Centripetal force stunt driver problem

AI Thread Summary
To determine the radius of curvature at which a stunt driver’s car begins to skid, the centripetal force equation is applied. Given the mass of the car is 2545 kg and the frictional force is 1.75x10^4 N while traveling at 24 m/s, the equation Fc=(m)(v)^2/r is used. Rearranging the equation leads to 1.75x10^4 N = 1,465,920/r, resulting in a calculated radius of curvature of 83.8 m. The solution demonstrates the relationship between speed, mass, and friction in determining skidding conditions. Understanding these dynamics is crucial for stunt driving safety and performance.
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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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