# Centrifugal acceleration over a changing curve?

1. Jul 4, 2010

### 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.

Best

Az

Here 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

2. Jul 4, 2010

### HallsofIvy

The "velocity" vector is, of course, the derivative of the position vector with respect to time. The "acceleration" vector is the derivative of the velocity vector with respect to time. Resolve that acceleration vector in to a component in the direction of the tangent to the curve at that point and a component perpendicular to the tangent. The component perpendicular to the tangent to the curve, multiplied by the mass, is the "centrifugal force" vector.

The component parallel to the tangent is the one responsible for changing the speed of the object.

3. Jul 4, 2010

### azaharak

Yes I know this very well, that is how one can derive the centrifugal force around a circle.

The problem is determining what the centrifugal force should be over a particular curve, say y=x^2, given that you know your speed while traversing the curve.