Problem 120 - University Physics with Modern Physics

Question

A small remote-controlled car with mass 1.60 \(\mathrm{kg}\) moves at a constant speed of \(v=12.0 \mathrm{m} / \mathrm{s}\) in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 \(\mathrm{m}(\) Fig. \(\mathrm{P} 5.120) .\) What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point \(A\) (at the bottom of the vertical circle) and (b) point \(B(\) at the top of the vertical circle)?

Step-by-Step Solution

Answer

At point A, the normal force is 61.76 N; at point B, it's 30.40 N.

Step 1: Determine forces at point A (bottom of the circle)

The car experiences gravitational force and a normal force from the walls at the bottom of the circle. The normal force acts upwards and must counteract the gravitational force and provide the centripetal force for circular motion. - Gravitational force, \( F_g = mg = 1.60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 15.68 \, \text{N} \).- Centripetal force needed, \( F_c = \frac{mv^2}{r} = \frac{1.60 \, \text{kg} \times (12.0 \, \text{m/s})^2}{5.00 \, \text{m}} = 46.08 \, \text{N} \).The normal force \( N_A \) equals the gravitational force plus the centripetal force:\[ N_A = F_g + F_c = 15.68 \, \text{N} + 46.08 \, \text{N} = 61.76 \, \text{N}. \]

Step 2: Determine forces at point B (top of the circle)

At the top of the circle, the gravitational force and the normal force both act downwards and contribute to providing the required centripetal force.- Gravitational force, \( F_g = 15.68 \, \text{N} \). - Centripetal force needed, \( F_c = 46.08 \, \text{N} \).The normal force \( N_B \) equals the centripetal force needed minus the gravitational force because both forces can combine to meet the required centripetal force:\[ N_B = F_c - F_g = 46.08 \, \text{N} - 15.68 \, \text{N} = 30.40 \, \text{N}. \]

Key Concepts

Normal Force

The normal force is an essential concept when considering circular motion. It refers to the force exerted by a surface (in this case, the walls of the cylinder) to support the car and keep it moving in a circle. At point A, which is the bottom of the circle, the normal force acts upwards against gravity. This is because gravity pulls down on the car, and the surface pushes up.
The normal force is not just counteracting gravity; it also provides the additional centripetal force necessary for circular motion. This means at the bottom of the circle, the normal force is stronger as it has to balance both gravity and the centripetal force required to keep the car on its circular path.

• At point A, the normal force is the sum of the gravitational force and centripetal force: 61.76 N.
• At point B, however, the scenario flips. Here, the normal force and gravitational pull both act downwards, contributing together to provide the required centripetal force.
• The normal force at the top, therefore, just fills in the gap to reach the needed centripetal force: 30.40 N.

Understanding how normal force works differently at various points in a circular path is key to mastering circular motion dynamics.

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. Without this force, an object would move off in a straight line due to inertia. When the remote-controlled car travels around the inside of the cylinder, this inward force is what keeps the car in its circular trajectory.
To calculate the centripetal force required, we use the formula: \[ F_c = \frac{mv^2}{r} \] where: \( m \) is the mass of the car, \( v \) is the speed, and \( r \) is the radius of the circle.
For the car:

• Mass \( m = 1.60 \; \text{kg} \)
• Speed \( v = 12.0 \; \text{m/s} \)
• Radius \( r = 5.00 \; \text{m} \)

With these values, the centripetal force is calculated to be 46.08 N. This force is constant for both points A and B but interacts differently with the other forces acting on the car, such as gravity and normal force.

Gravitational Force

Gravitational force always acts downward, that is, toward the center of the Earth. This force is constant and is calculated by the formula: \[ F_g = mg \] where \( g \) is the acceleration due to gravity, approximately 9.8 \( \text{m/s}^2 \). For our car with mass \( m = 1.60 \; \text{kg} \), the gravitational force \( F_g = 15.68 \; \text{N} \).
At any given point in the circle, gravity's pull remains the same in magnitude but interacts differently with other forces.

• At point A, gravity acts downward while normal force acts upward, both playing a role in ensuring the car's circular motion by meeting the centripetal force requirement together.
• At point B, both gravity and normal force point downwards, effectively adding up to provide the necessary centripetal force.

Understanding gravitational force in circular motion contexts is crucial because it consistently influences how other forces, such as the normal and centripetal forces, are applied or calculated in the system.