Chapter 13, example problems: x (cm) 10.0


 William Carter
 10 months ago
 Views:
Transcription
1 Chapter 13, example problems: (13.04) Reading Fig (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm. (Don t just say 10!) (c) The period T is 16 s. (d) The angular frequency ω is: 2π/ T = 2π/ (16 s) = (π/8) s 1 = s 1. Better yet, the unit should be given as rad/s, since 2π is actually 2π rad. x (cm) t(s) (13.14) Object undergoing SHM. T = s, and A = m. At t = 0, object is at x = 0. At t = s, x =? In general, x = A cos (ωt + φ). But since at t = 0, object is at x = 0, we must take φ = ± π/2, and obtain x = ±A sin (ωt). (The sign can not be decided since we do not know whether the objected started by moving to the right or left.) Thus at t = s, we have x = ± m sin ((2π/1.200 s) (0.480 s)) = ± m. That is, the object is m away from the equilibrium position. (13.28) 5.20 kg box attached to an ideal horizontal spring with k = 375 N/m. Inside the box, a stone with m = 3.44 kg. Oscillating with A = 7.50 cm. When at maximum speed, the stone is plucked vertically out of the box without touching the box. (a) T new = 2π / ω max = 2π / k/m new = 2π / (375 N/m / 5.20 kg) 1/2 = s. (b) The plucking does not exert any horizontal force to the motion of the box. The empty box will still have the maximum velocity of the original motion, which is v max = ω A = k/m A = (375 N/m / 8.64 kg) 1/ cm = 49.4 cm/s. But its angular frequency has changed to k/m new = (375 N/m / 5.20 kg) 1/2 = /s. Hence its new amplitude is A new = v max /ω new = 49.4 cm/s / /s = cm. (c) The new period is shorter, since it is now a lighter mass moving under the influence of the same spring. (The same spring force produces a larger acceleration on the lighter mass at any given displacement. So the frequency should be higher, and the period shorter.) (13.44) A simple pendulum on Mars: On earth its period is 1.60 s. On Mars where g = 3.71 m/s 2, the new period is: T Mars = 2π / g Mars /l = ( g Earth / g Mars) (2π / g Earth /l ) = ( 9.80 / 3.71) 1.60 s = 2.60 s. That is, the period is longer on Mars, because lower gravitational acceleration on Mars makes the swinging motion slower. (13.52) 1.80 kg monkey wrench pivoted m from its center of mass. Allowed to swing as a physical pendulum. Period of smallangle oscillation = s. (a) Moment of inertia of the wrench:
2 I = mgl / ω 2 = 1.80 kg 9.80 m/s m /(2π / s) 2 = kg m 2, (where use has been made that ω = mgl/i for a physical pendulum at smallangle oscillation, because the gravitational torque is τ = mgl sinθ, so for smallangle oscillations for which sinθ can be approximated by θ, the effective torque constant K is mgl. Thus ω = K/I becomes mgl/i. ) (b) Initial displacement = θ (t = 0) = rad. Want dθ /d t at θ = 0, i.e., ( dθ /d t ) max. We use energy conservation: (1/2) K θ max 2 = (1/2) I ( dθ /d t ) max 2 [which is the angular analog of (1/2) k x max 2 = (1/2) m v 2 max. The left hand side is at both extremes of the motion, where the velocity is zero, whereas the right hand side is at the center of the motion, where the displacement is zero.] Hence ( dθ /d t ) max = θ max (K/ I) 1/2 = ωθ max = (2π / s) rad = rad/s. (13.68) Before the small block slips, the frequency of the oscillation is: f = (1/2π) k/(m + M). Looking at the small block alone, the only horizontal force acting on it is the static frictional force, the maximum value of k which is: μ s N = μ s mg, (because the upward normal force N acing on the small block by the large block below it should cancel the downward weight of the small block, mg.) For slip to not occur to the small block, this maximum frictional force must be equal to its mass m times its maximum acceleration, or Aω 2. That is: μ s mg = maω 2 = ma [k/(m + M)], giving A = μ s (m+m)g/k. μ s frictionless m M (13.70) Rocket accelerating upward at 4.00 m/s 2 from launchpad on earth. A small 1.50 kg ball hangs from the ceiling inside the rocket by a light 1.10m wire. The ball is displaced by 8.50 from vertical and released. (i) Inside the rocket, the effective g is 9.80 m/s m/s 2 = m/s 2. Why? Because when the ball is in its equilibrium position below the hanging point of the ceiling, and T is the tension in the wire, which pulls the ball upward, then the Newton s second law gives T mg = m 4.00 m/s 2, giving T = m m/s 2. Thus the period of the oscillation is: 2π L/g eff = 2π 1.10 m/13.80 m/s 2 = 1.77 s. Remarks: (a) Einstein s equivalence principle precisely refers to this situation: In a frame which is accelerating upward with an acceleration a, any mass m behaves as if it has received an additional downward force equal to its mass m times a. Adding it to the downward weight mg of the mass m if it is near the surface of the earth, we find that the total force acting on the mass is m (g + a), as if g has been changed to (g + a). (b) Both tension and period are denoted by T. To avoid confusing we have not used T to denote the period here, so T stands for tension only here. In another problem in this chapter we will use T to denote period.
