A 300. kg roller coaster car is traveling at a constant speed of 15 m/s over a hill with a radius of curvature of 30. m.



b. What is the normal force acting on the car at the top of the hill?

c. Assume that the car is at the intersection of the radius and the hill and that a frictional force exists between the car and the hill. Draw a force diagram showing all forces.

2250 N

10. A child twirls his yo-yo horizontally about his head rather than using it properly. The yo-yo has a mass of 0.250 kg and is attached to a string 0.800 m long.


a. If the yo-yo makes a complete revolution each second, what tension must exist in the string?

b. To find the normal force acting on the car at the top of the hill, we need to consider the forces acting on the car. At the top of the hill, the car is at the highest point of its circular path. This means that the net force acting on the car must be directed towards the center of the circle.

The weight of the car acts vertically downwards and can be calculated using the formula:
Weight = mass * acceleration due to gravity
Weight = 300 kg * 9.8 m/s²
Weight = 2940 N

At the top of the hill, the normal force and the weight force are equal in magnitude but opposite in direction. This means the normal force is also 2940 N.

c. The force diagram for the roller coaster car at the intersection of the radius and the hill will include:

1. Weight (2940 N): This force acts vertically downwards due to the gravitational pull on the car.

2. Normal force (2940 N): This force acts perpendicular to the surface of the hill, pushing the car towards the center of the circular path.

3. Frictional force: This force acts parallel to the hill's surface and opposes the motion of the car, preventing it from sliding down the hill. The magnitude of the frictional force depends on the coefficient of friction between the car and the hill's surface.

Since we don't have the coefficient of friction or any other information, we cannot accurately determine the magnitude or direction of the frictional force.

To find the normal force acting on the roller coaster car at the top of the hill, we need to analyze the forces acting on the car.

First, let's draw a force diagram showing all the forces involved:

1. Gravitational force (Weight): This force acts vertically downwards and is equal to the mass of the car multiplied by the acceleration due to gravity (9.8 m/s^2).

2. Normal force: This force acts perpendicular to the surface of the hill and counterbalances the gravitational force. It is also the force that prevents the car from sinking into the surface of the hill.

3. Centripetal force: At the top of the hill, the car is moving in a circle of radius 30 m. Therefore, there is an inward force called the centripetal force, which is provided by the normal force.

4. Frictional force: We are assuming that there is a frictional force acting between the car and the hill. This force opposes the motion of the car and acts parallel to the surface of the hill. It is given by the coefficient of friction multiplied by the normal force.

Given that the car is traveling at a constant speed, the net force acting on it must be zero. Therefore, the sum of the gravitational force, the centripetal force, and the frictional force should add up to zero.

Now, let's solve for the normal force:

Since the normal force and the gravitational force are acting perpendicular to each other, we can apply the concepts of circular motion to find the normal force.

The centripetal force is given by the equation:

Centripetal force = (mass × velocity^2) / radius

Given:
Mass = 300 kg
Velocity = 15 m/s
Radius = 30 m

Centripetal force = (300 kg × (15 m/s)^2) / 30 m = (300 × 225) / 30 = 2250 N

According to the force diagram, the normal force and the centripetal force are equal and opposite, so the normal force at the top of the hill is also 2250 N.

Therefore, the answer to part (b) is: The normal force acting on the car at the top of the hill is 2250 N.