Is Radial Acceleration Always Downward in Circular Motion?

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Homework Help Overview

The discussion revolves around the concept of radial acceleration in circular motion, particularly in the context of a train on a curved track. Participants are exploring the implications of forces acting on the train as it approaches the point of leaving the tracks.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between normal force and radial acceleration, questioning the signs of acceleration and the implications of forces in a free body diagram (FBD). There are attempts to clarify the direction of radial acceleration and its relationship to gravity and normal force.

Discussion Status

The discussion is ongoing, with participants raising questions about the interpretation of the problem and the physical setup. Some guidance has been offered regarding the direction of forces and accelerations, but there is no explicit consensus on the correct approach or interpretation of the problem.

Contextual Notes

Participants express uncertainty about the problem's wording, particularly regarding the scenario described as the "bottom of a hill" and its implications for the train's motion. There are indications of missing information that may affect the analysis.

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Jason03 said:
Heres the problem I was working on

http://i14.photobucket.com/albums/a322/guitaristx/train.jpg


Heres what I came up with so far...
http://i14.photobucket.com/albums/a322/guitaristx/trainwork.jpg

I feel like I am missing some numbers...any suggestions?

Consider the situation when the train is just about to "fly off" and leave the tracks. In that case, the normal force on the train is zero. This should alllow you to solve for rho.

EDIT: Question b does not make sense to me...What do they mean by "bottom of a hill"? Maybe we are to assume that we have now a track curved upward with the same radius?
 
ok i got

A = N-g

A = 0 - 9.8m/s^2

so

r = 106^2/-9.8 = -1148 m

does that make sense for the first question?
 
Jason03 said:
ok i got

A = N-g

A = 0 - 9.8m/s^2

so

r = 106^2/-9.8 = -1148 m

does that make sense for the first question?

Well a radius can't be negative.

Be careful, you are using

\sum F_y = m a_y

so your a_y should be - v^2/ r not v^2/r.
Other than that it looks fine
 
how is Velocity negative when gravity is acting downward in the y direction?
 
when you solve A-sub-n you get -9.8m/s^2 however A-sub-n in the FBD is acting upward. So how do you know which sign to use?
 
Jason03 said:
how is Velocity negative when gravity is acting downward in the y direction?

I never said that a velocity is negative. I am talking about the y component of the acceleration. The radial (or centripetal) acceleration a_rad always points to ward the center of the circle. In your example, the center of the circle is below the train. Therefore

a_y = - a_{Rad} = - \frac{v^2}{R}
 
ok I see what your saying, than my FBD would be wrong? I have the A-rad pointing upward.

In terms of Normal and Tangential accerlation, when I draw the FBD if the normal Accerleration is upward, does that mean the Normal Force is also upwards?
 
Jason03 said:
ok I see what your saying, than my FBD would be wrong? I have the A-rad pointing upward.

In terms of Normal and Tangential accerlation, when I draw the FBD if the normal Accerleration is upward, does that mean the Normal Force is also upwards?

I am not sure what you mean by "normal acceleration". If we have circular motion, there is a radial accceleration and there is a tangential acceleration. The radial acceleration is toward the center of the circle. So downward in your diagram. The normal force is always perpendicualr and away from the surface so it's upward.
 

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