3 (ii) The amplitude of the oscillation is still 1.10 m (8.50 π radians / 180 ) = m, independent of the acceleration of the rocket. (13.80) A 40.0 N force stretches a vertical spring m. (a) Find m suspended from it to get a period of 1.00 s. Force constant of the spring k = 40.0 N / m = 160 N/m. T = 2π (m / k) 1/2 = 1.00 s. m = k (T / 2π ) 2 = 160 N/m (1.00 s / 2π ) 2 = 4.05 kg. (c) A = m, T = 1.00 s. Find x(t) and direction of motion at t = 0.35 s after the mass m has passed the equilibrium position, moving downward. We take downward as positive for x, measured from the new equilibrium position (which is m below the lower end of the unstretched spring), then x(t) = A sin (ωt) = A sin (2π f t) = A sin (2π t / T ) = m sin (2π 0.35 s / 1.00 s) = m. [Note that the argument of the sine function is in radians as revealed by the 2π factor. The formula ω = 2π f = 2π / T gives the angular frequency ω in radians per second.] As for the direction of motion at this t, we notice that 0.35 s / 1.00 s = 0.35 < 0.5 but > Thus the argument of the sine function has exceeded π / 2 but is still less than π. Thus the sine function has reached its first maximum and is now coming down. It means that the mass m is now moving toward the origin (i.e., the equilibrium position). You can also answer this question by computing the velocity, but it will take more time. (d) Find the force F acting on the mass when it is m below the equilibrium position moving upward. We use F = k x = 160 N/m 0.03 m = 4.80 N. The negative sign shows that this force is pointing upward, not because the motional velocity is pointing upward (given in the problem), but rather because x is positive. (The restoring force in a simple harmonic oscillator is always trying to bring the mass back to the origin.) 2 [Note that you can also use F = m a and a = ω x to do this problem, and the same answer will be obtained. (13.90) Model the leg of a T. rex. as two uniform rods, each 1.55 m long, joined rigidly end to end. The lower rod has mass M and the upper rod has mass 2M. The composite opject is pivoted about the top of the upper rod. Compute the oscillation period of this object for small amplitude oscillation. Compare results with that of Example We need to first find the moment of inertia and centerofmass location of this object. The total moment of inertia about P : P I = [(1/12)(2M) L 2 + (2M) (L/2) 2 ] + θ [(1/12)(M) L 2 + (M) (3L/2) 2 ] = (2/3)ML 2 + (14/6)ML 2 = 3ML 2 Let the center of mass of this composite object be located at a distance d from the pivot point P along the rod. Then an upward force of 3Mg at this point can support the whole object. Thus the 2Mg Mg
