Distance traveled in a circular motion

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SUMMARY

The discussion centers on calculating the distance a car travels around a circular curve before skidding, given a radius of 31.7 m, a constant tangential acceleration of 2.97 m/s², and a coefficient of friction of 0.851. The initial attempt at solving the problem incorrectly considered only the tangential acceleration, neglecting the combined effect of both tangential and centripetal accelerations. The correct approach requires integrating the centripetal acceleration into the frictional force equation to determine the total distance traveled before skidding, which is calculated to be 44.55 m.

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  • Understanding of Newton's laws of motion
  • Knowledge of centripetal acceleration and its formula
  • Familiarity with friction coefficients and their application in physics
  • Basic algebra for solving equations
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Homework Statement



A car starts from rest and accelerates around a flat curve of radius R = 31.7 m. The tangential component of the car’s acceleration remains constant at at = 2.97 m/s2, while the centripetal acceleration increases to keep the car on the curve as long as possible. The coefficient of friction between the tires and the road is μk = 0.851. What distance does the car travel around the curve before it begins to skid? (Be sure to include both the tangential and centripetal components of the acceleration.)


Homework Equations



F=(mv^2)/r
F=ma
F=μN


The Attempt at a Solution


mv^2/r=μN
mv^2/r=μmg
μg=v^2/r
v^2=rμg
v^2=v(initial)^2+2ax
rμg=2ax
x=rμg/(2a)
x=(31.7)(.851)(9.81)/(2(9.81))
x=44.55m
This is how i attempted the problem, but am apparently incorrect. What am i doing wrong?
 
Physics news on Phys.org
The static friction that acts at the tyres has to cover the total accelerating force which includes both centripetal and tangential acceleration.

ehild
 

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