## The Centripetal Force needed

What is the Centripetal Force needed to swing the rear end of a car a distance of 13.87 feet? The car is a front wheel drive, 15 feet long and it weighs 2400 pounds. Presumably, the front end of the car maintained its straight uphill path with a grade of 7% travelling at 40 m/h? The road pavement is average.
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Hi Amamma1! Welcome to PF!
 Quote by Amamma1 What is the Centripetal Force needed to swing the rear end of a car a distance of 13.87 feet? The car is a front wheel drive, 15 feet long and it weighs 2400 pounds. Presumably, the front end of the car maintained its straight uphill path with a grade of 7% travelling at 40 m/h? The road pavement is average.
Sorry, but your question makes no sense.

Can you please tell us exactly what happened, in what conditions, and why you're asking?

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## The Centripetal Force needed

Hi Ray

I don't understand why you're bringing up centripetal force and compression.

You were edging out when your front nudged the rear wing of the other car.

Even at 5 mph, that creates a considerable torque on the car, which is perfectly capable of rotating it.
 Thank you very much for your reply. Thank you for correcting my use of the wrong terminology. But shouldn't this still leave some significant tell-Tale sign on the fiberglass skin of my bumper? Also, would it be possible that a 53 degree rotation of the entire car could take place yet without altering its original path? I understand that the pull of front wheel drive car will correct its direction if the rear end slips but in this case the whole car was presumably rotated. Also, is it possible that the rear tires could get dragged sideways for 13.87 feet yet don't leave any skid marks? I appreciate your help. Thanks, Ray
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor The original forward momentum at 40 mph will not be affected. There will be an additional sideways component of momentum which will obviously be much less. The total momentum (and therefore the direction of motion) will hardly be affected. Skid marks depend on a lot of things, including the road surface. And yes, bumpers are designed to cope with light impacts.
 Wow. I obviously was way off bases. One last question, was I correct in using the equation S=r.θ to calculate the length of the arc? Where the radius r=15 feet (the length of the 1999 Saturn). θ is the central angle, measured in radians, that is formed as the rear end of the Saturn swung from the initial position to the final position. I calculated the value of the central angle through trigonometry in degrees, then convert it into radians. The original straight forward path of the car--the adjacent, the final position of her car after the rear end swung--the hypogenous which is 15 feet and that is the length of the car, and the vertical displacement of the rear end of the car --in this case is the width of the lane which is 12 feet-- together form a right triangle. Any of the above assumptions is wrong? Thanks again. Ray

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 Quote by Amamma1 … was I correct in using the equation S=r.θ to calculate the length of the arc?
The arc of what?

(and for what purpose?)

The rear of the decelerating car won't be moving in a circle.

Ray, you're losing touch with reality …

the equations are just maths, the reality is the facts "on the ground", and the facts you need are either missing (eg skid marks) or of no quantitative use (eg the damage to the bumper).

(and unless you have a degree in physics, you yourself can't present such evidence in court anyway, you'd need an accident examiner to give "expert evidence")
 I really do appreciate your advice. I am just trying to come to term with the whole thing. I witnessed the whole thing with my own eyes and I do agree with you that "The rear of the decelerating car won't be moving in a circle." I know for a fact that the other car didn't move an inch laterally, certainly not as she described it "and now the rear end of my car is in the other lane". Her statement does imply that the rear end of the car swept an arc. This kind of audacity in front of a judge was something new to me and I am trying to come to term with it. What puzzles me now is that your last answer above negated your earlier answer when you said "Even at 5 mph, that creates a considerable torque on the car, which is perfectly capable of rotating it." I will appreciate your input. Thanks, Ray

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