4 counterclockwise torque by this force at this point about P can balance the two clockwise torques about P, one by the weight 2Mg at (L/2) from P, and one by weight Mg at (3L/2) from P. That is, (3Mg) d sin θ = (2Mg) (L/2) sin θ + (Mg) (3L/2) sin θ giving d = (5/6) L. [This answer is independent of θ, so one could let θ = 90.] We are now ready to compute the oscillation period of this composite object: T = 2π [I / (3M)gd ] 1/2 = 2π (L 2 / gd ) 1/2 = 2π [1.55 m / 9.80 m/s 2 (5/6)] 1/2 = 2.74 s. Example found a period of 2.90 s by assuming that the object is a uniform rod of the same length 2L = 3.10 m. Clearly, by shifting the mass distribution toward the pivot point, the period becomes shorter. This is clearly right, since if the whole mass is concentrated at the lower end (farthest from the pivot point), we would get T = 2π [2L / g ] 1/2 = 3.53 s! Note that in all three cases, the answer does not depend on the total mass, only its distribution. (13.96) 0.100m m m m m P 1 P 2 Both springs are stretched from their natural length m to m so their free ends can be attached to the points P 1 and P 2. (a) If k 1 = 2.00 N/m and k 2 = 6.00 N/m, find the new equilibrium position of the block of mass m. Let the new equilibrium position be a distance x to the right of the original position. Then the (left) spring 1 is stretched by m + x, and the (right) spring 2 is stretched by m x. The force pulling the block to the left by the spring 1 is therefore F 1 = k 1 (0.100 m + x), and the force pulling the block to the right by the spring 2 is therefore F 2 = k 2 (0.100 m x). These two force should cancel each other for the new equilibrium position. Hence we obtain the equation: 2.00 N/m (0.100 m + x) = 6.00 N/m (0.100 m x). Recombining terms, we obtain N N = (2.00 N/m N/m) x. Solving for x and we obtain x = 0.05 m. Thus the new equilibrium position of the block is 0.05 m to the right of its initial position. (b) Find the period of oscillation of the block if it is slightly displaced from this new equilibrium position and released. Let the block be displaced by a distance y to the right of this new equilibrium position. Then the spring 1 is stretch by a total distance (0.100 m + x + y), and the spring 2 is stretched by a total distance (0.100 m x y). The force pulling the block to the left by the spring 1 is therefore F 1 = k 1 (0.100 m + x + y), and the force pulling the block to the right by the spring 2 is therefore F 2 = k 2 (0.100 m x y). These two forces no longer cancel, and the net force to the
5 right (i.e., in the direction of the displacement y) is F 2 F 1 = (k 1 + k 2 ) y, showing that the effective force constant for this oscillatory motion is (k 1 + k 2 ). Thus the period of this oscillatory motion is: T = (2π) m / ( k 1 + k 2 ) = (2π) kg / 8.00 N/m = s.
Physics 41 HW Set 1 Chapter 15
Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,
More informationPhysics 211 Week 12. Simple Harmonic Motion: Equation of Motion
Physics 11 Week 1 Simple Harmonic Motion: Equation of Motion A mass M rests on a frictionless table and is connected to a spring of spring constant k. The other end of the spring is fixed to a vertical
More informationSimple Harmonic Motion
Simple Harmonic Motion Restating Hooke s law The equation of motion Phase, frequency, amplitude Simple Pendulum Damped and Forced oscillations Resonance Harmonic Motion A lot of motion in the real world
More informationChapter 24 Physical Pendulum
Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...
