Learning Objective 1
- 1. Simple Pendulum: A simplependulum in physics101 isa mass that is suspended from a string or rod of negligible mass. It is a resonantsystem with a single resonantfrequency. The period of a simple pendulum can me represented by the following equation: -The Motion of a simple pendulum is like simple harmonic motion in that the equation for the angular displacement is: -One of the most important concepts about these simple pendulum systems is the fact that when a hanging mass is displaced from its equilibrium point the restoring force, which brings back the mass to the center, is represented as: Due to: Try the three practice problems to help your understanding of how these simple pendulums work. http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html
- 2. 1) Based on conservation of energy laws, what are the kinetic and potential energies of the simple pendulems shown? Potential Energy: . Kinetic Energy: 25.0j . Potential Energy: 40.0j . Kinetic Energy: . Potential Energy: 10.0j . Kinetic Energy: 70.0j . Potential Energy: 0j . Kinetic Energy: . 2) Starting on the red dot on the position (x) vs. time graph, sketch the position (x), velocity (v), and accelleration(a) graphs of a simple pendulum PRACTICE:
- 3. 3) A bell’s clapper (the metal ball that swings inside a bell) of mass 0.6Kg is attached to a 2.8m long pole which makes a simple pendulum. A) What is the period of the pendulum? B) What is the maximum speed of the metal ball if it is pulled back to an angle of pi/10 radians and then released?
- 4. SOLUTIONS: Question 1: A) PE: 0j KE: 80.0j B) PE: 55.0j KE: 25.0j C) PE: 40.0j KE: 40.0j D) PE: 10.0j KE: 70.0j Question 2: Question 3: L Cos(Θ) The maxvelocity will be at the pointin which there is minimal potential energy and all kinetic. Therefore, when the pendulumis at the bottom of its swing, it will be movingthe fastest. The potential Energy that the pendulum had when it is pulled up is converted into kinetic. Therefore if wecan solve for h wewill be able to determinethe total energy in the system then use that to solve for Vmax =1.64m/s