A Closed Form Solution of the Run-Time of a Sliding Bead along a Freely Hanging Slinky

Similar documents
DEPARTMENT OF PHYSICS. University at Albany State University of New York. Comprehensive Field Examination. Classical. Monday, May 21, 2018

Assignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class

arxiv: v2 [physics.class-ph] 13 Nov 2011

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial

Lectures Chapter 10 (Cutnell & Johnson, Physics 7 th edition)

= o + t = ot + ½ t 2 = o + 2

CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017

2 We alsohave a second constant of the motion, the energy, H = 2 (M + m) _X 2 +mr cos # _X _ #+ 2 mr2 _ # 2 mgr cos # = constant : For small displacem

Another Method to get a Sine Wave. X = A cos θ V = Acc =

Question 1: A particle starts at rest and moves along a cycloid whose equation is. 2ay y a

Physics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1

Exam II Difficult Problems

Physics A - PHY 2048C

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:

Instructions: (62 points) Answer the following questions. SHOW ALL OF YOUR WORK. A B = A x B x + A y B y + A z B z = ( 1) + ( 1) ( 4) = 5

Classical Mechanics and Statistical/Thermodynamics. August 2018

Classical Mechanics Comprehensive Exam Solution

Chapter 14 Periodic Motion

Oscillatory Motion and Wave Motion

Periodic Motion. Periodic motion is motion of an object that. regularly repeats

COMPLETE ALL ROUGH WORKINGS IN THE ANSWER BOOK AND CROSS THROUGH ANY WORK WHICH IS NOT TO BE ASSESSED.

Unit 7: Oscillations

Chapter 8. Rotational Equilibrium and Rotational Dynamics

Oscillations. PHYS 101 Previous Exam Problems CHAPTER. Simple harmonic motion Mass-spring system Energy in SHM Pendulums

Chapter 14: Periodic motion

What is a Force? Free-Body diagrams. Contact vs. At-a-Distance 11/28/2016. Forces and Newton s Laws of Motion

Physics 53 Summer Exam I. Solutions

Physics 2210 Homework 18 Spring 2015

2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity

Lecture 9: Eigenvalues and Eigenvectors in Classical Mechanics (See Section 3.12 in Boas)

Constrained motion and generalized coordinates

16.07 Dynamics Final Exam

Lagrangian Dynamics: Generalized Coordinates and Forces

Name: Date: Period: AP Physics C Rotational Motion HO19

PHY 001 (Physics I) Lecture 7

Physics for Scientists and Engineers 4th Edition, 2017

Physics 101 Discussion Week 12 Explanation (2011)

Chapter 8 Lecture. Pearson Physics. Rotational Motion and Equilibrium. Prepared by Chris Chiaverina Pearson Education, Inc.

Solution Set Two. 1 Problem #1: Projectile Motion Cartesian Coordinates Polar Coordinates... 3

Lagrange s Equations of Motion and the Generalized Inertia

Oscillatory Motion SHM

Physics 351, Spring 2015, Homework #6. Due at start of class, Friday, February 27, 2015

Chapter 10. Rotation

Physics 351, Spring 2017, Homework #3. Due at start of class, Friday, February 3, 2017

Slide 1 / 70. Simple Harmonic Motion

Do not fill out the information below until instructed to do so! Name: Signature: Student ID: Section Number:

Oscillations. Oscillations and Simple Harmonic Motion

PHYSICS 221 SPRING EXAM 2: March 30, 2017; 8:15pm 10:15pm

PH1104/PH114S - MECHANICS

AP Physics 1. April 11, Simple Harmonic Motion. Table of Contents. Period. SHM and Circular Motion

PHY2053 General Physics I

28. Pendulum phase portrait Draw the phase portrait for the pendulum (supported by an inextensible rod)

Chapter 9-10 Test Review

Chapter 15 Periodic Motion

11. (7 points: Choose up to 3 answers) What is the tension,!, in the string? a.! = 0.10 N b.! = 0.21 N c.! = 0.29 N d.! = N e.! = 0.

Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!

Simple Harmonic Motion Concept Questions

Afternoon Section. Physics 1210 Exam 2 November 8, ! v = d! r dt. a avg. = v2. ) T 2! w = m g! f s. = v at v 2 1.

