Circular motion of a car and tires

In summary, to safely negotiate an unbanked circular turn at a reduced maximum static frictional force, the driver must decrease the car's speed by a factor of the square root of 3. This is found by comparing the ratio of the required centripetal forces at the original and reduced maximum frictional forces.
  • #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
 
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  • #2
Hint: How does the required centripetal force relate to the car's speed?
 
  • #3
Doc Al said:
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
xtrmk said:
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
Doc Al said:
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
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].
 

1. What is circular motion in relation to a car and its tires?

Circular motion refers to the movement of a car and its tires along a curved path. This is often seen when a car is turning or going around a corner.

2. How does circular motion affect a car's tires?

Circular motion can cause a car's tires to experience a centripetal force, which is a force directed towards the center of the circular path. This force helps to keep the car and its tires moving along the curved path.

3. What factors influence the circular motion of a car and its tires?

The speed of the car, the radius of the circular path, and the mass of the car are all factors that can influence the circular motion of a car and its tires. The greater the speed and mass, or the smaller the radius, the stronger the centripetal force will be.

4. How does friction play a role in circular motion of a car and its tires?

Friction between the tires and the road is necessary for a car to maintain circular motion. The tires must have enough grip on the road to prevent slipping and maintain the necessary centripetal force.

5. Can circular motion be dangerous for a car and its tires?

If a car is traveling at high speeds or taking sharp turns, circular motion can be dangerous if the tires do not have enough friction to maintain the necessary centripetal force. This can lead to loss of control and potential accidents. It is important for drivers to follow speed limits and take turns at appropriate speeds to ensure safe circular motion.

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