What is the Coefficient of Static Friction for a Car on a Circular Track?

In summary, The car leaves the track because the centripetal force is greater than the static friction force.
  • #1
vijay123
122
0
hey guys, jus wunna see how u solve this question.

-a car traveling on a flat(unbanked) circular track accelerates uniformly from rest with a tangetial acceleration of 1.70m/s(squared). the car makes it one fourth of the way around the circle before it skids off the track. cetermine the coefficient of static friction between the car and the track from this data.
 
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  • #2
vijay123 said:
hey guys, jus wunna see how u solve this question.

-a car traveling on a flat(unbanked) circular track accelerates uniformly from rest with a tangetial acceleration of 1.70m/s(squared). the car makes it one fourth of the way around the circle before it skids off the track. cetermine the coefficient of static friction between the car and the track from this data.
You will have to show us what you have done so far to analyse the problem.

What is the condition for the car leaving the track?

AM
 
  • #3
thats what the question says...i don't understand your question.
i think the condition is that centripetal force is exerted in one side and static friction is at the other side...now i think that the centripetal force outdid the static frcition.so we know that tangential acceleration is 1.7 and we also know that the displacement of theta is pie divided 4 radians...since the displacemn is 45degrees.
 
  • #4
Do you know how the static friction Force depends on the coefficient?

What "formula" do you know for the "centripetal force"? What causes it?

What's different about the later part (car NOT travel in circle) and earlier?
 
  • #5
fine..i ll state down everythn...
static friction is equal to mu*mg.
centripetal froce is equal to mv(sqaured) divided by radius.
now they have given us tangetial acceration...i knoiw that tangentail acceleration is equal to radius * angular acceleration.
herfore...(omega)sqaured=2*angular acceleration*displacement...were omega is angular velocity, and displacement is theta which is 45degress or pie/4 rad. therofre...ifrom the info, we could find omega in terms of radius.now mv(squared)/r is simply m*(omega)(sqaured)*radius. hence if we would equate the equation on static friction to this equation...we would get an answer...i got an answer with was nearly half of the actual ans.
my ans was 0.24 but the actual one is 0.57(approx there). yea...so i wunted to see were i went wrong...(this is a 9th grade problem...which i am in)(chapter is on rotation motion)
 
  • #6
You did all the Physics right ...
but 1/4 of the way around a circular track is 90 degrees ... pi/2 .
 
  • #7
ooooooooo...my god!
how silly of me...thanks a lot for correcting me on that...lol...to about an hour pondering were i have gone wrong...lol...thanks a lot...really been a pleasure meeting you
 
  • #8
but..my answers is again a bit off...according to my calculations, the coffeicinet of frcition is 0.545 but the actual ans. is 0.57...(think dat made any difference?)
 
  • #9
Well, I can't check your calculations because you never told me the radius.
did you round off the angular accel? did you round off pi?

Wait ... the car also has tangential acceleration, as well as centripetal a.

You have to find the magnitude of the TOTAL acceleration (Pythagoras).
 
  • #10
k thx...i ll try that way...
the question does not state the radius anyway...i think they get canceled way if you do the final calculations...thx for the advice
 
  • #11
yay...i got the answer...thx a lot lightgrav...your da man!
 
  • #12
vijay123 said:
yay...i got the answer...thx a lot lightgrav...your da man!
Since the condition for leaving the track is: centripetal force = static friction force.

[tex]mv^2/R = m(at)^2/R = \mu mg[/tex]

(1) [tex]\mu = a^2t^2/gR[/tex]

So we need to find [itex]t^2/R[/itex].

[tex]\theta = \pi/2 = \int \omega dt = \int \alpha t dt = \frac{1}{2}\alpha t^2[/tex]

Since [itex]a = \alpha R[/itex]:

[tex]\frac{1}{2}\alpha t^2 = \frac{at^2}{2R} = \pi/2[/tex]

So:

[tex]t^2/R = \pi/a[/tex]

Substituting into (1):

[tex]\mu = a\pi/g[/tex]

[tex]\mu = 1.7*3.14/9.8 = .545[/tex]

AM
 
  • #13
hey mason...but the answer is 0.571...i too got that ans. but not with calculus...but actually, you would have to find the magnitude of acceleration and not radial accelerations...thats wut ma calculations show me...but the correct ans. is 0.571.
 
