Module 22: Simple Harmonic Oscillation and Torque

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1 Modue : Simpe Harmonic Osciation and Torque.1 Introduction We have aready used Newton s Second Law or Conservation of Energy to anayze systems ike the boc-spring system that osciate. We sha now use torque and the rotationa equation of motion to study osciating systems ike penduums or torsiona springs.. Simpe Penduum A penduum consists of an object hanging from the end of a string or rigid rod pivoted about the point S. The object is pued to one side and aowed to osciate. If the object has negigibe size and the string or rod is massess, then the penduum is caed a simpe penduum. The force diagram for the simpe penduum is shown in Figure.1. Figure.1 A simpe penduum. The string or rod exerts no torque about the pivot point S. The weight of the object has radia ˆr - and ˆ! - components given by and the torque about the pivot point S is given by m! g = mg(cos! ˆr " sin! ˆ!) (..1)! S =! r S, m " m! g = ˆr " m g(cos# ˆr $ sin# ˆ#) = $ m g sin# ˆk (..) and so the component of the torque in the z -direction (into the page in Figure.1 for! positive, out of the page for! negative) is (! S ) z = "mg sin#. (..3) The moment of inertia of a point mass about the pivot point S is 1/8/1 1

2 IS = m. (..4) From the rotationa dynamica equation is (! S ) z = I S " # I S d $ dt %mg sin$ = m d $ dt. (..5) Thus we have the equation of motion for the simpe penduum, d! g = " sin!. (..6) dt When the ange of osciation is sma, then we can use the sma ange approximation the rotationa dynamica equation for the penduum becomes sin! "! ; (..7) d! g " #!. (..8) dt This equation is simiar to the object-spring simpe harmonic osciator differentia equation from (add reference), d x dt k =! x, (..9) m which describes the osciation of a mass about the equiibrium point of a spring. Reca that in (add reference), the anguar frequency of osciation was given by k! spring =. (..1) m By comparison, the frequency of osciation for the penduum is approximatey with period g! penduum ", (..11) 1/8/1

3 T! = #!. (..1) " g penduum A procedure for determining the period for arger anges is given in Appendix.A..3 Physica Penduum A physica penduum consists of a rigid body that undergoes fixed axis rotation about a fixed point S (Figure.). The gravitationa force acts at the center of mass of the physica penduum. Suppose the center of mass is a distance from the pivot point S. Figure. Physica penduum. The anaysis is neary identica to the simpe penduum. The torque about the pivot point is given by! S =! r S, " m! g = ˆr " m g(cos# ˆr $ sin# ˆ#) = $ m g sin# ˆk. (.3.1) Foowing the same steps that ed from Equation (..) to Equation (..6), the rotationa dynamica equation for the physica penduum is ( ) " = I # = I S z S S d! $ mg sin! = IS. dt d! dt (.3.) Thus we have the equation of motion for the physica penduum, 1/8/1 3

4 d! mg = "!. (.3.3) dt sin IS As with the simpe penduum, for sma anges sin! "! and Equation (.3.3) reduces to the simpe harmonic osciator equation with anguar frequency and period mg! penduum " (.3.4) IS T physica! IS = #!. (.3.5) " m g penduum It is sometimes convenient to express the moment of inertia about the pivot point in terms of and I using the parae axis theorem (add ink), with ds,!, S I = I + m, with the resut T I physica "! +. (.3.6) g m g Thus, if the object is sma in the sense that I! m, the expressions for the physica penduum reduce to those for the simpe penduum. Note that this is not the case shown in Figure Exampe: Osciating rod A physica penduum consists of a uniform rod of ength d and mass m pivoted at one end. The penduum is initiay dispaced to one side by a sma ange! and reeased from rest. You can then approximate sin! "! (with! measured in radians). Find the period of the penduum. We sha find the period of the penduum using two different methods. 1. Appying the torque equation about the pivot point. 1/8/1 4

5 . Appying the energy equation. Appying the torque equation about the pivot point. With our choice of rotationa coordinate system the anguar acceeration is given by! = d " dt ˆk. (.3.7) The force diagram on the penduum is shown beow. In particuar, there is an unknown pivot force, the gravitationa force acting at the center of mass of the rod. The torque about the pivot point is given by! P =! r P, " m! g. (.3.8) The rod is uniform, therefore the center of mass is a distance d / from the pivot point. The gravitationa force acts at the center of mass, so the torque about the pivot point is given by! P = (d / )ˆr " mg(#sin$ ˆ$ + cos ˆr) = #(d / )mg sin$ ˆk. (.3.9) The rotationa dynamica equation (torque equation) is Therefore! P = I P! ". (.3.1)!(d / )mg sin" ˆk = I P d " dt ˆk. (.3.11) When the ange of osciation is sma, then we can use the sma ange approximation 1/8/1 5

