Physics 151: Lecture 32, Pg 1 Physics 151: Lecture 32 Today’s Agenda l Topics çThe Pendulum – Ch. 15 çPotential energy and SHM

Physics 151: Lecture 32, Pg 2 Simple Harmonic Motion Review The most general solution is x = Acos(  t +  ) where A = amplitude  = frequency  = phase constant l For a mass on a spring çThe frequency does not depend on the amplitude !!! l The oscillation occurs around the equilibrium point where the force is zero! l Energy is a constant, it transfers between potential and kinetic. See text: 15.1 to 15.3 k x m 0

Physics 151: Lecture 32, Pg 3 Lecture 32, Act 1 Simple Harmonic Motion l You have landed your spaceship on the moon and want to determine the acceleration due to gravity using a simple pendulum of length 1.0 m. If he period of the pendulum is 5.0 s what is the value of g on the moon ? a)1.3 m/s 2. b)1.6 m/s 2. c)0.80 m/s 2. d)0.63 m/s 2. e)2.4 m/s 2.

Physics 151: Lecture 32, Pg 4 General Physical Pendulum Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small  is (sin  )  = -Mgd -MgR   d Mg z-axis R x CM where  =  0 cos(  t +  ) See text: 15.5  

Physics 151: Lecture 32, Pg 5 Lecture 32, Act 2 Physical Pendulum l A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail.  What is the angular frequency of oscillation of the hoop for small displacements ? ( I CM = mR 2 for a hoop) (a) (b) (c) D pivot (nail)

Physics 151: Lecture 32, Pg 6 Lecture 32, Act 2 Solution l The angular frequency of oscillation of the hoop for small displacements will be given by pivot (nail) R cm x Use parallel axis theorem: I = I cm + mR 2 m = mR 2 + mR 2 = 2mR 2 So

Physics 151: Lecture 32, Pg 7 Torsion Pendulum Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. l The wire acts like a “rotational spring”. çWhen the object is rotated, the wire is twisted. This produces a torque that opposes the rotation.  In analogy with a spring, the torque produced is proportional to the displacement:  = -k  I wire   See text: 15.5 See figure 13.15

Physics 151: Lecture 32, Pg 8 Torsion Pendulum... Since  = -k   = I  becomes I wire   Similar to “mass on spring”, except I has taken the place of m (no surprise) See text: 15.4 See figure where

Physics 151: Lecture 32, Pg 9 Lecture 32, Act 3 Period l All of the following pedulum bobs have the same mass. Which pendulum rotates the fastest, i.e. has the smallest period? (The wires are identical) RRRR A) B) C) D)

Physics 151: Lecture 32, Pg 10 Lecture 32, Act 5 Period l Check each case. A) B) C) R D) The biggest is (D), the smallest moment of inertia

Physics 151: Lecture 32, Pg 11 Energy in SHM l For both the spring and the pendulum, we can derive the SHM solution using energy conservation. l The total energy (K + U) of a system undergoing SMH will always be constant! l This is not surprising since there are only conservative forces present, hence energy is conserved. -AA0 s U U K E See text: 15.3 Animation

Physics 151: Lecture 32, Pg 12 SHM and quadratic potentials l SHM will occur whenever the potential is quadratic. l Generally, this will not be the case: l For example, the potential between H atoms in an H 2 molecule looks something like this: -AA0 x U U K E See text: Fig U x

Physics 151: Lecture 32, Pg 13 SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for small displacements, the potential IS quadratic: U x U(x) = U(x 0 ) + U(x 0 ) (x- x 0 ) + U  (x 0 ) (x- x 0 ) U(x) = 0 (since x 0 is minimum of potential) x0x0 U x Define x = x - x 0 and U(x 0 ) = 0 Then U(x) = U  (x 0 ) x 2 See text: Fig. 15.6

Physics 151: Lecture 32, Pg 14 SHM and quadratic potentials... U x x0x0 U x U(x) = U  (x 0 ) x 2 Let k = U  (x 0 ) Then: U(x) = k x 2 SHM potential !! See text: Fig. 15.6

Physics 151: Lecture 32, Pg 15 Recap of today’s lecture l Chapter 15 – Pendula çSimple Pendulum çPhysical Pendulum çTorsional Pendulum l Next time: Damped and Driven Oscillations