SlideShare a Scribd company logo
1 of 3
Download to read offline
Solution
Week 85 (4/26/04)
Tower of cylinders
Both cylinders in a given row move in the same manner, so we may simply treat
them as one cylinder with mass m = 2M. Let the forces that the boards exert on
the cylinders be labelled as shown. “F” is the force from the plank below a given
cylinder, and “G” is the force from the plank above it.
F
G
F
G
n
an
αn
n
n+1
n+1
Note that by Newton’s third law, we have Fn+1 = Gn, because the planks are
massless.
Our strategy will be to solve for the linear and angular accelerations of each
cylinder in terms of the accelerations of the cylinder below it. Since we want to solve
for two quantities, we will need to produce two equations relating the accelerations
of two successive cylinders. One equation will come from a combination of F = ma,
τ = Iα, and Newton’s third law. The other will come from the nonslipping condition.
With the positive directions for a and α defined as in the figure, F = ma on the
nth cylinder gives
Fn − Gn = man, (1)
and τ = Iα on the nth cylinder gives
(Fn + Gn)R =
1
2
mR2
αn =⇒ Fn + Gn =
1
2
mRαn (2)
Solving the previous two equations for Fn and Gn gives
Fn =
1
2
man +
1
2
mRαn ,
Gn =
1
2
−man +
1
2
mRαn . (3)
But we know that Fn+1 = Gn. Therefore,
an+1 +
1
2
Rαn+1 = −an +
1
2
Rαn. (4)
We will now use the fact that the cylinders don’t slip with respect to the boards.
The acceleration of the board above the nth cylinder is an − Rαn. But the accel-
eration of this same board, viewed as the board below the (n + 1)st cylinder, is
an+1 + Rαn+1. Therefore,
an+1 + Rαn+1 = an − Rαn. (5)
1
Eqs. (4) and (5) are a system of two equations in the two unknowns, an+1 and αn+1,
in terms of an and αn. Solving for an+1 and αn+1 gives
an+1 = −3an + 2Rαn,
Rαn+1 = 4an − 3Rαn. (6)
We can write this in matrix form as
an+1
Rαn+1
=
−3 2
4 −3
an
Rαn
. (7)
We therefore have
an
Rαn
=
−3 2
4 −3
n−1
a1
Rα1
. (8)
Consider now the eigenvectors and eigenvalues of the above matrix. The eigenvectors
are found via
−3 − λ 2
4 −3 − λ
= 0 =⇒ λ± = −3 ± 2
√
2. (9)
The eigenvectors are then
V+ =
1√
2
, for λ+ = −3 + 2
√
2,
V− =
1
−
√
2
, for λ− = −3 − 2
√
2. (10)
Note that |λ−| > 1, so λn
− → ∞ as n → ∞. This means that if the initial (a1, Rα1)
vector has any component in the V− direction, then the (an, Rαn) vectors will head
to infinity. This violates conservation of energy. Therefore, the (a1, Rα1) vector
must be proportional to V+.1 That is, Rα1 =
√
2a1. Combining this with the fact
that the given acceleration, a, of the bottom board equals a1 + Rα1, we obtain
a = a1 +
√
2a1 =⇒ a1 =
a
√
2 + 1
= (
√
2 − 1)a. (11)
Remark: Let us consider the general case where the cylinders have a moment of inertia
of the form I = βMR2
. Using the above arguments, you can show that eq. (7) becomes
an+1
Rαn+1
=
1
1 − β
−(1 + β) 2β
2 −(1 + β)
an
Rαn
. (12)
And you can show that the eigenvectors and eigenvalues are
V+ =
√
β
1
, for λ+ =
√
β − 1
√
β + 1
,
V− =
√
β
−1
, for λ− =
√
β + 1
√
β − 1
. (13)
1
This then means that the (an, Rαn) vectors head to zero as n → ∞, because |λ+| < 1. Also,
note that the accelerations change sign from one level to the next, because λ+ is negative.
2
As above, we cannot have the exponentially growing solution, so we must have only the V+
solution. We therefore have Rα1 = a1/
√
β. Combining this with the fact that the given
acceleration, a, of the bottom board equals a1 + Rα1, we obtain
a = a1 +
a1
√
β
=⇒ a1 =
√
β
1 +
√
β
a. (14)
You can verify that all of these results agree with the β = 1/2 results obtained above.
Let’s now consider a few special cases of the
λ+ =
√
β − 1
√
β + 1
(15)
eigenvalue, which gives the ratio of the accelerations in any level to the ones in the next
level down.
• If β = 0 (all the mass of a cylinder is located at the center), then we have λ+ = −1.
In other words, the accelerations have the same magnitudes but different signs from
one level to the next. The cylinders simply spin in place while their centers remain
fixed. The centers are indeed fixed, because a1 = 0, from eq. (14).
• If β = 1 (all the mass of a cylinder is located on the rim), then we have λ+ = 0. In
other words, there is no motion above the first level. The lowest cylinder basically
rolls on the bottom side of the (stationary) plank right above it. Its acceleration is
a1 = a/2, from eq. (14).
• If β → ∞ (the cylinders have long massive extensions that extend far out beyond the
rim), then we have λ+ = 1. In other words, all the levels have equal accelerations. This
fact, combined with the Rα1 = a1/
√
β ≈ 0 result, shows that there is no rotational
motion at any level, and the whole system simply moves to the right as a rigid object
with acceleration a1 = a, from eq. (14).
3