  • #14
but thanks for the calculus method..it makes more sense...
 
  • #15
lightgrav said:
You have to find the magnitude of the TOTAL acceleration (Pythagoras).
I don't think you need to take the NET acceleration,
rather you need to use the fact that at the instant of slipping, max static frictional force ( which provides for rotation ) equals centripetal force, and the tangential acceleration has to be used separately as AM pointed out .
And as far as I can see AM's solution is absolutely correct ( as it usually is :) )

Arun
__________________________________________________________

Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe.
Albert Einstein
 
  • #16
that what my previous calculation did but my answer was just like mason's one. but when i flipped through the ans. book, it gave another. so i tried to find the magnitude of roational and tangential acceleration and then equated it to the force of friction. i got the correct ans.
i know it doesn't make sense but i don't have another way of doing it.
any suggestions. ans. is 0.57
 
  • #17
arunbg said:
I don't think you need to take the NET acceleration,
rather you need to use the fact that at the instant of slipping, max static frictional force ( which provides for rotation ) equals centripetal force, and the tangential acceleration has to be used separately as AM pointed out .
Yes, you do need to consider NET acceleration. Realize that the static friction provides the total force, not just the centripetal component.
 
  • #18
..

thanks a lot for that confirmation doc al.
i seem to get the concept now.
 
  • #19
Doc Al said:
Yes, you do need to consider NET acceleration. Realize that the static friction provides the total force, not just the centripetal component.
So where does it say that this is not a rocket powered car?:rolleyes:

If you assume the car accelerates due to tire friction, the expression for [itex]\mu[/itex] is:

[tex]\mu = \sqrt{a^2\pi^2 + a^2}/g = .571[/tex]

AM
 
  • #20
lol...i understand you too...if the car did not have wheels, then your ans. would have been true.
but too bad, this is not that roket space car in sci fi.
 
  • #21
Of course , Doc thanks for the correction .
What was I thinking !( rams head against wall )
I just proved my sig, the latter part that is :D

_______________________________________________

Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.
Albert Einstein
 
  • #22
Andrew Mason said:
So where does it say that this is not a rocket powered car?:rolleyes:
Good point! :smile: May as well have those rockets provide the centripetal force as well. :wink:
 
  • #23
haha..i like you guys:tongue2:
 
  • #24
vijay123 said:
lol...i understand you too...if the car did not have wheels, then your ans. would have been true.
but too bad, this is not that roket space car in sci fi.
If it has wheels but the friction between the wheels and the road does not provide the forward acceleration (e.g a rocket or jet engine), the tire friction provides only the centripetal acceleration.

Doc Al said:
Good point! :smile: May as well have those rockets provide the centripetal force as well. :wink:
Rocket steering wastes energy!

AM
 
  • #25
i dun understand wut you r trying to convey...wut do you mean by the friction provides only centripetal a.?
 
  • #26
In an ordinary car (no rockets :smile: ) the only force available to accelerate the car is friction; that friction will accelerate the car both tangentially and centripetally. But Andrew gave an example of a rocket-powered car where the rocket provides the tangential force, leaving friction to handle the centripetal force.
 
  • #27
ok...i get it...
 

1. What is rotational motion?

Rotational motion is the movement of an object around a fixed point, known as the axis of rotation. In the case of a car, the wheels rotating around the axle would be an example of rotational motion.

2. How does rotational motion affect a car's movement?

Rotational motion plays a crucial role in a car's movement. The rotational motion of the wheels allows the car to move forward or backward, while the rotational motion of the steering wheel enables the car to turn left or right.

3. What factors affect the rotational motion of a car?

Several factors can affect the rotational motion of a car, such as the mass and size of the wheels, the amount of force applied to the wheels, and the friction between the wheels and the ground. The car's speed and direction can also impact its rotational motion.

4. How is rotational motion different from translational motion?

Rotational motion involves an object rotating around a fixed point, while translational motion is the movement of an object from one point to another in a straight line. In the case of a car, the rotational motion of the wheels allows it to move, while the translational motion of the car is its movement in a straight line.

5. How can we measure the rotational motion of a car?

The rotational motion of a car can be measured using various methods such as calculating the angular velocity of the wheels, measuring the torque applied to the wheels, or using sensors to track the rotation of the wheels. These measurements can help determine the car's speed, acceleration, and other parameters related to rotational motion.

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