# Rotational motion of a car

1. Jul 1, 2006

### vijay123

hey guys, jus wunna see how u solve this question.

-a car travelling 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.

2. Jul 1, 2006

### Andrew Mason

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. Jul 1, 2006

### vijay123

thats what the question says....i dont 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. Jul 1, 2006

### lightgrav

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. Jul 1, 2006

### vijay123

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. Jul 1, 2006

### lightgrav

You did all the Physics right ...
but 1/4 of the way around a circular track is 90 degrees ... pi/2 .

7. Jul 1, 2006

### vijay123

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. Jul 1, 2006

### vijay123

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. Jul 1, 2006

### lightgrav

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. Jul 2, 2006

### vijay123

k thx....i ll try that way....
the question does not state the radius anyway....i think they get cancelled way if you do the final calculations...thx for the advice

11. Jul 2, 2006

### vijay123

12. Jul 2, 2006

### Andrew Mason

Since the condition for leaving the track is: centripetal force = static friction force.

$$mv^2/R = m(at)^2/R = \mu mg$$

(1) $$\mu = a^2t^2/gR$$

So we need to find $t^2/R$.

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

Since $a = \alpha R$:

$$\frac{1}{2}\alpha t^2 = \frac{at^2}{2R} = \pi/2$$

So:

$$t^2/R = \pi/a$$

Substituting into (1):

$$\mu = a\pi/g$$

$$\mu = 1.7*3.14/9.8 = .545$$

AM

13. Jul 2, 2006

### vijay123

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. Jul 2, 2006

### vijay123

but thx for the calculus method..it makes more sense...

15. Jul 2, 2006

### arunbg

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
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16. Jul 2, 2006

### vijay123

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 doesnt make sense but i dont have another way of doing it.
any suggestions. ans. is 0.57

17. Jul 2, 2006

### Staff: Mentor

Yes, you do need to consider NET acceleration. Realize that the static friction provides the total force, not just the centripetal component.

18. Jul 2, 2006

### vijay123

..

thanks a lot for that confirmation doc al.
i seem to get the concept now.

19. Jul 3, 2006

### Andrew Mason

So where does it say that this is not a rocket powered car?

If you assume the car accelerates due to tire friction, the expression for $\mu$ is:

$$\mu = \sqrt{a^2\pi^2 + a^2}/g = .571$$

AM

20. Jul 3, 2006

### vijay123

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.