More Related Content

What's hot

The Laplace Transform of Modeling of a Spring-Mass-Damper System
The Laplace Transform of Modeling of a Spring-Mass-Damper System The Laplace Transform of Modeling of a Spring-Mass-Damper System
The Laplace Transform of Modeling of a Spring-Mass-Damper System Mahmoud Farg
 
Midterm assign 2
Midterm assign 2Midterm assign 2
Midterm assign 2meezanchand
 
Laplace transformation
Laplace transformationLaplace transformation
Laplace transformationWasim Shah
 
Damped force vibrating Model Laplace Transforms
Damped force vibrating Model Laplace Transforms Damped force vibrating Model Laplace Transforms
Damped force vibrating Model Laplace Transforms Student
 
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...Waqas Afzal
 
Applications laplace transform
Applications laplace transformApplications laplace transform
Applications laplace transformMuhammad Fadli
 
Chapter 2 laplace transform
Chapter 2 laplace transformChapter 2 laplace transform
Chapter 2 laplace transformLenchoDuguma
 
Fundamentals of electromagnetics
Fundamentals of electromagneticsFundamentals of electromagnetics
Fundamentals of electromagneticsRichu Jose Cyriac
 
Laplace Transform and its applications
Laplace Transform and its applicationsLaplace Transform and its applications
Laplace Transform and its applicationsDeepRaval7
 
Laplace Transformation & Its Application
Laplace Transformation & Its ApplicationLaplace Transformation & Its Application
Laplace Transformation & Its ApplicationChandra Kundu
 
Laplace transform and its application
Laplace transform and its applicationLaplace transform and its application
Laplace transform and its applicationmayur1347
 

What's hot (19)

The Laplace Transform of Modeling of a Spring-Mass-Damper System
The Laplace Transform of Modeling of a Spring-Mass-Damper System The Laplace Transform of Modeling of a Spring-Mass-Damper System
The Laplace Transform of Modeling of a Spring-Mass-Damper System
 
Inverse Laplace Transform
Inverse Laplace TransformInverse Laplace Transform
Inverse Laplace Transform
 
Inverse laplace transforms
Inverse laplace transformsInverse laplace transforms
Inverse laplace transforms
 
Midterm assign 2
Midterm assign 2Midterm assign 2
Midterm assign 2
 
Laplace transformation
Laplace transformationLaplace transformation
Laplace transformation
 
Damped force vibrating Model Laplace Transforms
Damped force vibrating Model Laplace Transforms Damped force vibrating Model Laplace Transforms
Damped force vibrating Model Laplace Transforms
 
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...
laplace transform and inverse laplace, properties, Inverse Laplace Calculatio...
 
