College Physics 1: Lecture 21 - Solving Circular Dynamics Problems

In Lecture 21 of College Physics 1, the focus is on solving circular dynamics problems. The lecture begins by revisiting the concept of circular motion, where an object moves in a circular path at a constant speed but with a continuously changing direction, resulting in acceleration. The necessity of a centripetal force to keep the object in its circular path is emphasized. The lecture's goal is to teach problem-solving strategies for forces in circular motion, including drawing free body diagrams and applying Newton's laws. A unique aspect of these problems is the orientation of the x-axis, which always points toward the center of the circle, differing from the standard Cartesian coordinate system.

The lecture outlines a problem-solving strategy for circular dynamics problems, which involves drawing a free body diagram with a twist: the x-axis is always drawn pointing toward the center of the circle, regardless of the orientation of the motion. This approach simplifies the math involved in solving the problems. The strategy also includes writing Newton's laws for the forces in the x and y directions. It is highlighted that in circular motion problems, the x equation often simplifies to mv^2/r, derived from F=ma, where 'a' is the centripetal acceleration v^2/r. The y direction typically results in a net force of zero, as the forces balance out. The importance of determining the speed first, either through force equations or circular kinematics, is stressed as it is essential for solving other aspects of the problem.

The first example problem involves a father spinning his 20 kg child on a 5 kg cart using a 2-meter rope with a tension of 100 N. The goal is to find the time taken for one complete rotation, which translates to finding the period of the motion. The problem-solving strategy is applied by drawing a free body diagram, identifying the forces acting on the child (weight, normal force, and tension), and setting up Newton's laws accordingly. The tension force in the x direction is equated to mv^2/r, and the normal force equals the weight in the y direction. The speed of the child is calculated using the given tension, radius, and total mass. With the speed determined, the period is then found using the relationship between speed, distance (2πr), and time (period).

The second example addresses a level curve with a 150-meter radius and seeks to determine the maximum speed at which a car can safely traverse the curve on a rainy day with a static friction coefficient (μs) of 0.4. The problem introduces the concept of static friction as the force that prevents the car from sliding off the road. Newton's second law is applied in both the x and y directions. The x direction involves static friction as the centripetal force, while the y direction confirms that the normal force equals the weight. The challenge is to find the maximum speed without causing the car to skid. The solution involves using the friction equation (μs * N) and the normal force (N = mg) to solve for the speed, which turns out to be approximately 24 m/s.

The lecture concludes with a review of free body diagrams for various circular motion scenarios, emphasizing the correct identification of forces acting on objects in motion. Three questions are presented, each with a different scenario: a pendulum swinging back and forth, a car coasting over a hill, and a roller coaster car at the top of a loop-de-loop. For each scenario, the correct free body diagram is chosen based on the forces present, such as weight, tension, and normal force. The importance of recognizing that there is no additional force propelling the object in the direction of motion is highlighted, and the correct diagrams are identified for each scenario.

A special case is discussed involving a roller coaster car at the top of a loop-de-loop. The normal force is highlighted as being downward, contrary to the common misconception that it acts upward. This is due to the surface of the track being above the car, which means the normal force acts in the opposite direction to the weight, both pointing downward. This scenario is set up to lead into the next lecture, which will delve deeper into apparent forces in circular motion.