Curvilinear Motion of a particle

  • #1
3,003
2

Homework Statement


A particle moves along the curve [itex]y(x)=x^2-4[/itex] 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
[tex]|a|=\sqrt{a_n^2+a_t^2}[/tex]



The Attempt at a Solution


I have assumed that 'constant speed' means [tex]v_t=5 \Rightarrow a_t=0[/tex]

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

Thanks,
Casey
 

Answers and Replies

  • #2
3,003
2
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 [tex]a_n=\frac{[1+\frac{dy}{dx}^2]^{3/2}}{\frac{d^2y}{dx^2}}[/tex] 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.
 
  • #3
3,003
2
Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?
 
  • #4
3,003
2
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)
 
  • #5
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
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.
 
Last edited:
  • #6
3,003
2
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)?
 
  • #7
3,003
2
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...
 
  • #8
662
0
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?
 
  • #9
3,003
2
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.
 
  • #10
662
0
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 ... :uhh:
 
  • #11
3,003
2
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?
 
  • #12
Can I jump in? :smile:
First of all constant velocity means [itex]u^2=u_x^2+u_y^2=C[/itex]. Thus first take the derivative of [itex]y(t)=x(t)^2-4[/itex] to find the relation bewteen [itex]u_x,\,u_y[/itex] and plug it in [itex]u^2[/itex]
 
  • #13
3,003
2
Can I jump in? :smile:
First of all constant velocity means [itex]u^2=u_x^2+u_y^2=C[/itex]. Thus first take the derivative of [itex]y(t)=x(t)^2-4[/itex] to find the relation bewteen [itex]u_x,\,u_y[/itex] and plug it in [itex]u^2[/itex]
You lost me. I am given y as a function of x... not time. i'll look it over some more, though.
 
  • #14
The [itex]y,\,x[/itex] are actually [tex]y(t),\,x(t)[/itex]. Can you take the derivative of [itex]y(t)=x(t)^2-4[/itex] with respect to [itex]t[/itex]?

Hint. [itex]u_x=\dot{x}(t),u_y=\dot{y}(t)[/itex]
 
  • #15
3,003
2
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?
 
  • #16
Ok! :smile:

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

[tex]u^2=u_x^2+u_y^2[/tex]

Do you follow up to now?
 
  • #17
3,003
2
Ok! :smile:

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.
 
  • #18
The partile moves on the parabola [itex]y=x^2-4[/itex] doesn't it?
That means that it's coordinates [itex](x,y)[/itex] are not constant thus they vary with time. Correct?
 
  • #19
3,003
2
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 [tex]\frac{d}{dx}(y)=v(x)[/tex] and

[tex]v=\frac{dx}{dt}[/tex]

I can do this: [tex]\frac{dy}{dt}=\frac{dx}{dt}*\frac{dy}{dx}[/tex]?

Or am getting close?
 
  • #20
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 [itex]H[/itex] with horizontial velocity [itex]u_o[/itex] then it will moves both horizontal and vertical, i.e. [itex]x(t),y(t)[/itex] and the orbit of the stone is
[tex]y=H-\frac{g}{2\,u_o^2}\,x^2[/tex]
Your "stone" is moving on the orbit
[tex]y=x^2-4[/tex]

Is it clear now?
 
  • #21
3,003
2
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?
 
  • #22
Let's take two points of the parabola, e.g. [tex](x_1,y_1)=(0,-4),\,(x_2,y_2)=(1,-3)[/tex]. Let the particle is at the 1st point at some time [tex]t_1[/tex], isn't it obvious that it would be at the 2nd point at a different time [tex]t_2[/tex]?
 
  • #23
3,003
2
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.
 
  • #24
3,003
2
Anyway, I am off to bed. thanks for your help thus far Rainbow!! I do appreciate it!

Casey
 
  • #25
Ok! I will post some hints for your original post, and I hope they will help
 

Related Threads on Curvilinear Motion of a particle

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
12K
  • Last Post
Replies
1
Views
502
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
5
Views
470
Replies
2
Views
562
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
Top