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The Centripetal Force needed

by Amamma1
Tags: centripetal, force
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Amamma1
#1
May10-12, 09:49 PM
P: 5
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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tiny-tim
#2
May11-12, 08:11 AM
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Hi Amamma1! Welcome to PF!
Quote Quote by Amamma1 View Post
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?
Amamma1
#3
May12-12, 10:48 AM
P: 5
Thank you very much for your help. Not too long ago I was involved in a very minor brush and go contact with another vehicle. I was driving a 1999 Dodge Ram 1500, which weighs about 4500 pounds. The other vehicle involved was a 1999 Saturn Sport Coupe, which is 15 feet long and it weighs 2400 pounds. I did not have a chance to examine the extent of the damage that the 1999 Saturn sustained since the driver was driving much faster and by the time she stopped she was about 40 feet ahead of me. However, in court, this lady claimed that the severity of the impact caused the rear end of her Saturn to swing (clockwise) from the lane she was driving at all the way to the next lane while the front of her car continued driving the original lane. At the time I didn't hear her make this claim at court, therefore, I didn't get the chance to challenge it. Obviously the judge must had had found her claim to be credible since he found me 100% at fault. Later, I got shocked when I read it in the transcripts of the trial. Since then, I have been trying to put everything in perspective so I can present it as evidence with my petition to reopen the case.
Knowing the fact that the minimum width of a lane is 12 feet, I calculated the sweeping arc, based on her scenario, to be around 13.87 feet and the angle allegedly her car swung to be around 53 degrees. Although this scenarios is probable, in this particular case all facts at hand makes it implausible. The first fact is, my word against hers, at that moment my pickup was barely moving faster than 5 m/h. However, from the transcript of her testimony, it was established that after the contact between our two vehicles and until she stopped 40 feet later, she continued to drive in her original lane in an un deviated straight line. Logic dictates that after her car was forced to swing 53 degrees towards the cars to her right, her car should have T-boned some of these cars to her right as she continued driving a distance of about 40 feet. Also, logic dictates that somekind of skid mark should have been generated by her rear tires especially knowing the fact that the grade of the road was about 7% uphill which should have caused more of the weight of her car to be shifting to the rear tires of her car. But the most important fact in all of this is fact that my pickup only received a very minor paint scuff mark on the very edge of the fiberglass cover of the front bumper. This alone makes her scenario implausible. Logic and science dictates that for her alleged scenario to happen, it requires a compression force of some significance. However, for a compression force of some significance to be generated it requires the collision of two structural systems. This compression force that is strong enough to swing the rear end of a 2400 pounds car a distance of 13.87 feed should also be strong enough to leave some tell-tale signs on my pickup. All that presumable stress on my fiberglass skin, which got sandwitched between these two compression forces, yet the fiberglass skin of my bumper didn't shatter, crack, not even gouged. Not a single mounting and supporting component behind the fiberglass skin of my bumper got shattered, bent, not even cracked.
Any scientific help from members will be greatly appreciated. Thanks, Ray

tiny-tim
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May12-12, 12:00 PM
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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.
Amamma1
#5
May12-12, 12:26 PM
P: 5
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
tiny-tim
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May12-12, 12:41 PM
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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.
Amamma1
#7
May12-12, 01:18 PM
P: 5
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
tiny-tim
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May13-12, 02:49 AM
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Quote Quote by Amamma1 View Post
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")
Amamma1
#9
May13-12, 07:06 AM
P: 5
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
tiny-tim
#10
May13-12, 08:06 AM
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Quote Quote by Amamma1 View Post
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."
Even if your car loses only 0.5 mph (that's 10% of the momentum), and the other car is only half the mass, you'd expect it to get about 1 mph sideways, most of it at the rear where you hit it.


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