Inward Acceleration for Ununiform Slope

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Homework Help Overview

The discussion revolves around finding the apparent instantaneous radius "r" of an object moving along a non-uniform slope, specifically described by the function y=x^2-x. Participants are exploring the concept of inward acceleration in relation to this path.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the nature of the acceleration being discussed, questioning whether it refers to normal acceleration or another component. There are discussions about the relationship between tangential speed and inward acceleration, as well as the implications of the non-circular nature of the path.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's assumptions. Some guidance has been offered regarding the components of acceleration, but no consensus has been reached on the specific approach to determine the radius "r" for the given path.

Contextual Notes

There is an emphasis on the complexity of the path being non-circular, which may affect the calculations and assumptions regarding acceleration. Participants are also considering the definitions and components of acceleration in their discussions.

Mechanics
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I started doing some physics today for the first time since school got out a few months ago. The period of inactivity was painfully apparent throughout the attempt.

I'm trying to find the apparent instantenous "r" of an object moving along an ununiform slope where "r" is the radius of a circle that would have the same inward acceleration. Example of possible path: y=x^2-x.
 
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I assume you are referring to the radius of a circle that would have the same acceleration as described by the equations of rotational motion? Is the example function representative of the displacement the velocity or the acceleration?
 
The function describes the vertical displacement, y, compared to the horizontal displacement, x.
 
Would'nt it be

[tex]r = \frac{V_{tan}^2}{a_{in}}[/tex]

where one uses the tangential speed in the calculation?
 
Last edited:
I'm trying to find the equivalent "r" in terms of x and y for the acceleration. Yes, if it was a circle the acceleration would be v^2/r...but this surface we're talking about is not as nice as a circle.
 
What is this "inward" acceleration you are talking about? I thought you were talking about the normal acceleration component. The other component that goes with it is the tangential acceleration component. That is the components normal and tangential to the object's path.

The normal acceleration component is directed towards the inner curvature of the path and is given by

[tex]a_n = \frac{v^2}{\rho}[/tex]

where [tex]\rho[/tex] is the curvature of the path. Do you need to determine it for a given path?
 

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