Course Review. Physics 2210 Fall Semester 2014

Torque and Simple Harmonic Motion

Physics 2211 ABC Quiz #3 Solutions Spring 2017

PH1104/PH114S MECHANICS

= y(x, t) =A cos (!t + kx)

Final Exam April 26, 2016

Two-Dimensional Rotational Kinematics

Phys 7221 Homework # 8

(a) Sections 3.7 through (b) Sections 8.1 through 8.3. and Sections 8.5 through 8.6. Problem Set 4 due Monday, 10/14/02

AP Physics. Harmonic Motion. Multiple Choice. Test E

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

PHY321 Homework Set 10

7 Pendulum. Part II: More complicated situations

Rotational motion problems

PHYSICS 149: Lecture 21

Chapter 4. Oscillatory Motion. 4.1 The Important Stuff Simple Harmonic Motion

Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)

ME 274 Spring 2017 Examination No. 2 PROBLEM No. 2 (20 pts.) Given:

a) Calculate the moment of inertia of the half disk about the z-axis. (The moment of inertia of a full disk

Physics Courseware Physics I Energy Conservation

Chapter 14 Oscillations. Copyright 2009 Pearson Education, Inc.

Force, Energy & Periodic Motion. Preparation for unit test

Solution to phys101-t112-final Exam

Preliminary Examination - Day 2 May 16, 2014

Physics Mechanics. Lecture 32 Oscillations II

Physics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top

PHYSICS. Chapter 12 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.

Final Test for PHYS 2130 section 091 Date: 2 nd May 2007

PROBLEM 2 10 points. [ ] increases [ ] decreases [ ] stays the same. Briefly justify your answer:

b) 2/3 MR 2 c) 3/4MR 2 d) 2/5MR 2

t = g = 10 m/s 2 = 2 s T = 2π g

Dynamics Review Checklist

Review for 3 rd Midterm

Chapter 14. Oscillations. Oscillations Introductory Terminology Simple Harmonic Motion:

31 ROTATIONAL KINEMATICS

Unforced Oscillations

= 2 5 MR2. I sphere = MR 2. I hoop = 1 2 MR2. I disk

AP Physics C : Mechanics Course Syllabus Instructor: Mr. Ronald J. Maniglia

Chapter 12. Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx

Physics A - PHY 2048C

UPDATED: NOVEMBER A particle of mass m and energy E moving in one dimension comes in from - and encounters the repulsive potential:

PHYS 1114, Lecture 33, April 10 Contents:

Transcription:

A Closed Form Solution of the Run-Time of a Sliding Bead along a Freely Hanging Slinky Haiduke Sarafian The Pennsylvania State University York, PA 17403, USA has2@psu.edu Abstract. The author has applied Lagrangian formalism to explore the kinematics of a bead sliding along a frictionless, freely hanging vertical Slinky. For instance, we derived a closed analytic equation for the runtime of the bead as a function of the traversed coil number. We have applied Mathematica to animate the 3-dimensional motion of the bead. The derived run-time is incorporated within the animation to clock the bead s actual motion. With the help of Mathematica we have solved the inverse run-time equation and have expressed the traversed coil number as a function of the run-time. The latter is applied to further the analysis of the problem conducive to analytic time-dependent equations for the bead s vertical position, its falling speed and its falling acceleration, and its angular velocity about the symmetry axis of the Slinky. It is also justified that a Slinky is a device capable of converting the gravitational potential energy of a sliding bead into pure rotational energy. 1 Introduction A Slinky is a massive, soft spring-like object and has curious dynamic and static features. By hanging masses to a freely, vertically suspended Slinky and setting it in motion, the authors of [1] have investigated some of its dynamic features. The Slinky s static characteristics have been studied in [2]. In both references, analytically, it is proven how the Slinky s own weight contributes to the uneven spacing of the adjacent coils along a vertically hung Slinky. The proven equations match the intuitive expectations the spacing between the adjacent coils for the coils closer to the support are wider than the ones at the bottom. Furthermore, the lower coils are less slanted verses the ones closer to the support. In fact, the bottom coils are almost horizontal. To incorporate these known characteristic features and to broaden the scope of the Slinky s related issues, we studied a kinematic problem. We considered the effects of the uneven spacing of the Slinky s coils to calculate the run-time of a sliding bead under the gravity pull. We have shown, although the calculation of the run-time of a sliding bead along a theoretical, massless evenly-spaced vertically hung spring is trivial, it is not so for a real Slinky. We were able to solve the Slinky problem exactly and derived an analytic closed form equation to express the run-time of a sliding bead as a function of the traversed coil number. M. Bubak et al. (Eds.): ICCS 2004, LNCS 3039, pp. 319 326, 2004. c Springer-Verlag Berlin Heidelberg 2004

