康涅狄格州大学：《普通物理》（英文版） Lecture 22 The Simple Pendulum

Physics 121: Lecture 22 Today's Agenda Announcements No class next weel Homework 9: due Friday Nov 18@ 6: 00 PM Chap.9:#1,7,9,13,17,23,24,27,29,37,39,and40. Today 's topics SHM Pendulum Damped oscillations Resonance Waves Physics 121: Lecture 22, Pg

Physics 121: Lecture 22, Pg 1 Physics 121: Lecture 22 Today’s Agenda Announcements No class next week Homework 9: due Friday Nov. 18 @ 6:00 PM. Chap. 9: # 1, 7, 9, 13, 17, 23, 24, 27, 29, 37, 39, and 40. Today’s topics SHM Pendulum Damped oscillations Resonance Waves

The simple Pendulum A pendulum is made by suspending a mass m at the end of a string of length L Find the frequency of oscillation for small displacements 0 L mg Physics 121: Lecture 22, Pg 2

Physics 121: Lecture 22, Pg 2 The Simple Pendulum A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.  L m mg z

Aside: sin e and cos e for small e A Taylor expansion of sin 0 and cos 0 about 0=0 gives sine=0 and cos0=1 3!5! So forθ<<1,sin≈0 and cOS0≈1 Physics 121: Lecture 22, Pg 3

Physics 121: Lecture 22, Pg 3 Aside: sin  and cos  for small  A Taylor expansion of sin  and cos  about  = 0 gives: ... 3! 5! sin 3 5 −  +   =  − ... 2! 4! cos 1 2 4 −  +  and  = − So for  <<1, sin   and cos 1

The Simple Pendulum. Recall that the torque due to gravity about the rotation(z) axis Isτ=la=-mgd d=Lsin0≈ Le for smal|0 soτ=-mgL0 melo= mLa Butτ=lo,/=mL2 a=-00 where VL m Same equation as for simple harmonic motion d 0=0ocos(ot+φ) mg Physics 121: Lecture 22, Pg 4

Physics 121: Lecture 22, Pg 4 The Simple Pendulum... Recall that the torque due to gravity about the rotation (z) axis is  = Ia = -mgd. d = Lsin   L for small  so  = -mg L But  = Ia  I = mL2  L d m mg z  a 2 − mgL = mL a   2 = −  = g L where Same equation as for simple harmonic motion !  = 0 cos(t + )

The rod pendulum a pendulum is made by suspending a thin rod of length L and mass M at one end Find the frequency of oscillation for small displacements X CM Physics 121: Lecture 22, Pg 5

Physics 121: Lecture 22, Pg 5 The Rod Pendulum A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements.  L mg z x CM

The rod Pendulum The torque about the rotation(z) axis is t=-mgd =-mgL2sine x-mgL210 for small 0 n this case I=。mL Soτ= la becomes L2 mg-0=mla XCM dL g a=-@ 0 where 2L Physics 121: Lecture 22, Pg 6

Physics 121: Lecture 22, Pg 6 The Rod Pendulum... The torque about the rotation (z) axis is  = -mgd = -mg{L/2}sin  -mg{L/2} for small  In this case So  = Ia becomes  d L mg z L/2 xCM I = 1 3 2 mL  a 2 3 1 2 mL L − mg = a   2 = −  = 3 2 g L where d I

Lecture 22: Act 1 Period has the same period as the rod pendulum? so that What length do we make the simple pendulum R 3 a R R Physics 121: Lecture 22, Pg7

Physics 121: Lecture 22, Pg 7 Lecture 22: Act 1 Period (a) (b) (c) What length do we make the simple pendulum so that it has the same period as the rod pendulum? LR LS S LR 3 2 S LR L = 2 3 L = LS = LR

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 Z-aXIs The torque about the rotation(z) axis for small e is(sin0≈0) d20 T=-Mgd≈-MgR MgRe=/ at e CM d20 o 0 where IgR 0=θcos(ot+d Physics 121: Lecture 22, Pg 8

Physics 121: Lecture 22, Pg 8 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 xCM d dt 2 2  2 = −   = MgR I where  = 0 cos(t + ) 2 2 dt d MgR I  −  =  a  

Lecture 22: Act 2 Physical 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 D (b) D VD 2D Physics 121: Lecture 22, Pg 9

Physics 121: Lecture 22, Pg 9 Lecture 22: Act 2 Physical 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) (a) (b) (c)  = g D  = 2g D  = g 2D D pivot (nail)

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 this produces a torque that he object is rotated, the wire is 0 twisted T (L opposes the rotation In analogy with a spring, the torque produced is proportional to the displacement: t=-k0 Physics 121: Lecture 22, Pg 10

Physics 121: Lecture 22, Pg 10 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 I wire  