Curvilinear Motion

Saladsamurai

1. The problem statement, all variables and given/known data
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.



2. Relevant equations radius of curvature
[tex]|a|=\sqrt{a_n^2+a_t^2}[/tex]



3. 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

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 [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.

Saladsamurai

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

Saladsamurai

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

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

Saladsamurai

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 [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]

Saladsamurai

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 [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]

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. [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?

Saladsamurai

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.