320 H. Sarafian We have applied Mathematica [3] to animate the 3-dimensional motion of a bead. We considered a typical Slinky and have applied its geometrical lengths to clock the run-time of a sliding bead. The numeric values of the run-time were embedded in the 3-d animation so that one can visually correspond the actual movement of the bead to its run-time and the traversed coil number. Because of the length limitation of the article the Mathematica code is not included, the oral presentation will feature the animation. To further our analysis with the help of Mathematica we have solved the inverse run-time equation, we expressed the traversed coil number as a function of the run-time. The detail of the procedure and the specific use of Mathematica in achieving this goal is given in section 4. The latter is applied to further the analysis of the problem conducive to analytic time-dependent equations for the bead s vertical position, its falling speed and its falling acceleration, and its angular velocity about the symmetry axis of the Slinky. For comprehensive understanding, the derived equations are plotted v.s. time. 2 The Physics and the Analysis of the Problem We denote the number of Slinky s coils by N, the radius of the circular coils by R, its un-stretched axial length when laid on a level table by L 0 and its stretched length when freely suspended vertically by L. Figure 1 depicts one such Slinky. We have applied Mathematica s ParamtericPlot3D command to display the Slinky. The origin of a right-handed Cartesian coordinate system is set at the bottom of the Slinky with the z-axis pointing to the top of the page. The first coil, n =0,isatthez = 0 and the top coil, N = 35, is at the support. The height of the individual coil is measured from the bottom, the data is shown in Fig. 2. The size of the data points indicates the accuracy of the measurements along the vertical axis. According to [1,2] the height of the nth coil is given by; z n = an + bn 2 with a = L0 L L0 N and b = N. In Fig. 2, z 2 n, is shown by the solid line it perfectly fits the data. The position vector of a bead in the aforementioned Cartesian coordinate system is r = {x(t),y(t),z(t)} and can be written as r = {R cos(2πn ),Rsin(2πn ),z n } where n, is the number of the traversed coils and equals n = N n. The kinetic energy T = 1 2 mν2 and the potential energy V = mgz n of a freely released bead of mass m in terms of the traversed coil number n are: T = 1 2 m n 2 {c 2 +[a +2b(N n )] 2 } (1) V = mg[a(n n )+b(n n ) 2 ] (2) where n = d dt n, c =2πR and g is the gravity.

A Closed Form Solution of the Run-Time 321 Fig. 1. The display of a typical Slinky with specs of N = 35, R =5.0cm, L 0 =7.0cm and L = 172.0cm z,m 1.75 1.5 1.25 1 0.75 0.5 0.25 5 10 15 20 25 30 35 n Fig. 2. The dots are the data and the solid line is z n = an + bn 2 d The Euler-Lagrange equation, dt T and V subject to (1) and (2) is, L n = L n for Lagrangian L = T V with n {c 2 +[a +2b(N n )] 2 } 2b n 2 [a +2b(N n )] g[a +2b(N n )]=0. (3) To solve (3) we introduce ξ = a +2b(N n ). In terms of ξ, (3) becomes, ξ(c 2 + ξ 2 )+ ξ 2 ξ + ν 2 ξ = 0 (4)

322 H. Sarafian Here, ν 2 =2bg and assumes the speedsquared dimension. To solve (4) we set, η = ξ 2. In terms of η and ξ, (4) yields η 2 ξ (c2 + ξ 2 )+ξ(η + ν 2 )=0 (5) By separating the variables of (5) and integrating both sides of the resulting equation with appropriate limits we arrive at η dη ξ 0 η + ν 2 = 2ξdξ ξ 0 c 2 + ξ 2 (6) here, ξ 0 = a +2bN. The integration of (6) yields, η = ν 2 ξ 0 2 ξ 2 c 2 + ξ 2 (7) In (7) we replace η with ( dξ dt )2, and by rearranging the terms and integrating the result, we arrive at t = 1 ν ξ ξ 0 c 2 + ξ 2 ξ 2 dξ (8) 0 ξ2 this yields t = 1 c ν 2 + ξ 2 0 E(arccos( a +2bn ξ ), 0 ξ 0 c2 + ξ ) (9) 2 0 E(δ, r) is the Elliptic integral of the second kind [4]. Equation (9)is the run-time; i.e. it is the time a bead starting from the top coil takes to traverse to the nth coil. We have noticed that (8) is a convoluted analog of the kinematics of one dimensional uniform motion. I.e. (8) can be viewed as l = νt, with l being the Slinky s characteristic length given by ξ l = w(c, ξ 0,ξ)dξ (10) ξ 0 In other words, the bead slides along the characteristic length, l, given by the weighted differential length dξ. The weight, w, is defined by c w(c, ξ 0,ξ)= 2 + ξ 2 ξ 2 (11) 0 ξ 2 As mentioned earlier, the bead s characteristic speed ν is ν = 2gb. Intuitively, one would expect the run-time of a bead, for a skinny Slinky to be the same as the run-time of a freely falling bead released at the same height. To verify this, in (9) we set R = 0, this yields,

