Lect22 - Physics 211 Lecture 22 Today's Concepts Simple Harmonic Motion Motion of a Pendulum Physics 211 Lecture 22 Slide 1 Physics Is the final

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Unformatted text preview: Physics 211 Lecture 22 Today's Concepts: Simple Harmonic Motion: Motion of a Pendulum Physics 211 Lecture 22, Slide 1 Physics Is the final going to be evenly divided into the various topics throughout the Is course or will it focus on the newer material? The lack of a prelecture. I'm like a lost soul out in the wilderness with nothing but the cruel sun to beat down upon me. Woe is me! Will I never know the true nature of the pendulum? Excuse me while I cry in the corner. im disappointed that the pre-lecture is not ready...they really help me disappointed understand I don't know why, but angular anything seems confusing to me. I'm not sure how we determine if we're supposed to use sin or cos. Can we postpone one of the homework assignments until next week if at least 85% of us get the clicker questions right? HOW MUCH MORE DO WE HAVE TO LEARN?! The part about our final being at 8am. What is up with that? I mean, can't we just have it at night like our hour exams? Or in the afternoon? Why my roommate keeps on ramming Grape Fantas into my mini fridge is way into more confusing than anything that we've covered this year So this neutron walks into a bar and asks, "How much for a drink?" The bartender replies, "For you, no charge." Torsion Pendulum Torsion τ = Iα −κθ d 2θ I2 dt wire wire d 2θ = −ω 2θ dt 2 ω= κ θ τ I I θ(t ) = θ max cos(ωt + φ) Physics 211 Lecture 22, Slide 3 Preflight A torsion pendulum is used as the timing element in a clock as shown. The speed of the clock is adjusted by changing the distance of two small disks from the rotation axis of the pendulum. If we adjust the disks so that they are closer to the rotation axis, the clock runs A) Faster B) Slower Small disks Physics 211 Lecture 22, Slide 4 If we adjust the disks so that they are closer to the rotation axis, the clock runs A) Faster ω= B) Slower κ I A) The moment of inertia decreases, so the angular frequency increases, which makes the period shorter and thus the clock faster. B) Smaller moment of inertia means it takes less torque to turn the pendulum and results in a shorter period.. Physics 211 Lecture 22, Slide 5 Pendulum pivot τ = Iα For small θ small θ xcm cm − Mgxcm − MgRcmθ d 2θ I2 dt MgRcm d 2θ =− θ 2 dt I dθ = −ω 2θ dt 2 2 ω= Rcm MgRcm I θ R xcm = c-length ar Rθ Physics 211 Lecture 22, Slide 6 The Simple Pendulum pivot pivot The simple case θ Rcm θ cm L M M ω= MgRcm I ω= MgL = 2 ML g L The general case Physics 211 Lecture 22, Slide 7 Preflight A simple pendulum is used as the timing element in a clock as shown. An adjustment screw is used to make the pendulum shorter (longer) by moving the weight up (down) along the shaft that connects it to the pivot. If the clock is running too fast, the weight needs to be moved A) Up B) Down Adjustment screw Physics 211 Lecture 22, Slide 8 If the clock is running too fast, the weight needs to be moved A) Up B) Down ω= g L A) since T= (L/G)^1/2 decreasing the length will shorten the period. B) The longer the length of a pendulum, the longer the period. Physics 211 Lecture 22, Slide 9 The Stick Pendulum pivot θ Rcm cm MgRcm ω= I M L L 2 1 ML2 3 ω= g 2 L 3 2 L 3 Same period Physics 211 Lecture 22, Slide 10 Preflight Case 1 Case 2 m m m In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to the center of the same stick. In which case is the period of the pendulum the longest? A) Case 1 B) Case 2 C) Same Physics 211 Lecture 22, Slide 11 Case 1 Case 2 1 L 2 L m ω= g 2 L 3 ω= m g 1 L 2 In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to a string of length L/2? In which case is the period of the pendulum longest? A) Case 1 B) Case 2 C) Same Physics 211 Lecture 22, Slide 12 m Suppose you start with 2 different pendula, one having period T1 and the other having period T2. T2 m T1 m T1 > T2 Now suppose you make a new pendulum by hanging the first two from the same pivot and gluing them together. What is the period of the new pendulum? m A) T1 B) T2 C) In between Physics 211 Lecture 22, Slide 13 Physics Case 1 Case 2 m m m In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to the center of the same stick. In which case is the period of the pendulum the longest? A) Case 1 B) Case 2 C) Same Physics 211 Lecture 22, Slide 14 m m Lets compare L mg 2 ω= m MgRcm for each case I L 2mg 2 Physics 211 Lecture 22, Slide 15 m m Lets compare 12 mL 3 ω= m MgRcm for each case I 12 42 2 mL + mL = mL 3 3 1 21 25 2 mL + mL = mL 3 2 6 2 (A) (B) 12 7 ⎛L⎞ 2 mL + m⎜ ⎟ = mL (C) 3 ⎝ 2 ⎠ 12 So we can work out ω= Case 2 Case 1 ω= MgRcm I g 2 L 3 m m g ω= 7 L 12 m In which case is the period longest? A)Case 1 B)Case 2 C)They are the same Physics 211 Lecture 22, Slide 17 The small angle approximation R xcm = c-length ar Rθ % difference between θ and sinθ θ d 2θ = −ω2 sin θ - Exact expression dt 2 Angle (degrees) 13 15 17 13 1 5 17 sin θ = θ − θ + θ − θ + ... = θ + θ− θ + ... 3! 5! 7! 6 120 5040 Physics 211 Lecture 22, Slide 18 Act A pendulum is made by hanging a thin hoola-hoop of pendulum diameter D on a small nail. on What is the angular frequency of oscillation of the hoop for small displacements? (ICM = mR2 for a hoop) mR for ω= g D (B) ω = 2g D (C) ω = pivot (nail) g 2D (A) D Physics 211 Lecture 22, Slide 19 The angular frequency of oscillation of the hoop for small mgRcm ω= displacements will be given by I Use parallel axis theorem: I = Icm + mR2 = mR2 + mR2 = 2mR2 ω= mgR = 2 2 mR g = 2R pivot (nail) g D R So ω= g D x cm m Physics 211 Lecture 22, Slide 20 ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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