Laplace transformation
Laplace transformationLaplace transformation
Laplace transformation
 
Laplace transformation
Laplace transformationLaplace transformation
Laplace transformation
 
Applications laplace transform
Applications laplace transformApplications laplace transform
Applications laplace transform
 
Chapter 2 laplace transform
Chapter 2 laplace transformChapter 2 laplace transform
Chapter 2 laplace transform
 
Laplace transforms
Laplace transformsLaplace transforms
Laplace transforms
 
Chapter 16 1
Chapter 16 1Chapter 16 1
Chapter 16 1
 
Fundamentals of electromagnetics
Fundamentals of electromagneticsFundamentals of electromagnetics
Fundamentals of electromagnetics
 
Convex hull
Convex hullConvex hull
Convex hull
 
Laplace Transform and its applications
Laplace Transform and its applicationsLaplace Transform and its applications
Laplace Transform and its applications
 
convex hull
convex hullconvex hull
convex hull
 
Laplace Transformation & Its Application
Laplace Transformation & Its ApplicationLaplace Transformation & Its Application
Laplace Transformation & Its Application
 
Laplace transform and its application
Laplace transform and its applicationLaplace transform and its application
Laplace transform and its application
 

Viewers also liked (20)

Sol56
Sol56Sol56
Sol56
 
Bab ii sis hukum & perad nas
Bab ii sis hukum & perad nasBab ii sis hukum & perad nas
Bab ii sis hukum & perad nas
 
iPharma Connect - strategy design for mobile
iPharma Connect  - strategy design for mobileiPharma Connect  - strategy design for mobile
iPharma Connect - strategy design for mobile
 
Başak Ateş
Başak AteşBaşak Ateş
Başak Ateş
 
Candan Erçetin
Candan ErçetinCandan Erçetin
Candan Erçetin
 
Building an event/conference website like FUDCon.in
Building an event/conference website like FUDCon.inBuilding an event/conference website like FUDCon.in
Building an event/conference website like FUDCon.in
 
Sol59
Sol59Sol59
Sol59
 
Sol78
Sol78Sol78
Sol78
 
Sol86
Sol86Sol86
Sol86
 
99k House Competition (bpm Entry 08)
99k House Competition (bpm Entry 08)99k House Competition (bpm Entry 08)
99k House Competition (bpm Entry 08)
 
Sol65
Sol65Sol65
Sol65
 
Sol68
Sol68Sol68
Sol68
 
Sol79
Sol79Sol79
Sol79
 
Sol55
Sol55Sol55
Sol55
 
Sol82
Sol82Sol82
Sol82
 
Sol73
Sol73Sol73
Sol73
 
Sol63
Sol63Sol63
Sol63
 
Sol87
Sol87Sol87
Sol87
 
Sol69
Sol69Sol69
Sol69
 
Información
InformaciónInformación
Información
 

Similar to Sol85 (20)

Solution baupc 2004
Solution baupc 2004Solution baupc 2004
Solution baupc 2004
 
Solution baupc 2003
Solution baupc 2003Solution baupc 2003
Solution baupc 2003
 
Problem baupc 2003
Problem baupc 2003Problem baupc 2003
Problem baupc 2003
 
Sol33
Sol33Sol33
Sol33
 
Math IA
Math IAMath IA
Math IA
 
Dynamics slideshare
Dynamics slideshareDynamics slideshare
Dynamics slideshare
 
Dynamics problems
Dynamics problemsDynamics problems
Dynamics problems
 
Sol31
Sol31Sol31
Sol31
 
Sol31
Sol31Sol31
Sol31
 
L^2_vs_Lz_Lx_Ly_Consistency_check.pdf
L^2_vs_Lz_Lx_Ly_Consistency_check.pdfL^2_vs_Lz_Lx_Ly_Consistency_check.pdf
L^2_vs_Lz_Lx_Ly_Consistency_check.pdf
 
Sol68
Sol68Sol68
Sol68
 
Linear Algebra Gauss Jordan elimination.pptx
Linear Algebra Gauss Jordan elimination.pptxLinear Algebra Gauss Jordan elimination.pptx
Linear Algebra Gauss Jordan elimination.pptx
 
Permutations 2020
Permutations 2020Permutations 2020
Permutations 2020
 
Binomial Theorem, Recursion ,Tower of Honai, relations
Binomial Theorem, Recursion ,Tower of Honai, relationsBinomial Theorem, Recursion ,Tower of Honai, relations
Binomial Theorem, Recursion ,Tower of Honai, relations
 