A Closed Form Solution of the Run-Time 323 t,s 3 2.5 2 1.5 1 0.5 5 10 15 20 25 30 35 n Fig. 3. The run-time of a bead v.s. the coil number n. The specs of the Slinky are the ones used in Fig. 1 t = 1 ν ξ 0E(arccos( a +2bn ), 1) (12) ξ 0 On the other hand, since. L 0 L, (12) simplifies further, 2L t g E(arccos( n ), 1) (13) N For n = 0, (13) yields the run-time of a bead traversing the entire length of the stretched Slinky, L. The E in (13) for n = 0 yields, E( π 2, 1) = 1 and yields the shortest run-time, 2L t (14) g Equation (14) is identical to the time of flight of a free falling object, L = 1 2 gt2, released at height L. Equation (14) for L = 172.0cm, the length of a hanging Slinky gives t =0.592s and matches the numeric value of (9) for small values of R, e.g. R =0.005cm. It is instructive to display the run-time t, given by (9) v.s. the coil number n. For the Slinky on hand, this is shown in Fig 3. 3 Corollary Topics of Interest It is curious to find out at any given time how high the bead is from the bottom of the Slinky, how fast the bead is falling and its falling acceleration. To address

324 H. Sarafian n 35 30 25 20 15 10 5 0.5 1 1.5 2 2.5 3 t,s Fig. 4. The plots of the coil number n v.s. t. The solid line is the fitted polynomial these questions, there is a need to solve (9) for t, this is problematic. However, we pursued the goal graphically. First, by interchanging the horizontal and vertical axes in Fig. 3, we re-plot the data points. We then apply Mathematica s Fit command to fit the data with a suitable polynomial. Figure 4 displays the output. For the Slinky on hand the fitted function is an incomplete third order polynomial with the given fitted coefficients n(t) =35.190 4.925t 2 +0.381t 3.By substituting n(t) inz n we evaluate the vertical height of the bead, z(t), its falling speed, ż(t) and its falling acceleration z(t). These quantities are plotted in Fig. 5. The ordinate of Fig. 5, is calibrated in MKS units, and hence, m, m s, m s are 2 to be used to read the height, the velocity and the acceleration. According to Fig. 5, the bead starts off with an initial zero velocity and of about 1 m s acceleration. In 1.6s, it reaches its maximum, 0.83 m 2 s velocity and acquires zero acceleration. In according to Fig. 3, t =1.6s, corresponds to n = 24, that is the bead reaches its maximum velocity after traversing 11 coils. From this point on, it takes another 1.4s to decelerate to the bottom. We also noted, a Slinky is a device capable of converting the gravitational potential energy to a pure rotational energy. This is because the bottom coils are horizontal and the bead upon reaching the bottom is to circulate about the Slinky s symmetry axis with no further fall. We quantify our claim by applying the conservation of energy principle to the two ends of the Slinky; the potential energy at the top and the rotational kinetic energy at the bottom. That is (PE) i =(KE) f rot, this gives mgl = 1 2 Iω2 max, here, I = mr 2 is the moment of inertia of the bead about the Slink s symmetry axis and ω max is its maximum angular velocity. Solving this equation for ω max yields ω max = R 1 2gL. For the specs of the Slinky on hand this yields

A Closed Form Solution of the Run-Time 325 1.5 1 0.5 0.5 1 1.5 2 2.5 3 t,s -0.5-1 Fig. 5. The plot of z(t)(solid-line),the falling velocity,ż(t)(short dashed-line) and the falling acceleration z(t) (long dashed-line) w, rad s 120 100 80 60 40 20 0.5 1 1.5 2 2.5 3 t,s Fig. 6. Plot of angular velocity, ω, v.s. t. The Slinky s specs are the ones used in Fig. 1 ω max = 116 rad s. On the other hand, the angular velocity of the bead is ω(t) = d dt {2π[N n(t)]}. The quantity in the braces is the traversed azimuthal angle about the symmetry axis of the Slinky and n(t) is the aforementioned fitted polynomial. Figure 6 displays ω(t) v.s. time. The maximum value of ω max (t) at the end of the run is 120 rad rad s, this is in good agreement with the predicted, 116 s.

326 H. Sarafian References 1. Thomas C. Heard, Neal D. Newby, Jr. : Behavior of a soft spring. Am. J. Phys. 45 (1977) 1102 1106 2. French, A. P. : The Suspended Slinky - A Problem in Static Equilibrium. The Physics Teacher 32(1994) 244 245 3. Wolfram, S. The Mathematica book. New 4th edn. Cambridge Press (1999). 4. Gradshteyn, I. S., Ryzhik, I. M. : Table of Integrals, Series and Products, 2nd edn. Academic Press, p.276 (1980)