Module 22: Simple Harmonic Oscillation and Torque 22.1 Introduction We have already used Newton’s Second Law or Conservation of Energy to analyze systems like the bloc-spring system that oscillate. We shall now use torque and the rotational equation of motion to study oscillating systems like pendulums or torsional springs. 22.2 Simple Pendulum A pendulum consists of an object hanging from the end of a string or rigid rod pivoted about the point S . The object is pulled to one side and allowed to oscillate. If the object has negligible size and the string or rod is massless, then the pendulum is called a simple pendulum. The force diagram for the simple pendulum is shown in Figure 22.1. Figure 22.1 A simple pendulum. The string or rod exerts no torque about the pivot point S . The weight of the object has ˆ radial rˆ - and ! - components given by ! mg mg(cos rˆ sin ˆ) (22.2.1) = ! " ! ! and the torque about the pivot point S is given by ! ! ! ! = r " mg = l rˆ " m g(cos# rˆ $ sin# #ˆ) = $l m g sin# kˆ (22.2.2) S S , m and so the component of the torque in the z -direction (into the page in Figure 22.1 for ! positive, out of the page for ! negative) is (! S )z = "mgl sin# . (22.2.3) The moment of inertia of a point mass about the pivot point S is 12/28/2010 1 2 IS = ml . (22.2.4) From the rotational dynamical equation is d 2$ (! ) = I " # I S z S S dt 2 (22.2.5) 2 2 d $ %mgl sin$ = ml 2 . dt Thus we have the equation of motion for the simple pendulum, d 2! g = " sin! . (22.2.6) dt 2 l When the angle of oscillation is small, then we can use the small angle approximation sin! " ! ; (22.2.7) the rotational dynamical equation for the pendulum becomes d 2! g " # ! . (22.2.8) dt 2 l This equation is similar to the object-spring simple harmonic oscillator differential equation from (add reference), d 2 x k = ! x , (22.2.9) dt 2 m which describes the oscillation of a mass about the equilibrium point of a spring. Recall that in (add reference), the angular frequency of oscillation was given by k ! = . (22.2.10) spring m By comparison, the frequency of oscillation for the pendulum is approximately g ! " , (22.2.11) pendulum l with period 12/28/2010 2 2! l T = # 2! . (22.2.12) "pendulum g A procedure for determining the period for larger angles is given in Appendix 22.A. 22.3 Physical Pendulum A physical pendulum consists of a rigid body that undergoes fixed axis rotation about a fixed point S (Figure 22.2). The gravitational force acts at the center of mass of the physical pendulum. Suppose the center of mass is a distance lcm from the pivot point S . Figure 22.2 Physical pendulum. The analysis is nearly identical to the simple pendulum. The torque about the pivot point is given by ! ! ! ! = r " mg = l rˆ " m g(cos# rˆ $ sin# #ˆ) = $l m g sin# kˆ . (22.3.1) S S ,cm cm cm Following the same steps that led from Equation (22.2.2) to Equation (22.2.6), the rotational dynamical equation for the physical pendulum is d 2! (" S ) = IS# = IS 2 z dt (22.3.2) d 2! $mgl sin! = I . cm S dt 2 Thus we have the equation of motion for the physical pendulum, 12/28/2010 3 2 d ! mglcm 2 = " sin! . (22.3.3) dt IS As with the simple pendulum, for small angles sin! "! and Equation (22.3.3) reduces to the simple harmonic oscillator equation with angular frequency mg lcm !pendulum " (22.3.4) IS and period 2! IS Tphysical = # 2! . (22.3.5) "pendulum m g lcm It is sometimes convenient to express the moment of inertia about the pivot point in terms of lcm and Icm using the parallel axis theorem (add link), with dS ,cm ! lcm , 2 IS = Icm + mlcm , with the result lcm Icm Tphysical " 2! + . (22.3.6) g m g lcm 2 Thus, if the object is “small” in the sense that Icm ! mlcm , the expressions for the physical pendulum reduce to those for the simple pendulum. Note that this is not the case shown in Figure 22.11. 22.3.1 Example: Oscillating rod A physical pendulum consists of a uniform rod of length d and mass m pivoted at one end. The pendulum is initially displaced to one side by a small angle and released !0 from rest. You can then approximate sin! " ! (with ! measured in radians). Find the period of the pendulum. We shall find the period of the pendulum using two different methods. 