A car is safely negotiating an unbanked circular turn at a speed of 34 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 3. If the car is to continue safely around the curve, to what speed must the driver slow the car???

To determine the new speed at which the car can safely negotiate the turn after encountering a wet patch that reduces the maximum static friction force, we can use the principles of circular motion and friction. Let's break this down step by step.

Understanding the Forces at Play

When a car travels around a circular path, it requires a centripetal force to keep it moving in that circle. This force is provided by the friction between the tires and the road. The maximum static frictional force can be calculated using the formula:

• F_friction = μ_s * N

Where:

• F_friction is the maximum static frictional force.
• μ_s is the coefficient of static friction.
• N is the normal force, which for a flat surface is equal to the weight of the car (mg).

Initial Conditions

Initially, the car is traveling at a speed of 34 m/s. The centripetal force required to keep the car moving in a circle is given by:

• F_c = (m * v^2) / r

Where:

• F_c is the centripetal force.
• m is the mass of the car.
• v is the speed of the car.
• r is the radius of the turn.

For the car to safely navigate the turn, the maximum static frictional force must be equal to or greater than the required centripetal force:

• F_friction ≥ F_c

Impact of the Wet Patch

Now, when the car hits the wet patch, the maximum static frictional force is reduced by a factor of 3. This means:

• F_friction_new = F_friction / 3

To maintain safety, we need to find the new speed (v_new) that satisfies:

• F_friction_new ≥ (m * v_new^2) / r

Setting Up the Equation

Substituting the new frictional force into the inequality gives us:

• F_friction / 3 ≥ (m * v_new^2) / r

Since we know that initially:

• F_friction = (m * v^2) / r

We can substitute this into our new equation:

• (m * v^2) / (3 * r) ≥ (m * v_new^2) / r

Notice that the mass (m) and radius (r) cancel out, simplifying our equation to:

• v^2 / 3 ≥ v_new^2

Calculating the New Speed

Now, we can solve for v_new:

• v_new^2 ≤ v^2 / 3
• v_new ≤ v / √3

Substituting the initial speed (34 m/s) into the equation:

• v_new ≤ 34 / √3

Calculating this gives:

• v_new ≤ 34 / 1.732 ≈ 19.6 m/s

Final Thoughts

To safely navigate the turn after encountering the wet patch, the driver must reduce the speed to approximately 19.6 m/s. This ensures that the reduced frictional force is still sufficient to provide the necessary centripetal force for safe turning.