How Slow Must the Car Go to Stay Safe on a Wet, Unbanked Curve?

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To safely navigate an unbanked circular turn at a wet patch in the road, the driver must reduce the car's speed from 17 m/s to 10.83 m/s due to a decrease in maximum static friction by a factor of three. The calculations involve using the centripetal force equation and the relationship between frictional force and speed. The radius of the turn is determined to be approximately 123.23 m based on the initial safe speed. Ultimately, the reduced speed ensures that the car can maintain traction and safely negotiate the curve. The driver must adhere to this new speed limit to avoid losing control.
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I'm setting this problem up, but I'm now stuck.

A car is safely negotiating an unbanked circular turn at a speed of 17m/s. The maximum static frictional force act 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 made a free body diagram and came up with f net equations. I have fnetx=1/3usg=v^2/r. I have fnety=N-mg. For fnetx, I further get 3.267m/s^2us=v^2/r. And this is where I get stuck. I'm not sure what I do to find the speed without knowing the radius.
 
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The magnitude of the initial frictional force satisfies:
F=m\frac{v_{0}^{2}}{R}

Afterwards, we have:
\frac{1}{3}F=m\frac{v_{1}^{2}}{R}

By division, we get:
\frac{1}{3}=(\frac{v_{1}}{v_{0}})^{2}
 


To find the speed at which the driver must slow the car, you can use the equation for centripetal force: Fc = mv^2/r. In this case, the force acting on the car is the maximum static frictional force, which is now 1/3 of its original value. So the equation becomes: 1/3usmg = mv^2/r.

Since we don't know the radius, we can use the fact that the car is safely negotiating the turn at a speed of 17m/s, which means that the centripetal force at this speed is equal to the maximum static frictional force. So we can write: usmg = mv^2/r.

By substituting this into the previous equation, we get: 1/3(usmg) = mv^2/r.

Now we can cancel out the mass (m) on both sides and rearrange the equation to solve for v: v = √[(1/3usg)r].

Since we still don't know the radius, we can use the fact that the car is safely negotiating the turn at a speed of 17m/s to find the radius. At this speed, the centripetal force is equal to the maximum static frictional force, so we can write: usmg = mv^2/r.

By substituting the given values, we get: (1/3)(0.7)(9.8)(m) = (m)(17)^2/r.

Solving for r, we get: r = (17)^2/2.333 = 123.23m.

Now we can plug this value of r into the previous equation to find the speed at which the driver must slow the car: v = √[(1/3)(0.7)(9.8)(123.23)] = 10.83m/s.

Therefore, the driver must slow the car to a speed of 10.83m/s to safely navigate the unbanked curve with the reduced maximum static frictional force.
 
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