More informationHooke s Law. Spring. Simple Harmonic Motion. Energy. 12/9/09 Physics 201, UWMadison 1
Hooke s Law Spring Simple Harmonic Motion Energy 12/9/09 Physics 201, UWMadison 1 relaxed position F X = kx > 0 F X = 0 x apple 0 x=0 x > 0 x=0 F X =  kx < 0 x 12/9/09 Physics 201, UWMadison 2 We know
More informationPHYS2020: General Physics II Course Lecture Notes Section VII
PHYS2020: General Physics II Course Lecture Notes Section VII Dr. Donald G. Luttermoser East Tennessee State University Edition 4.0 Abstract These class notes are designed for use of the instructor and
More informationAP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
More informationAdvanced Higher Physics: MECHANICS. Simple Harmonic Motion
Advanced Higher Physics: MECHANICS Simple Harmonic Motion At the end of this section, you should be able to: Describe examples of simple harmonic motion (SHM). State that in SHM the unbalanced force is
More informationLecture Presentation Chapter 14 Oscillations
Lecture Presentation Chapter 14 Oscillations Suggested Videos for Chapter 14 Prelecture Videos Describing Simple Harmonic Motion Details of SHM Damping and Resonance Class Videos Oscillations Basic Oscillation
More informationPeriodic Motion or Oscillations. Physics 232 Lecture 01 1
Periodic Motion or Oscillations Physics 3 Lecture 01 1 Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium Stable Equilibrium Unstable Equilibrium A necessary requirement
More informationPhysics 1022: Chapter 14 Waves
Phys 10: Introduction, Pg 1 Physics 10: Chapter 14 Waves Anatomy of a wave Simple harmonic motion Energy and simple harmonic motion Phys 10: Introduction, Pg Page 1 1 Waves New Topic Phys 10: Introduction,
More informationSimple Harmonic Motion Concepts
Simple Harmonic Motion Concepts INTRODUCTION Have you ever wondered why a grandfather clock keeps accurate time? The motion of the pendulum is a particular kind of repetitive or periodic motion called
More information226 Chapter 15: OSCILLATIONS
Chapter 15: OSCILLATIONS 1. In simple harmonic motion, the restoring force must be proportional to the: A. amplitude B. frequency C. velocity D. displacement E. displacement squared 2. An oscillatory motion
More informationA B = AB sin(θ) = A B = AB (2) For two vectors A and B the cross product A B is a vector. The magnitude of the cross product
1 Dot Product and Cross Products For two vectors, the dot product is a number A B = AB cos(θ) = A B = AB (1) For two vectors A and B the cross product A B is a vector. The magnitude of the cross product
More informationturntable in terms of SHM and UCM: be plotted as a sine wave. n Think about spinning a ball on a string or a ball on a
RECALL: Angular Displacement & Angular Velocity Think about spinning a ball on a string or a ball on a turntable in terms of SHM and UCM: If you look at the ball from the side, its motion could be plotted
More informationSIMPLE HARMONIC MOTION
PERIODIC MOTION SIMPLE HARMONIC MOTION If a particle moves such that it repeats its path regularly after equal intervals of time, its motion is said to be periodic. The interval of time required to complete
More informationPractice Test SHM with Answers
Practice Test SHM with Answers MPC 1) If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? (There could be more than one
More informationLABORATORY 9. Simple Harmonic Motion
LABORATORY 9 Simple Harmonic Motion Purpose In this experiment we will investigate two examples of simple harmonic motion: the massspring system and the simple pendulum. For the massspring system we
More informationSolution Derivations for Capa #10
Solution Derivations for Capa #10 1) The flywheel of a steam engine runs with a constant angular speed of 172 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel
More informationSpring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations
Spring Simple Harmonic Oscillator Simple Harmonic Oscillations and Resonance We have an object attached to a spring. The object is on a horizontal frictionless surface. We move the object so the spring
More informationIMPORTANT NOTE ABOUT WEBASSIGN:
Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationChapter 07: Kinetic Energy and Work
Chapter 07: Kinetic Energy and Work Conservation of Energy is one of Nature s fundamental laws that is not violated. Energy can take on different forms in a given system. This chapter we will discuss work
More informationPhysics 231 Lecture 15
