Safely Negotiating Unbanked Curves: Slowing Car Speed for Safety

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A car negotiating an unbanked circular turn at 23 m/s faces a reduction in maximum static friction due to a wet patch, decreasing it by a factor of three. To maintain safety while navigating the curve, the driver must adjust the speed. The relationship between speed, friction, and radius is expressed as μ_s(g) = v^2/r. When static friction is reduced, the equation must be balanced by adjusting the speed, leading to the conclusion that the new speed must be v' such that v'^2 = v^2/3. The discussion emphasizes that the radius remains constant, and the focus is solely on determining the new safe speed.
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A car is safely negotiating an unbanked circular turn at a speed of 23 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static friction 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?

i have it so where mu_s = v^2/r but I don't know if that's right, help please! thanks :-)
 
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quickclick330 said:
A car is safely negotiating an unbanked circular turn at a speed of 23 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static friction 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?

i have it so where mu_s = v^2/r but I don't know if that's right, help please! thanks :-)
I think you mean mu_s(g) = v^2/r. But in any case, if mu_s is reduced by 3, then v^2 must be reduced by what factor to keep the car from slipping?
 
3/r maybe?
 
The radius of the turn is constant. If you divide \mu_s by 3, what must you do to balance your equation?
 
then you must divide the other side of the equation to balance it, so would it be ((v^2/r)/3)?
 
Exactly. But the only thing you're allowed to change is v.
What should you do to v to get the whole expression to equal {\frac{v^2}{3r}?
 
I'm not sure...not having the radius measurement is throwing me off
 
It doesn't matter what the radius is. It doesn't change.
You have \frac{v^2}{r}
You need to divide all of that by three, but you're NOT allowed to touch r. All you're allowed to change is v. What might you do to v so that

\frac{v^2}{3r}=\frac{(v')^2}{r} ?
 
I tried to divide by 3 again...what is v prime??
 
  • #10
quickclick330 said:
I tried to divide by 3 again...what is v prime??

The speed the driver must slow down to in order to continue along the curve safely. This is what you are trying to solve for.
 
  • #11
quickclick330 said:
I tried to divide by 3 again...what is v prime??

v prime is your new velocity. The one that makes the origianl exprssion (v^2/r) get divided by 3.
 

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