Acceleration and radius of curvature of a particle

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Discussion Overview

The discussion revolves around finding the acceleration and radius of curvature for a particle defined by parametric equations at a specific time, t=2. It includes aspects of mathematical reasoning and technical explanation related to the motion of the particle.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the parametric equations for the motion of a particle and seeks to find acceleration and radius of curvature.
  • Another participant emphasizes the importance of accounting for the normal unit vector (en) in the calculation of radius of curvature.
  • A different participant shares equations for the unit vectors et and en, suggesting that the angle θ can be used to derive en.
  • There is a discussion about the relationship between the components of acceleration and the normal component equated to v²/ρ.
  • One participant expresses uncertainty about the necessity of dv/dt in the calculations, indicating a potential misunderstanding of its role.
  • A later reply indicates that a solution was found by determining en as the unit vector perpendicular to et, and using the dot product to find the normal component of acceleration.

Areas of Agreement / Disagreement

Participants generally agree on the approach to finding the radius of curvature and the role of unit vectors, but there are differing views on the necessity of certain calculations, such as dv/dt. The discussion reflects a collaborative effort to clarify the problem without reaching a definitive consensus on all aspects.

Contextual Notes

Some participants express confusion regarding the role of unit vectors and the calculations involved, indicating potential limitations in their understanding of the mathematical framework. The discussion also highlights the dependency on specific definitions and interpretations of the equations presented.

ehilge
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Homework Statement


The motion of a particle is defined as:

x=[(t-4)3/6]+t2

y=t3/6-(t-1)2/4

Find the acceleration and radius of curvature at t=2

Homework Equations



a=(dv/dt)(et)+(v2/[tex]\rho[/tex])(en)

where et and en are the tangential and normal unit vecotrs to the curve and [tex]\rho[/tex] is the radius of curvature.


The Attempt at a Solution



So I was able to get the acceleration by differentiating both of the equations twice and plugging in 2. The x component becomes 0 and the y is +1.5m/s. I'm stumped on the radius though. I differentiated to find dv/dt and v and tried solving for rho but it didn't work. That would have been essentially ignoring et and en since they are unit vectors, so the length is on it they shouldn't affect the outcome, right? Any ideas on how to solve the problem. Also, I think the way the equation is written with the unit vectors in there somewhat confuses me too so if there's anything that could help clear up any confusion with the equation that'd be great too.
Thanks!
 
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You do need to account for those unit vectors, at least the en one since that is where the radius of curvature comes into play.

Think of a as having a normal and tangential component (instead of x and y components). The normal component may be equated to v2/ρ, according to your equation.

If you can come up with an expression for en, that may help get you going.
 
the equations I have in my notes are

en = -sin[tex]\vartheta[/tex]i + cos[tex]\vartheta[/tex]j

et = cos[tex]\vartheta[/tex]i + sin[tex]\vartheta[/tex]j

where [tex]\vartheta[/tex] would be the angle between the tangential vector and the x axis

I worked out [tex]\vartheta[/tex] = 14 by taking the inverse tan (dy/dt)/(dx/dt).
And then in the original equation, v is the length of <dx/dt, dy/dt> at t=2, and then I'm not exactly sure what dv/dt is. I was thinking that would be the final acceleration, but that would be what is on the left side of the equation so that doesn't make sense. Either way, I have feeling there is an easier way to do all this, any thoughts?
 
I'm not sure of an easier way here.

I agree with the θ = 14° result, which can be used to determine en. Can you now find the component of a in the direction of en? That result will be equivalent to v2/ρ, so you could then solve for ρ.

dv/dt will not be needed.
 
ok, I was able to talk with my classmates today and figure out the problem without having to use theta. en can be found by taking the unit vector of the vector perpendicular to et. et is found by differentiating the equation and finding vx and vy. Then a = <0,1.5> is dotted with en to get an. Finally, set an equal to v2/rho and solve.
Thanks for your help!
 
You're welcome, glad it worked out. :smile:
 

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