Physics 31 ecture 15 Main points of today s lecture: Simple harmonic motion Mass and Spring Pendulum Circular motion T 1/f; f 1/ T; ω πf for mass and spring ω x Acos( ωt) v ωasin( ωt) x ax ω Acos( ωt)
More informationSimple Harmonic Motion(SHM) Period and Frequency. Period and Frequency. Cosines and Sines
Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position position of the natural length of a spring Amplitude maximum displacement Period and Frequency Period (T) Time for one complete
More informationSimple Harmonic Motion
Simple Harmonic Motion 9M Object: Apparatus: To determine the force constant of a spring and then study the harmonic motion of that spring when it is loaded with a mass m. Force sensor, motion sensor,
More informationC B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a
More informationLab 5: Conservation of Energy
Lab 5: Conservation of Energy Equipment SWS, 1meter stick, 2meter stick, heavy duty bench clamp, 90cm rod, 40cm rod, 2 double clamps, brass spring, 100g mass, 500g mass with 5cm cardboard square
More informationPHY121 #8 Midterm I 3.06.2013
PHY11 #8 Midterm I 3.06.013 AP Physics Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension
More informationLecture 15. Torque. Center of Gravity. Rotational Equilibrium. Cutnell+Johnson:
Lecture 15 Torque Center of Gravity Rotational Equilibrium Cutnell+Johnson: 9.19.3 Last time we saw that describing circular motion and linear motion is very similar. For linear motion, we have position
More informationLab M1: The Simple Pendulum
Lab M1: The Simple Pendulum Introduction. The simple pendulum is a favorite introductory exercise because Galileo's experiments on pendulums in the early 1600s are usually regarded as the beginning of
More informationUNIT 14: HARMONIC MOTION
Name St.No.  Date(YY/MM/DD) / / Section UNIT 14: HARMONIC MOTION Approximate Time three 100minute sessions Back and Forth and Back and Forth... Cameo OBJECTIVES 1. To learn directly about some of the
More informationPhysics 1120: Simple Harmonic Motion Solutions
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Physics 1120: Simple Harmonic Motion Solutions 1. A 1.75 kg particle moves as function of time as follows: x = 4cos(1.33t+π/5) where distance is measured
More informationPhysics 271, Sections H1 & H2 Thursday, Nov 20, 2014
Physics 271, Sections H1 & H2 Thursday, Nov 20, 2014 Problems #11 Oscillations 1) Consider a mass spring system, with mass M and spring constant k. We put a mass m on top of the mass M. The coefficient
More informationMECHANICS IV  SIMPLE HARMONIC MOTION
MIVp.1 A. OSCILLATIONS B. SIMPLE PENDULUM C. KINEMATICS OF SIMPLE HARMONIC MOTION D. SPRINGANDMASS SYSTEM E. ENERGY OF SHM F. DAMPED HARMONIC MOTION G. FORCED VIBRATION A. OSCILLATIONS A toandfro
More informationphysics 111N rotational motion
physics 111N rotational motion rotations of a rigid body! suppose we have a body which rotates about some axis! we can define its orientation at any moment by an angle, θ (any point P will do) θ P physics
More informationLesson 04: Newton s laws of motion
www.scimsacademy.com Lesson 04: Newton s laws of motion If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics, before reading this lesson.
More informationSIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE
MISN026 SIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE SIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE by Kirby Morgan 1. Dynamics of Harmonic Motion a. Force Varies in Magnitude and Direction................
More informationVersion PREVIEW Practice 8 carroll (11108) 1
Version PREVIEW Practice 8 carroll 11108 1 This printout should have 12 questions. Multiplechoice questions may continue on the net column or page find all choices before answering. Inertia of Solids
More informationChapter 8 Rotational Motion
Chapter 8 Rotational Motion Textbook (Giancoli, 6 th edition): Assignment 9 Due on Thursday, November 26. 1. On page 131 of Giancoli, problem 18. 2. On page 220 of Giancoli, problem 24. 3. On page 221
More informationAP Physics  Chapter 8 Practice Test
AP Physics  Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on
More information1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. t + ") # x (t) = A! n. t + ") # v(0) = A!
1.1 Using Figure 1.6, verify that equation (1.1) satisfies the initial velocity condition. Solution: Following the lead given in Example 1.1., write down the general expression of the velocity by differentiating
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationMechanics 1. Revision Notes
Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....