6 sin! "!. (.3.1) Then the torque equation becomes d! (d / )mg +! = (.3.13) dt I P which is the simpe harmonic osciator equation. The anguar frequency of osciation for the penduum is approximatey! " (d / )mg I P. (.3.14) The moment of inertia of a rod about the end point P is I P = (1/ 3)md anguar frequency is therefore the! " (d / )mg (3 / )g = (1 / 3)md d (.3.15) with period T =! #! d " 3 g. (.3.16) Appying the energy equation. Take the zero point of gravitationa potentia energy to be the point where the center of mass of the penduum is at its owest point, that is,! =. When the penduum is at an ange! the potentia energy is 1/8/1 6

7 d U = m g ( 1" cos! ). (.3.17) The kinetic energy of rotation about the pivot point is The mechanica energy is then 1 K I! rot = p. (.3.18) d 1 E = U + Krot = m g ( 1# cos! ) + I p", (.3.19) with I P = (1/ 3) md. There are no non-conservative forces acting, so the mechanica energy is constant, and therefore its time derivative is zero = de = m g d sinè d! + I d p" ". (.3.) dt dt dt Reca that! = d" / dt and d! / dt d " / dt = so Eq. (.3.) becomes d d! = m g sin!" + I p". (.3.1) dt Therefore two soutions,! =, in which the same remains at the bottom of the swing and d d! = m g sin! + I p (.3.) dt Using the sma ange approximation, we have the simpe harmonic osciator equation (Eq. (.3.13)) d! m g( d / ) +! =. (.3.3) dt I p.3. Exampe: Torsiona Osciator Soution A disk with moment of inertia about the center of mass I rotates in a horizonta pane. It is suspended by a thin, massess rod. If the disk is rotated away from its equiibrium position by an ange!, the rod exerts a restoring torque given by! = $ "#. At t =, 1/8/1 7

8 the disk is reeased from rest at an anguar dispacement of!. Find the subsequent time dependence of the anguar dispacement! ( t). Soution: Choose a coordinate system such that ˆk is pointing upwards, then the anguar acceeration is given by! = d " dt ˆk. (.3.4) The torque about the center of mass is given in the statement of the probem as a restoring torque! = "#$ ˆk. (.3.5) Therefore the ˆk -component of the torque equation!! = I! " is This is a simpe harmonic osciator equation with soution where the anguar frequency of osciation is given by The z -component of the anguar veocity is given by!"# = I d # dt. (.3.6)!(t) = Acos(" t) + Bsin(" t) (.3.7)! = " / I. (.3.8) d! dt (t) = "# Asin(# t) + # Bcos(# t). (.3.9) 1/8/1 8

9 The initia conditions at t =, are that! ( t = ) = A =!, and (d! / dt)(t = ) = " B =. Therefore!(t) =! cos( " / I t). (.3.3) Appendix.A: Higher-Order Corrections to the Period for Larger Ampitudes of a Simpe Penduum In Section. we found that using the sma ange approximation the period for a simpe penduum is (..1) T! = #!. " g penduum How good is this approximation? If the penduum is pued out to an initia ange! that is not sma (such that our first approximation sin! "! no onger hods) then our expression for the period is no onger vaid. We then woud ike to cacuate the firstorder (or higher-order) correction to the period of the penduum. Let s first consider the mechanica energy, a conserved quantity in this system. Choose an initia state when the penduum is reeased from rest at an ange! ; this need not be at time t =, and in fact ater in this derivation we see that it s inconvenient to choose this position to be at t =. Choose for the fina state the position and veocity of the bob at an arbitrary time t. Choose the zero point for the potentia energy to be at the bottom of the bob s swing (Figure.A.1). Figure.A.1 Energy states for a simpe penduum. The initia mechanica energy is E = K + U = m g (1 " cos! ). (.A.1) The tangentia veocity of the bob at an arbitrary time t is given by 1/8/1 9

10 d! v =, (.A.) dt tan and the kinetic energy at the fina state is The fina mechanica energy is then 1 1 " d # = tan = $ %! K f mv m & dt '. (.A.3) 1 " d! # E f = K f + U f = m% & + m g (1 $ cos! ). (.A.4) ' dt ( Since the tension in the string is aways perpendicuar to the dispacement of the bob, the tension does no work and mechanica energy is conserved, E f = E. Thus 1 m " d! % # $ dt & ' " d! % # $ dt & ' + m g (1( cos!) = m g (1( cos! ) = g (cos! ( cos! ). (.A.5) We can sove Equation (.A.5) for the anguar veocity as a function of!, d! g = cos! " cos!. (.A.6) dt Note that we have taken the positive square root, impying that d! / dt ". This ceary cannot aways be the case, and we shoud change the sign of the square root every time the penduum s direction of motion changes. For our purposes, this is not an issue. If we wished to find an expicit form for either!(t) or t (! ), we woud have to consider the signs in Equation (.A.6) more carefuy. Before proceeding, it s worth considering the difference between Equation (.A.6) and the equation for the simpe penduum in the simpe harmonic osciator imit, d! g!! = ". (.A.7) dt In both Equations (.A.6) and (.A.7) the ast term in the square root is proportiona to the difference between the initia potentia energy and the fina potentia energy. The 1/8/1 1

