Custom Problem Centrifugal Acceleration with varying radius

[azaharak]

Hi... I have kinda made this problem up myself. How would one calculate the effect of the centrifugal force over a curve. For instance the following problem is taken from Physics GRE test GR8677 Question 6, but I have modified the question.

The question states.

A particle is initially at rest at the top of a curved frictionless track. The x and y coordinates of the track are related in dimensionless units by y=(x^2)/4, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration?

The solutions is gx/sqrt(x^2+4), which is easily obtained by differentiating y with respect to x to determine the slope of the tangent line, also related to tan(theta). Now you know both the opp and adj sides, so you can form sin(theta) from this giving you the answer.Ok this is the tangential acceleration, but there must be some centrifugal acceleration which points perpendicular to the tangent line, and should decay because the radial length is getting larger. Any ideas in quantifying what the strength of the centrifugal acceleration is in terms of the position...

Or maybe, just determining the effective radius as a function of position.Thanks in advance.

Best

AzHere is my thinking. Similar to the derivation of the centrifugal acceleration, forgot about the fact that the blocks speed is increasing for the moment. I simply want to calculate the varying centrifugal force associated with the curved path. Consider a small time delta t, the vector change in velocity between time 0 and time dt should should point normal toward the surface. The speed we are assuming is the same, so the angle between the two velocity vectors is delta (theta).

Using the law of cosines, I can relate the change in the velocity vector (Delta V)^2= V^2 + V^2 - 2V^2 cos(delta theta), which obviously simplifies. Now I get that (Delta V) is 2*V*sqrt(1-cos(deta theta))

i have some ideas where to go next but I'm debating they will lead anywhere.

thanks again

Best

Az

 

[Cruikshank]

I calculated the radius of curvature, used conservation of energy to get v as a function of x, and simply used (v^2)/r. Unless I am missing a subtlety, you could imagine joining an osculating circle to the parabola at the object's current location; it would then be pursuing circular motion. I see no reason why entering the circular portion of track would cause a sudden change in acceleration.

 

[azaharak]

I calculated the radius of curvature, used conservation of energy to get v as a function of x, and simply used (v^2)/r. Unless I am missing a subtlety, you could imagine joining an osculating circle to the parabola at the object's current location; it would then be pursuing circular motion. I see no reason why entering the circular portion of track would cause a sudden change in acceleration.

Well... I guess your considering the original problem with gravity, that's the only way you could have determined v as a function of x. I was considering the problem without gravity assuming a constant velocity.

I'm not sure how you determine the radius of curvature since it should be increasing with x. There should be an acceleration that points perpendicular to the curve.

 

[aim1732]

For all it's worth:

Instantaneous radius of curvature of a trajectory:
ρ = [ 1+ (dy/dx)²]^1.5 /[d²y/dx²]

 

[Cruikshank]

I did assume gravity. Now I assume you mean constant speed, since constant velocity would not stay on the parabola. The previous comment has the curvature equation I used to give the radius of curvature as a function of x. With constant speed you simply put in (v^2)/r, unless I am missing something else.

 

[azaharak]

For all it's worth:

Instantaneous radius of curvature of a trajectory:
ρ = [ 1+ (dy/dx)²]^1.5 /[d²y/dx²]

Thanks aim, I think I figured it out last night, i was too tired to comment. Here is my reasoning.

Imaging traveling the following path, y=-sqrt(R^2 -x^2) +R, which is a curve starting from the origin that mimics a circle. Obviously here the radius of curvature is R. I seek to describe this in terms of the tangent line or tangent angle.

dy/dx=Tan(theta) = x*(R^2-x^2)^(-1/2)

If you solve for R in terms of everything else, R=xsqrt(1+cot(theta)^2)=x/sin(theta)

so this leads me to believe that the radius of curvature is given by R=x/sin(theta)

In terms of the original problem, dy/dx=x/2=tan(theta)

so sin(theta)=x/sqrt(x^2+4)

and using my formula, R= x/sqrt(x^2/ (x^2+4) )= sqrt(x^2+4)

It turns out that is is exactly what you get when you use your formula (aim). I hope this it not a mere coincidence.
----------------

As for Cruikshank, I'm sorry I mean the magnitude of the velocity which is the speed.

But you must understand that the radius is changing. V^2/R works fine but R is not constant.

Thanks everyone.---------------------------

Oops I spoke too soon. The two results are not the same, unless that power from (aim's) formula should be 1/2 instead of 3/2

sorry

 

[aim1732]

Here's the link to the formula:
http://en.wikipedia.org/wiki/Radius_of_curvature_(applications)"