Kinematics of a Particle Tangent-Normal components

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SUMMARY

The discussion focuses on the kinematics of a particle, specifically analyzing the tangent-normal components of acceleration. The key equations used include a = (v^2/p) en + (v*) et, where 'a' represents acceleration, 'v' is velocity, 'en' is the inward normal vector, 'et' is the unit vector along the tangent, 'v*' is the tangential acceleration, and 'p' denotes curvature. The participant concludes that since the velocity is constant, the tangential acceleration is 0.4g, and they must adjust the curvature by subtracting the car's dimensions to account for the center of gravity being above the road surface.

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  • Understanding of kinematic equations and their components
  • Familiarity with vector notation in physics
  • Knowledge of curvature in motion analysis
  • Basic principles of dynamics related to vehicles
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  • Learn about the effects of curvature on vehicle dynamics
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Homework Statement



http://img136.imageshack.us/img136/1930/carrs8.th.png

Homework Equations



a = (v^2/p) en + (v*)et
a- accel
v-velocity
en- inward normal
et- unit vector along tangent
v*- tangential acceleration
p- curvature

The Attempt at a Solution



FOr the first part I just need to find the velocity using the above equation. Since velocity is constant a =0. Therefore tangential accel (at) = 0.4g.
Then just solve for v. Now they gave us the dimensions of the car, so I guess I can't treat the car as a just a particle. The TA just told us to subtract the curvature from the given dimensions,BUT why are we doing that? :confused:

so curvature = 120m - 0.6
 
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I believe the adjustment is made because the center of gravity of the car is slightly above the road surface.
 

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