More informationMechanics Cycle 1 Chapter 13. Chapter 13
Chapter 13 Torque * How Shall a Force be Applied to Start a Rotation? The Torque Formula Systems in Equilibrium: Statics Glimpse of Newton s rotational second law: Q: Review: What causes linear acceleration
More informationp = F net t (2) But, what is the net force acting on the object? Here s a little help in identifying the net force on an object:
Harmonic Oscillator Objective: Describe the position as a function of time of a harmonic oscillator. Apply the momentum principle to a harmonic oscillator. Sketch (and interpret) a graph of position as
More informationPSI AP Physics I Rotational Motion
PSI AP Physics I Rotational Motion MultipleChoice questions 1. Which of the following is the unit for angular displacement? A. meters B. seconds C. radians D. radians per second 2. An object moves from
More informationRecitation 2 Chapters 12 and 13
Recitation 2 Chapters 12 and 13 Problem 12.20. 65.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her and the bridge (Figure P12.20. The unstretched length of the cord is 11.0 m.
More informationSpinning Stuff Review
Spinning Stuff Review 1. A wheel (radius = 0.20 m) is mounted on a frictionless, horizontal axis. A light cord wrapped around the wheel supports a 0.50kg object, as shown in the figure below. When released
More information8 SIMPLE HARMONIC MOTION
8 SIMPLE HARMONIC MOTION Chapter 8 Simple Harmonic Motion Objectives After studying this chapter you should be able to model oscillations; be able to derive laws to describe oscillations; be able to use
More informationF mg (10.1 kg)(9.80 m/s ) m
Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationCollege Physics 140 Chapter 4: Force and Newton s Laws of Motion
College Physics 140 Chapter 4: Force and Newton s Laws of Motion We will be investigating what makes you move (forces) and how that accelerates objects. Chapter 4: Forces and Newton s Laws of Motion Forces
More informationWave Motion. Solutions of Home Work Problems
Chapter 16 Wave Motion. s of Home Work Problems 16.1 Problem 16.17 (In the text book) A transverse wave on a string is described by the wave function [( πx ) ] y = (0.10 m) sin + 4πt 8 (a) Determine the
More informationSimple Harmonic Motion
Simple Harmonic Motion Simple harmonic motion is one of the most common motions found in nature and can be observed from the microscopic vibration of atoms in a solid to rocking of a supertanker on the
More informationRotation, Angular Momentum
This test covers rotational motion, rotational kinematics, rotational energy, moments of inertia, torque, crossproducts, angular momentum and conservation of angular momentum, with some problems requiring
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationPhysics 2101, First Exam, Fall 2007
Physics 2101, First Exam, Fall 2007 September 4, 2007 Please turn OFF your cell phone and MP3 player! Write down your name and section number in the scantron form. Make sure to mark your answers in the
More informationChapter 14 Wave Motion
Query 17. If a stone be thrown into stagnating water, the waves excited thereby continue some time to arise in the place where the stone fell into the water, and are propagated from thence in concentric
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More informationPhysics 271 FINAL EXAMSOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath
Physics 271 FINAL EXAMSOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath 1. The exam will last from 8:00 am to 11:00 am. Use a # 2 pencil to make entries on the answer sheet. Enter the following id information
More informationRotational Dynamics. Luis Anchordoqui
Rotational Dynamics Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation ( O ). The radius of the circle is r. All points on a straight line
More informationChapter 1. Oscillations. Oscillations
Oscillations 1. A mass m hanging on a spring with a spring constant k has simple harmonic motion with a period T. If the mass is doubled to 2m, the period of oscillation A) increases by a factor of 2.
