Question

A car speeds over the top of a hill. If the radius of curvature of the hill at the top is \(9.00 \mathrm{~m}\), how fast can the car be traveling and maintain constant contact with the ground?

Short answer

Answer: The maximum speed the car can travel over the hill while maintaining constant contact with the ground is approximately 9.06 m/s.

Step-by-step solution

Define the centripetal acceleration formula

Centripetal acceleration is the acceleration of an object moving in a circle at constant speed. The formula for centripetal acceleration is \(a_c = v^2/r\), where \(v\) is the speed of the object and \(r\) is the radius of the circle.

Set up the forces acting on the car

According to Newton's second law, the sum of forces acting on the car must be equal to the mass times the centripetal acceleration. There are two forces acting on the car in this scenario: gravity (\(F_g = m\cdot g\), where \(m\) is the mass of the car and \(g\) is the gravitational acceleration) and the contact force between the tires and the ground (\(F_n\)). At the top of the hill, these forces should be equal and opposite in the vertical direction for the car to maintain constant contact with the ground. As the car goes faster, the contact force will decrease, and when it reaches zero, the car will lose contact with the ground.

Write the equation for the contact force and centripetal acceleration

At the top of the hill, we can write the equation for the vertical forces acting on the car as: \(F_n = m\cdot g - m\cdot a_c\). Since the centripetal acceleration \(a_c = v^2/r\), we can rewrite the equation as: \(F_n = m\cdot g - m\cdot (v^2/r)\).

Find the critical speed when the contact force becomes zero

The car will lose contact with the ground when the contact force \(F_n\) becomes zero. We can find the critical speed \(v_{critical}\) by setting the contact force equation to zero and solving for \(v\): \(0 = m\cdot g - m\cdot (v_{critical}^2/r)\). Now, we can solve for \(v_{critical}\): \(v_{critical}^2 = r\cdot g\), so \(v_{critical} = \sqrt{r\cdot g}\).

Calculate the maximum speed

We are given the radius of curvature of the hill, \(r = 9.00 \mathrm{~m}\) and the gravitational acceleration, \(g = 9.81 \mathrm{~m/s^2}\). We can now plug these values into the equation for \(v_{critical}\) to find the maximum speed the car can travel: \(v_{critical} = \sqrt{9.00 \mathrm{~m} \cdot 9.81 \mathrm{~m/s^2}} \approx \boxed{9.06 \mathrm{~m/s}}\).

Key concepts

Uniform Circular Motion

When an object moves in a circle at a steady speed, it's experiencing uniform circular motion. This motion is unique because the speed of the object remains constant, but the velocity, which is speed in a given direction, changes due to the continuous change in direction. This means the object is constantly accelerating even though its speed doesn't change.

The acceleration that an object in uniform circular motion experiences is known as 'centripetal acceleration.' It always points towards the center of the circle and is responsible for changing the direction of the object's velocity. The formula for centripetal acceleration is given by \( a_c = \frac{v^2}{r} \), where \( v \) is the speed of the object and \( r \) is the radius of the circular path.

This acceleration is not felt as a change in speed but as a 'pull' towards the center of the circular path. Objects in uniform circular motion require a force to keep them moving in this path, and this force is known as 'centripetal force', which must be provided by some interaction like gravitational pull, friction, or tension.

Newton's Second Law

Newton's second law of motion is all about the relationship between an object's mass, the forces acting upon it, and the acceleration that it experiences. The law is commonly stated as \( F = m \cdot a \), where \( F \) represents the net force acting on the object, \( m \) is the mass of the object, and \( a \) is the acceleration produced.

Applying this law to uniform circular motion, we know the net force acting on an object moving in a circle at constant speed is the centripetal force required to keep it in motion. This force is needed to change the direction of the object, thereby causing the centripetal acceleration we mentioned earlier. It's important to note that while the speed remains constant in uniform circular motion, the acceleration that comes from the continuous change of direction means that there's a constant force being exerted; this force results from the interaction that causes the centripetal acceleration.

In the example of a car moving over the top of a hill, the forces acting on the car include gravitational force pulling it downward and the normal force from the road providing the necessary centripetal force to keep it in circular motion.

Gravitational Acceleration

Gravitational acceleration, often denoted as \( g \), is a measure of the acceleration due to gravity that an object experiences when it is near a massive body like the Earth. It is approximately \( 9.81 \text{ m/s}^2 \) on the surface of the Earth. This acceleration is what causes objects to fall towards the ground when they are dropped, and it acts on all objects equally regardless of their mass.

In our exercise, gravitational acceleration is what pulls the car towards the center of the Earth, which is also toward the center of its circular path over the hill. The ‘normal’ force, which the road applies to the car, counteracts some of this gravitational force. The net effect of the gravitational and normal forces determines whether the car maintains contact with the ground or loses it.

As the car speeds up, the requirement for centripetal force to keep it in circular motion increases, which at some high speed, equals the gravitational force. At this point, any additional speed would result in the car lifting off the ground because the gravitational force can no longer provide the necessary centripetal force for the circular motion. This is why our solution involves setting the gravitational force equal to the centripetal force required for circular motion to find the maximum speed the car can maintain to still remain in contact with the road.