1. Applying the torque equation about the pivot point. 12/28/2010 4 2. Applying the energy equation. Applying the torque equation about the pivot point. With our choice of rotational coordinate system the angular acceleration is given by 2 ! d " ˆ ! = 2 k . (22.3.7) dt The force diagram on the pendulum is shown below. In particular, there is an unknown pivot force, the gravitational force acting at the center of mass of the rod. The torque about the pivot point is given by ! ! ! ! = r " mg . (22.3.8) P P,cm The rod is uniform, therefore the center of mass is a distance d / 2 from the pivot point. The gravitational force acts at the center of mass, so the torque about the pivot point is given by ! ! = (d / 2)rˆ " mg(#sin$ $ˆ + cosrˆ) = #(d / 2)mg sin$ kˆ . (22.3.9) P The rotational dynamical equation (torque equation) is ! ! ! = I " . (22.3.10) P P Therefore 2 ˆ d " ˆ !(d / 2)mg sin" k = I P 2 k . (22.3.11) dt When the angle of oscillation is small, then we can use the small angle approximation 12/28/2010 5 sin! " ! . (22.3.12) Then the torque equation becomes d 2! (d / 2)mg + ! = 0 (22.3.13) dt 2 I P which is the simple harmonic oscillator equation. The angular frequency of oscillation for the pendulum is approximately (d / 2)mg ! " . (22.3.14) 0 I P The moment of inertia of a rod about the end point P is I (1/ 3)md 2 therefore the P = angular frequency is (d / 2)mg (3 / 2)g ! " = (22.3.15) 0 (1/ 3)md 2 d with period 2! 2 d T = # 2! . (22.3.16) " 3 g 0 Applying the energy equation. Take the zero point of gravitational potential energy to be the point where the center of mass of the pendulum is at its lowest point, that is, ! = 0 . When the pendulum is at an angle ! the potential energy is 12/28/2010 6 d U = m g (1" cos! ). (22.3.17) 2 The kinetic energy of rotation about the pivot point is 1 K = I ! 2 . (22.3.18) rot 2 p The mechanical energy is then d 1 2 E = U + Krot = m g (1# cos! )+ I p" , (22.3.19) 2 2 2 with IP = (1/ 3)md . There are no non-conservative forces acting, so the mechanical energy is constant, and therefore its time derivative is zero dE d d! d" 0 = = m g sinè + I " . (22.3.20) dt 2 dt p dt Recall that ! = d" / dt and d! / dt = d 2" / dt 2 so Eq. (22.3.20) becomes d d 2! 0 = m g sin!" + I " . (22.3.21) 2 p dt 2 Therefore two solutions, ! = 0 , in which the same remains at the bottom of the swing and d d 2! 0 = m g sin! + I (22.3.22) 2 p dt 2 Using the small angle approximation, we have the simple harmonic oscillator equation (Eq. (22.3.13)) d 2! m g(d / 2) 2 + ! = 0 . (22.3.23) dt I p 22.3.2 Example: Torsional Oscillator Solution A disk with moment of inertia about the center of mass Icm rotates in a horizontal plane. It is suspended by a thin, massless rod. If the disk is rotated away from its equilibrium position by an angle ! , the rod exerts a restoring torque given by ! cm = $"# . At t = 0 , 12/28/2010 7 the disk is released from rest at an angular displacement of !0 . Find the subsequent time dependence of the angular displacement ! (t) . Solution: Choose a coordinate system such that kˆ is pointing upwards, then the angular acceleration is given by 2 ! d " ˆ ! = 2 k . (22.3.24) dt The torque about the center of mass is given in the statement of the problem as a restoring torque ! ! = "#$ kˆ . (22.3.25) cm ! ! Therefore the kˆ -component of the torque equation ! = I " is cm cm d 2# !"# = Icm 2 . (22.3.26) dt This is a simple harmonic oscillator equation with solution (t) Acos( t) Bsin( t) (22.3.27) ! = "0 + "0 where the angular frequency of oscillation is given by ! = " / I . (22.3.28) 0 cm The z -component of the angular velocity is given by d! (t) = "#0 Asin(#0 t) + #0 Bcos(#0 t) . (22.3.29) dt 12/28/2010 8 The initial conditions at t 0, are that ! (t = 0) = A = ! , and (d / dt)(t 0) B 0 . = 0 ! = = "0 = Therefore !(t) = ! cos( " / I t) . (22.3.30) 0 cm Appendix 22.A: Higher-Order Corrections to the Period for Larger Amplitudes of a Simple Pendulum In Section 22.2 we found that using the small angle approximation the period for a simple pendulum is (22.2.12) 2! l T = # 2! .