More informationRotational Mechanics  1
Rotational Mechanics  1 The Radian The radian is a unit of angular measure. The radian can be defined as the arc length s along a circle divided by the radius r. s r Comparing degrees and radians 360
More information5.2 Rotational Kinematics, Moment of Inertia
5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia Name: 5.2 Rotational Kinematics, Moment of Inertia 5.2.1 Rotational Kinematics In (translational) kinematics, we started out with the position
More informationPES 2130 Fall 2014, Spendier Lecture 10/Page 1
PES 2130 Fall 2014, Spendier Lecture 10/Page 1 Lecture today: Chapter 16 Waves1 1) Wave on String and Wave Equation 2) Speed of a transverse wave on a stretched string 3) Energy and power of a wave traveling
More informationNewton s Law of Motion
chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating
More informationTHE NOT SO SIMPLE PENDULUM
INTRODUCTION: THE NOT SO SIMPLE PENDULUM This laboratory experiment is used to study a wide range of topics in mechanics like velocity, acceleration, forces and their components, the gravitational force,
More informationLinear Centripetal Tangential speed acceleration acceleration A) Rω Rω 2 Rα B) Rω Rα Rω 2 C) Rω 2 Rα Rω D) Rω Rω 2 Rω E) Rω 2 Rα Rω 2 Ans: A
1. Two points, A and B, are on a disk that rotates about an axis. Point A is closer to the axis than point B. Which of the following is not true? A) Point B has the greater speed. B) Point A has the lesser
More informationDetermination of Acceleration due to Gravity
Experiment 2 24 Kuwait University Physics 105 Physics Department Determination of Acceleration due to Gravity Introduction In this experiment the acceleration due to gravity (g) is determined using two
More informationboth double. A. T and v max B. T remains the same and v max doubles. both remain the same. C. T and v max
Q13.1 An object on the end of a spring is oscillating in simple harmonic motion. If the amplitude of oscillation is doubled, how does this affect the oscillation period T and the object s maximum speed
More informationChapter 18 Static Equilibrium
Chapter 8 Static Equilibrium 8. Introduction Static Equilibrium... 8. Lever Law... Example 8. Lever Law... 4 8.3 Generalized Lever Law... 5 8.4 Worked Examples... 7 Example 8. Suspended Rod... 7 Example
More informationSolutions to Homework Set #10 Phys2414 Fall 2005
Solution Set #0 Solutions to Homework Set #0 Phys244 Fall 2005 Note: The numbers in the boxes correspond to those that are generated by WebAssign. The numbers on your individual assignment will vary. Any
More informationSample Questions for the AP Physics 1 Exam
Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiplechoice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each
More informationPHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
More informationPhysics 53. Oscillations. You've got to be very careful if you don't know where you're going, because you might not get there.
Physics 53 Oscillations You've got to be very careful if you don't know where you're going, because you might not get there. Yogi Berra Overview Many natural phenomena exhibit motion in which particles
More informationSimple Harmonic Motion. Simple Harmonic Motion
Siple Haronic Motion Basic oscillations are siple haronic otion (SHM) where the position as a function of tie is given by a sign or cosine Eaples of actual otion: asses on springs vibration of atos in
More informationHomework #7 Solutions
MAT 0 Spring 201 Problems Homework #7 Solutions Section.: 4, 18, 22, 24, 4, 40 Section.4: 4, abc, 16, 18, 22. Omit the graphing part on problems 16 and 18...4. Find the general solution to the differential
More informationForces. Isaac Newton was the first to discover that the laws that govern motions on the Earth also applied to celestial bodies.
Forces Now we will discuss the part of mechanics known as dynamics. We will introduce Newton s three laws of motion which are at the heart of classical mechanics. We must note that Newton s laws describe
More informationPHYS101 The Laws of Motion Spring 2014
The Laws of Motion 1. An object of mass m 1 = 55.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m 2
More informationNewton s Laws of Motion
Section 3.2 Newton s Laws of Motion Objectives Analyze relationships between forces and motion Calculate the effects of forces on objects Identify force pairs between objects New Vocabulary Newton s first
More informationPhysics Exam 2 Formulas
INSTRUCTIONS: Write your NAME on the front of the blue exam booklet. The exam is closed book, and you may have only pens/pencils and a calculator (no stored equations or programs and no graphing). Show
More informationSimple Harmonic Motion
Simple Harmonic Motion Objective: In this exercise you will investigate the simple harmonic motion of mass suspended from a helical (coiled) spring. Apparatus: Spring 1 Table Post 1 Short Rod 1 Rightangled
More informationHW 7 Q 14,20,20,23 P 3,4,8,6,8. Chapter 7. Rotational Motion of the Object. Dr. Armen Kocharian
HW 7 Q 14,20,20,23 P 3,4,8,6,8 Chapter 7 Rotational Motion of the Object Dr. Armen Kocharian Axis of Rotation The radian is a unit of angular measure The radian can be defined as the arc length s along
More informationChapter 4 Newton s Laws: Explaining Motion
Chapter 4 Newton s s Laws: Explaining Motion Newton s Laws of Motion The concepts of force, mass, and weight play critical roles. A Brief History! Where do our ideas and theories about motion come from?!
