Centripetal force stunt driver problem

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SUMMARY

The discussion centers on calculating the radius of curvature required for a stunt driver to make a 2545 kg car skid while turning at a speed of 24 m/s. Using the formula for centripetal force, Fc = (m)(v)^2/r, and the given force of friction of 1.75 x 10^4 N, the radius of curvature is determined to be 83.8 m. The calculation involves rearranging the equation to find r, leading to the conclusion that the car will begin to skid at this radius.

PREREQUISITES
  • Centripetal force concepts
  • Basic physics of friction
  • Understanding of Newton's laws of motion
  • Algebraic manipulation of equations
NEXT STEPS
  • Study the effects of different friction coefficients on skid radius
  • Explore advanced applications of centripetal force in stunt driving
  • Learn about the physics of vehicle dynamics during turns
  • Investigate the role of tire composition in frictional force
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, stunt coordinators, and anyone interested in the dynamics of vehicle motion and safety during high-speed maneuvers.

xcortz
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Homework Statement


A stunt driver for a movie needs to make a 2545kg car skid on a large, flat, parking lot surface. The force of friction between the tires and the concrete surface is 1.75x10^4N and he is driving at a speed of 24m/s. As he turns more sharply, what radius of curvature will he reach when the car just begins to skid? (Ans: 83.8m)

Homework Equations


Fc=(m)(v)^2/r

The Attempt at a Solution


1.75x10^4N=(2545kg)(24m/s)^2/r
1.75x10^4N=1,465,920/r
0.011937m=r
 
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You solved for 1/r.
 

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