Physics 151: Lecture 32 Today’s Agenda

1. Physics 151: Lecture 32 Today’s Agenda • Topics • The Pendulum – Ch. 15 • Potential energy and SHM

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

3. Lecture 32, Act 1Simple Harmonic Motion • 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/s2. b) 1.6 m/s2. c) 0.80 m/s2. d) 0.63 m/s2. e) 2.4 m/s2.

4.   where  = 0 cos(t + ) See text: 15.5 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 z-axis R  x CM d Mg

5. Lecture 32, Act 2Physical Pendulum • 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 ? (ICM = mR2 for a hoop) pivot (nail) (a) (b) (c) D

6. So Lecture 32, Act 2Solution • The angular frequency of oscillation of the hoop for small displacements will be given by Use parallel axis theorem: I = Icm + mR2 = mR2 + mR2 = 2mR2 pivot (nail) cm x R m

7. wire   I See text: 15.5 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. • The wire acts like a “rotational spring”. • When 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 See figure 13.15

8. wire   where I See text: 15.4 Torsion Pendulum... • Since = -k=Ibecomes Similar to “mass on spring”, except I has taken the place of m (no surprise) See figure 13.15

9. R R R R Lecture 32, Act 3Period • 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) C) B) A) D)

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

11. U K E U s -A 0 A See text: 15.3 Energy in SHM • For both the spring and the pendulum, we can derive the SHM solution using energy conservation. • The total energy (K + U) of a system undergoing SMH will always be constant! • This is not surprising since there are only conservative forces present, hence energy is conserved. Animation

12. U U K E U x x -A 0 A See text: Fig. 15.6 SHM and quadratic potentials • SHM will occur whenever the potential is quadratic. • Generally, this will not be the case: • For example, the potential betweenH atoms in an H2 molecule lookssomething like this:

13. U(x) = U(x0 ) + U(x0 ) (x- x0 ) + U (x0 ) (x- x0 )2+.... U U x0 x Definex = x - x0 andU(x0 ) = 0 Then U(x) = U (x0 )x2 x  See text: Fig. 15.6 SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for smalldisplacements, the potential IS quadratic: U(x) = 0 (since x0 is minimum of potential)

14. See text: Fig. 15.6 SHM and quadratic potentials... U(x) = U (x0)x2 Letk = U (x0) Then: U(x) = kx2 U U x0 x x  SHM potential !!

15. Recap of today’s lecture • Chapter 15 – Pendula • Simple Pendulum • Physical Pendulum • Torsional Pendulum • Next time: Damped and Driven Oscillations