A car is safely negotiating an unbanked circular turn at a speed of 17.2 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?
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To determine the speed at which the car must slow down, we need to consider the change in the maximum static frictional force caused by the wet patch on the road.
Here's how you can solve this problem:
1. Identify the relevant equations:
- Centripetal force (Fc) = (m * v^2) / r, where m is the mass of the car, v is the velocity, and r is the radius of the circular turn.
- Maximum static frictional force (Ff) = μs * N, where μs is the coefficient of static friction and N is the normal force.
2. Determine the initial maximum static frictional force:
Since no value for μs or N is provided, we can assume that the maximum static frictional force is equal to the centripetal force in this case.
Fc = Ff
(m * v^2) / r = μs * N
3. Determine the final maximum static frictional force:
The problem states that the maximum static frictional force is reduced by a factor of three, which means the new maximum static frictional force is one-third of the initial maximum static frictional force.
Ff_final = Ff_initial / 3
4. Substitute the known values into the equations:
We can substitute the equations for Fc and Ff into the equation for the final maximum static frictional force:
(m * v^2) / r = (1/3) * (m * g)
Note that the mass of the car cancels out.
5. Solve for the final velocity:
Rearrange the equation to solve for v:
v^2 = (r / 3) * g
v = sqrt((r / 3) * g)
Now we can plug in the values you provided and solve for v.
Assuming a value of g (acceleration due to gravity) of approximately 9.8 m/s^2 and a given radius of the circular turn, you can calculate the final velocity by substituting these values into the derived equation.