More informationThis week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap.
This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension PHYS 2: Chap. 19, Pg 2 1 New Topic Phys 1021 Ch 7, p 3 A 2.0 kg wood box slides down a vertical
More informationPeople s Physics book 3e Ch 251
The Big Idea: In most realistic situations forces and accelerations are not fixed quantities but vary with time or displacement. In these situations algebraic formulas cannot do better than approximate
More informationPHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
More information9 ROTATIONAL DYNAMICS
CHAPTER 9 ROTATIONAL DYNAMICS CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by τ = Fl, where l is the lever arm. When a long pipe is slipped
More informationWaves I: Generalities, Superposition & Standing Waves
Chapter 5 Waves I: Generalities, Superposition & Standing Waves 5.1 The Important Stuff 5.1.1 Wave Motion Wave motion occurs when the mass elements of a medium such as a taut string or the surface of a
More informationHOOKE S LAW AND SIMPLE HARMONIC MOTION
HOOKE S LAW AND SIMPLE HARMONIC MOTION Alexander Sapozhnikov, Brooklyn College CUNY, New York, alexs@brooklyn.cuny.edu Objectives Study Hooke s Law and measure the spring constant. Study Simple Harmonic
More informationHOOKE'S LAW AND SIMPLE HARMONIC MOTION OBJECT
5 M19 M19.1 HOOKE'S LAW AND SIMPLE HARMONIC MOTION OBJECT The object of this experiment is to determine whether a vertical massspring system obeys Hooke's Law and to study simple harmonic motion. THEORY
More informationHW#4b Page 1 of 6. I ll use m = 100 kg, for parts bc: accelerates upwards, downwards at 5 m/s 2 A) Scale reading is the same as person s weight (mg).
HW#4b Page 1 of 6 Problem 1. A 100 kg person stands on a scale. a.) What would be the scale readout? b.) If the person stands on the scale in an elevator accelerating upwards at 5 m/s, what is the scale
More informationANGULAR POSITION. 4. Rotational Kinematics and Dynamics
ANGULAR POSITION To describe rotational motion, we define angular quantities that are analogous to linear quantities Consider a bicycle wheel that is free to rotate about its axle The axle is the axis
More informationTest  A2 Physics. Primary focus Magnetic Fields  Secondary focus electric fields (including circular motion and SHM elements)
Test  A2 Physics Primary focus Magnetic Fields  Secondary focus electric fields (including circular motion and SHM elements) Time allocation 40 minutes These questions were ALL taken from the June 2010
More informationRotational inertia (moment of inertia)
Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = Σ m i r i 2 or r i : the perpendicular distance between m i and the given rotation axis m 1 m 2 x 1 x 2 Moment
More informationNewton s Laws of Motion
Physics Newton s Laws of Motion Newton s Laws of Motion 4.1 Objectives Explain Newton s first law of motion. Explain Newton s second law of motion. Explain Newton s third law of motion. Solve problems
More informationMath Review: Circular Motion 8.01
Math Review: Circular Motion 8.01 Position and Displacement r ( t) : position vector of an object moving in a circular orbit of radius R Δr ( t) : change in position between time t and time t+δt Position
More informationPHY231 Section 1, Form B March 22, 2012
1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate
More informationName: Lab Partner: Section:
Chapter 10 Simple Harmonic Motion Name: Lab Partner: Section: 10.1 Purpose Simple harmonic motion will be examined in this experiment. 10.2 Introduction A periodic motion is one that repeats itself in
More informationf max s = µ s N (5.1)
Chapter 5 Forces and Motion II 5.1 The Important Stuff 5.1.1 Friction Forces Forces which are known collectively as friction forces are all around us in daily life. In elementary physics we discuss the
More informationPHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013
PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be
More information