Question

A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is \(\mu_{\mathrm{s}}=1.20 .\) If the radius of the curve is \(10.0 \mathrm{~m},\) what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

Short answer

Based on the provided solution, if a race car has a coefficient of static friction of 1.20 and is moving on a curve with a radius of 10.0 meters, its maximum speed to avoid sliding is approximately 10.9 meters per second.

Step-by-step solution

Write down the given values and variables

Given: - Coefficient of static friction, \(\mu_{\mathrm{s}} = 1.20\) - Radius of the curve, \(r = 10.0 \mathrm{~m}\) - Gravitational force, \(N = mg\) - Speed of the car, \(v\)

Equate the centripetal force and friction force

We know that the friction force is equal to the centripetal force to ensure the car does not slide. Thus, \(F_{\mathrm{f}} = F_{\mathrm{c}}\)

Substitute the expressions for friction force and centripetal acceleration

We have \(F_{\mathrm{f}} = \mu_{\mathrm{s}} \cdot N\) and \(F_{\mathrm{c}} = m \cdot a_{\mathrm{c}}\). We also know that \(a_{\mathrm{c}} = \frac{v^2}{r}\). Thus, \(\mu_{\mathrm{s}} \cdot N = m \cdot \frac{v^2}{r}\)

Substitute gravitational force for the normal force

Since the normal force is equal to the gravitational force, we have \(N = mg\). So, \(\mu_{\mathrm{s}} \cdot mg = m \cdot \frac{v^2}{r}\)

Solve for the speed, \(v\)

We can simplify the equation to solve for \(v\) as follows: \(v^2 = \mu_{\mathrm{s}} \cdot g \cdot r\) Now, taking the square root of both sides to find \(v\): \(v = \sqrt{\mu_{\mathrm{s}} \cdot g \cdot r}\)

Substitute the given values and calculate the maximum speed

Now, we substitute the given values of \(\mu_{\mathrm{s}}\) and \(r\) into the equation, along with the gravitational acceleration (\(g = 9.81 \mathrm{~m/s^2}\)): \(v = \sqrt{(1.20) \cdot (9.81 \mathrm{~m/s^2}) \cdot (10.0 \mathrm{~m})}\) Solving the equation, we get: \(v \approx 10.9 \mathrm{~m/s}\) So, the maximum speed at which the race car can turn without sliding is approximately \(10.9 \mathrm{~m/s}\).

Key concepts

Coefficient of Static Friction

Understanding the coefficient of static friction is crucial when analyzing how much force is needed to initiate movement between two objects at rest relative to each other. This coefficient, denoted as \(\mu_{\mathrm{s}}\), is a measure of the resistance to motion that needs to be overcome for an object to start sliding across a surface. In our scenario with the race car, \(\mu_{\mathrm{s}} = 1.20\) signifies a high level of friction between the tires and the track. This high friction is essential for the car to make a U-turn without skidding out of control.

Here are key points about the coefficient of static friction:

• It is a dimensionless value, meaning it has no unit.
• The value varies depending on the materials in contact. For race tires and a track, it’s typically high to prevent slipping.
• It does not depend on the surface area in contact or the mass of the vehicle.

The coefficient is directly involved in calculating the maximum speed the car can achieve in a turn without losing grip, which leads to our next section on centripetal force.

Centripetal Force

Moving to centripetal force, which is a term describing the force required to keep an object moving in a circular path at constant speed. It acts towards the center of the circle and is responsible for changing the direction of the object's velocity. To prevent the car from sliding, the static friction must provide a force equal to the required centripetal force.

Formulating centripetal force involves the mass of the object (\(m\)), the speed (\(v\)), and the radius of the circular path (\(r\)), given by the equation \(F_{\mathrm{c}} = m \cdot \frac{v^2}{r}\).

• The force increases with the square of the speed, meaning that doubling the speed requires four times the centripetal force.
• As the radius of the turn decreases, a greater force is needed to maintain the same speed.
• This force does not do work as it doesn't cause a change in the speed, only direction.

Centripetal force is what keeps the race car hugging the curve, and as we saw in the solution, it’s matched by the frictional force to ensure no sliding occurs.

Maximum Speed Without Sliding

Lastly, maximum speed without sliding is the highest speed at which the car can travel around the curve without its tires losing grip on the road surface. It's determined by the available static friction and the need for sufficient centripetal force to sustain circular motion.

In the given problem, we used the formula \(v = \sqrt{\mu_{\mathrm{s}} \cdot g \cdot r}\) where \(g\) is the acceleration due to gravity and \(\mu_{\mathrm{s}}\) and \(r\) are the coefficient of static friction and radius of the curve, respectively. This equation derives from matching the forces of friction and centripetal acceleration and simplifying. Knowing these values allows us to calculate the maximum safe speed to navigate the turn.

• The vehicle can go faster in wider turns (larger \(r\)) and with higher friction (larger \(\mu_{\mathrm{s}}\)).
• If the speed exceeds this maximum, the car will start to slide due to insufficient frictional force to provide the necessary centripetal force.
• Driving practices and tire conditions can also affect this maxim speed threshold.

The exercise demonstrates how physics principles apply to everyday situations like driving, providing critical insights into safe and efficient vehicle design and operation.