ps09sol - 11/7/2010 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 11/7/2010 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Problem Set 9: Practice Problems Exam 2 Problem 1: Simple Pendulum by Energy Method Solution A simple pendulum consists of a massless string of length l and a point like object of mass m is attached to one end. Suppose the string is fixed at the other end and is initially pulled out at an angle of ! from the vertical and released at rest. a. Use the fact that the energy is constant to find a differential equation describing how the second derivative of the angle ! the object makes with the vertical varies in time. b. Find an expression for the angular velocity of the object at the bottom of its swing. c. Now assume that the initial angle 1 rad ! << and thus use the small angle approximation sin ! ! " to show that the simple pendulum behaves like a simple harmonic oscillator. d. Also use the approximation 2 cos 1 2 ! ! " # , to find an expression for the angular velocity of the object at the bottom of its swing. Solution: We can use energy methods to find the differential equation describing the time evolution of the angle ! . Choose the coordinate system shown in the figure below. When the string is at an angle ! with respect to the vertical, the gravitational potential energy (relative to a choice of zero potential energy at the bottom of the swing where ! = ) is given by 11/7/2010 2 U = mgl 1 ! cos " ( ) (1.1) The component of the velocity of the object is given by v ! = l ( d ! / dt ) so the kinetic energy is K = 1 2 mv ! 2 = 1 2 m l d ! dt " # $ % & ' 2 . (1.2) The energy of the system is then E = K + U = 1 2 m l d ! dt " # $ % & ' 2 + mgl 1 ( cos ! ( ) (1.3) Since there is no non-conservative work, the energy is constant hence = dE dt = 1 2 m 2 l 2 d ! dt d 2 ! dt 2 + mgl sin ! d ! dt = ml 2 d ! dt d 2 ! dt 2 + g l sin ! " # $ % & ' . (1.4) There are two solutions to this equation, the first one d ! / dt = is the equilibrium solution, the angular speed is zero means the suspended object is not moving. The second solution is the one we are interested in d 2 ! dt 2 + g l sin ! = . (1.5) We can also find the component of the angular velocity at the bottom of the arc using energy methods. When we release the bob from rest, the energy is only potential energy ( ) 1 cos E U mgl ! = = " (1.6) When the bob is at the bottom of the arc, the only contribution to the energy is the kinetic energy given by 2 2 2 1 1 1 1 1 2 2 d K mv ml dt ! " # = = $ % & ' . (1.7) Since the energy is constant, we have that U = K 1 or 11/7/2010 2 U = mgl 1 ! cos " ( ) (1.1) The component of the velocity of the object is given by v ! = l ( d ! / dt ) so the kinetic energy is K = 1 2 mv ! 2 = 1 2 m l d !...
View Full Document

This note was uploaded on 04/13/2011 for the course PHYSICS 8.01 taught by Professor Guth during the Fall '09 term at MIT.

Page1 / 39

ps09sol - 11/7/2010 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online