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Circular motion of a car and tires

  • Thread starter xtrmk
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  • #1
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Homework Statement


A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static frictional force by a factor of three. If the car is to continue safely around the curve, to what speed must the driver slow the car?


Homework Equations





The Attempt at a Solution


Do I convert V to angular velocity and figure out the radius? Then ?_?
Do not know how to start, can someone help. Thanks
 

Answers and Replies

  • #2
Doc Al
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Hint: How does the required centripetal force relate to the car's speed?
 
  • #3
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Hint: How does the required centripetal force relate to the car's speed?
Centripetal force = mv^2 / r ?
But I don't have an r or m. and what do I do without a given coefficient of friction?

(Sorry i just came back to class after a sick break)

(edit)
Am i starting it right?
Fc = friction = mv^2/r

u(mg) = mv^2/r?
 
  • #4
Doc Al
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Centripetal force = mv^2 / r ?
Right.
But I don't have an r or m. and what do I do without a given coefficient of friction?
You won't need any of those. Hint: Think ratios.
Am i starting it right?
Fc = friction = mv^2/r
Yes.
 
  • #5
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Right.

You won't need any of those. Hint: Think ratios.

Yes.
Could you look at this:

[u(mg)]/3 = mv^2 / r

I divide by 3 because it was reduced by the factor of 3 ?
I canceled out the masses and:

3.266u = 441/r

do I solve for the radius? Then consider a case 2 and plug it in?
 
  • #6
Doc Al
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As I said earlier, you don't need to know (or try to solve for) m, r, or mu, since they don't change. The only things that change are the force and the speed. Think ratios.

Try this:
[tex]F_1 = m v_1^2/r[/tex]

[tex]F_2 = m v_2^2/r[/tex]

You are given [tex]F_1/F_2 = 3[/tex]; your job is to figure out [tex]v_1/v_2[/tex].
 

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