(t)} {dt} \hat{i} + \frac {dy(t)} {dt} \hat{J} [/tex]

you are given that [tex]|\dot{R}|[/tex] is constant (5[itex] \frac m s [/itex])

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

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

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.
 
  • #29
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Form the position vector:
[tex] R= x(t) \hat{i} + y(t) \hat{j} [/tex]
then get the velocity vector:
[tex] \dot{R} = \frac {dx(t)} {dt} \hat{i} + \frac {dy(t)} {dt} \hat{J} [/tex]

you are given that [tex]|\dot{R}|[/tex] is constant (5[itex] \frac m s [/itex])

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

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

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
 
  • #30
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Dummy variable? What dummy variable, I did not introduce any dummy variables, I expressed the problem in terms of, x,y and t.
 
  • #31
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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.
 
  • #32
Hootenanny
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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;

[tex]y\left(x(t)\right) = x^2(t)-4[/tex]

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.
 
  • #33
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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;

[tex]y\left(x(t)\right) = x^2(t)-4[/tex]

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

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

be written as [itex]y(x(t))=x^2(t)-4[/itex]

Am I correct so far?
 
  • #34
Hootenanny
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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

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

be written as [itex]y(x(t))=x^2(t)-4[/itex]

Am I correct so far?
You are indeed correct. Hence, we can express the position vector;

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

Do you follow, and can you now make the next step?
 
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  • #35
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You are indeed correct. Hence, we can express the position vector;

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

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

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

Thus, [tex]\vec{a}(t)=\frac{d\vec{v}}{dt}=[/tex] Oh God. . something is amiss. . . how do I differentiate dx/dt WRT t ?
 
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  • #36
Hootenanny
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Oh man...

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

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

So your velocity would become;

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

But either notation is equally correct, so just use whichever you feel most comfortable with.
 
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  • #37
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I am thinking product rule is needed for the j component?[tex]\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}[/tex]

Or am I way off the mark now?
 
  • #38
Hootenanny
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I am thinking product rule is needed for the j component?[tex]\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}[/tex]

Or am I way off the mark now?
Nope, your spot on again :approve:! So, now you need to find the maximum of this function...
 
  • #39
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Nope, your spot on again :approve:! 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:redface:
 
  • #40
Hootenanny
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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:redface:
No problem, it's a pleasure :smile:
 
  • #41
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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?
 
  • #42
Hootenanny
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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;

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

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

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

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;

[tex]\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[/tex]

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;

[tex]\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[/tex]
 

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