11 fina potentia energy for the two cases is potted in Figures.A. beow for #! < " <! on the eft and #! / < " <! / on the right (the vertica scae is in units of mg ). Figures.A. Potentia energies as a function of dispacement ange. It woud seem to be to our advantage to express the potentia energy for an arbitrary dispacement of the penduum as the difference between two squares. This is done by recaing the trigonometric identity with the resut that Equation (.A.6) may be re-expressed as 1! cos" = sin (" / ) (.A.8) d! dt = g (sin (! / ) " sin (! / )). (.A.9) (Note that using Equation (.A.8) is undoing one step of (.A.5).) Equation (.A.9) is separabe, d! (! )" (! ) sin / sin / = g dt (.A.1) What comes next may seem ike it s been pued out of a hat. The motivation is anaogous to for the simpe harmonic osciator. That is, rewrite Equation (.A.1) as d! (! ) (! ) sin / sin (! / ) 1" sin / = g dt. (.A.11) 1/8/1 11

12 The ratio sin(! / ) / sin(! / ) in the square root in the denominator wi osciate (but not with simpe harmonic motion) between! 1 and 1, and so we wi make the identification sin! = sin(" / ) sin(" / ). (.A.1) If! were proportiona to the time t, the motion woud be simpe harmonic motion. We don t expect this to be the case, but we might make some simpifications in doing the needed integration, and aow comparison to simpe harmonic motion. Let k = sin(! / ), so that sin! = k sin" cos! = $ 1#! ' % & sin ( ) 1 = (1# k sin ") 1. (.A.13) Pease be sure to note that here k = sin(! / ) is a dimensioness parameter of the system, not a spring constant. The integra in (.A.11) can then be rewritten as g dt $ $. (.A.14) k d! 1# sin " = From differentiating the first expression in Equation (.A.13), we have that 1! cos d! = k cos " d" cos" 1# sin " d! = k d" = k d" cos(! / ) 1# sin (! / ) (.A.15) 1# sin " = k d". 1# k sin " Substituting the ast equation in (.A.15) into the eft-hand side of the integra in (.A.14) yieds 1/8/1 1

13 k 1" sin! d! # d! = #. (.A.16) k 1" sin! 1" k sin! 1" k sin! Thus the integra in Equation (.A.14) becomes d! 1" k sin! = g dt # #. (.A.17) This integra is one of a cass of integras known as eiptic integras. We wi encounter a simiar integra when we sove the Keper Probem, where the orbits of an object under the infuence of an inverse square gravitationa force are determined. We find a power series soution to this integra by expanding the function (1! k sin ")!1 = 1+ 1 k sin " k 4 sin 4 " + ###. (.A.18) The integra in Equation (.A.17) then becomes " 1 3 # ' 8 ( 4 4 )% 1+ k sin! + k sin! +$$$ & d! = ) dt. (.A.19) Now et s integrate over one period. Set t = when! =, the owest point of the swing, so that sin! = and! =. One period T has eapsed the second time the bob returns to the owest point, or when! = ". Putting in the imits of the! -integra, we can integrate term by term, noting that g " 1 k sin " 1 #! d! = k 1 # (1$ cos(!)) d! = 1 k 1 %! $ sin(!) ( & ' ) * " (.A.) = 1 "k = 1 " sin +. Thus, from Equation (.A.19) we have that 1/8/1 13

14 We can now sove for the period, $ 1 3 % ) 8 *! T k sin k sin d dt ' + " + " +&&& ( " = #!! sin + + &&& = g g T, (.A.1) T # 1! $ & ' g ( 4 ). (.A.) = " 1+ sin + %%% If the initia ange is sma compared to 1 (measured in radians),! << 1, then sin (! / ) "! / 4 and the period is approximatey T # 1 1 $ #! " 1 1 $ % + = + " g & ' T & ' ( 16 ) ( 16 ), (.A.3) where T =! (.A.4) g is the period of the simpe penduum with the standard sma ange approximation. The first order correction to the period of the penduum is then 1 " T1 =! T. (.A.5) 16 Figure.A.3 beow shows the three functions given in Equation (.A.4) (the horizonta, or red pot if seen in coor), Equation (.A.3) (the midde, paraboic or green pot) and the numericay-integrated function obtained by integrating the expression in Equation (.A.17) (the upper, or bue pot) between! = and! = " as a function of k = sin(! / ). The pots demonstrate that Equation (.A.4) is a vaid approximation for sma vaues of!, and that Equation (.A.3) is a very good approximation for a but the argest ampitudes of osciation. The vertica axis is in units of / g. Note the dispacement of the horizonta axis. 1/8/1 14

15 Figure.A.3 Penduum Period Approximations as Functions of Ampitude. 1/8/1 15

16 MIT OpenCourseWare 8.1SC Physics I: Cassica Mechanics For information about citing these materias or our Terms of Use, visit: 1/8/1 16

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