AYUSH.pptx
AYUSH.pptxAYUSH.pptx
AYUSH.pptx
 
41050EPathshala_071720200709PM.pdf
41050EPathshala_071720200709PM.pdf41050EPathshala_071720200709PM.pdf
41050EPathshala_071720200709PM.pdf
 
Superposition of Harmonic oscillator-2.docx
Superposition of Harmonic oscillator-2.docxSuperposition of Harmonic oscillator-2.docx
Superposition of Harmonic oscillator-2.docx
 
Sol87
Sol87Sol87
Sol87
 
coordinate Geometry straight line
coordinate Geometry   straight linecoordinate Geometry   straight line
coordinate Geometry straight line
 
Lecture Notes: EEEC4340318 Instrumentation and Control Systems - System Models
Lecture Notes:  EEEC4340318 Instrumentation and Control Systems - System ModelsLecture Notes:  EEEC4340318 Instrumentation and Control Systems - System Models
Lecture Notes: EEEC4340318 Instrumentation and Control Systems - System Models
 

More from eli priyatna laidan

Up ppg daljab latihan soal-pgsd-set-2
Up ppg daljab latihan soal-pgsd-set-2Up ppg daljab latihan soal-pgsd-set-2
Up ppg daljab latihan soal-pgsd-set-2eli priyatna laidan
 
Soal up sosial kepribadian pendidik 5
Soal up sosial kepribadian pendidik 5Soal up sosial kepribadian pendidik 5
Soal up sosial kepribadian pendidik 5eli priyatna laidan
 
Soal up sosial kepribadian pendidik 4
Soal up sosial kepribadian pendidik 4Soal up sosial kepribadian pendidik 4
Soal up sosial kepribadian pendidik 4eli priyatna laidan
 
Soal up sosial kepribadian pendidik 3
Soal up sosial kepribadian pendidik 3Soal up sosial kepribadian pendidik 3
Soal up sosial kepribadian pendidik 3eli priyatna laidan
 
Soal up sosial kepribadian pendidik 2
Soal up sosial kepribadian pendidik 2Soal up sosial kepribadian pendidik 2
Soal up sosial kepribadian pendidik 2eli priyatna laidan
 
Soal up sosial kepribadian pendidik 1
Soal up sosial kepribadian pendidik 1Soal up sosial kepribadian pendidik 1
Soal up sosial kepribadian pendidik 1eli priyatna laidan
 
Soal sospri ukm ulang i 2017 1 (1)
Soal sospri ukm ulang i 2017 1 (1)Soal sospri ukm ulang i 2017 1 (1)
Soal sospri ukm ulang i 2017 1 (1)eli priyatna laidan
 
Soal perkembangan kognitif peserta didik
Soal perkembangan kognitif peserta didikSoal perkembangan kognitif peserta didik
Soal perkembangan kognitif peserta didikeli priyatna laidan
 
Soal latihan utn pedagogik plpg 2017
Soal latihan utn pedagogik plpg 2017Soal latihan utn pedagogik plpg 2017
Soal latihan utn pedagogik plpg 2017eli priyatna laidan
 
Bank soal pedagogik terbaru 175 soal-v2
Bank soal pedagogik terbaru 175 soal-v2Bank soal pedagogik terbaru 175 soal-v2
Bank soal pedagogik terbaru 175 soal-v2eli priyatna laidan
 

More from eli priyatna laidan (20)

Up ppg daljab latihan soal-pgsd-set-2
Up ppg daljab latihan soal-pgsd-set-2Up ppg daljab latihan soal-pgsd-set-2
Up ppg daljab latihan soal-pgsd-set-2
 
Soal utn plus kunci gurusd.net
Soal utn plus kunci gurusd.netSoal utn plus kunci gurusd.net
Soal utn plus kunci gurusd.net
 
Soal up sosial kepribadian pendidik 5
Soal up sosial kepribadian pendidik 5Soal up sosial kepribadian pendidik 5
Soal up sosial kepribadian pendidik 5
 
Soal up sosial kepribadian pendidik 4
Soal up sosial kepribadian pendidik 4Soal up sosial kepribadian pendidik 4
Soal up sosial kepribadian pendidik 4
 
Soal up sosial kepribadian pendidik 3
Soal up sosial kepribadian pendidik 3Soal up sosial kepribadian pendidik 3
Soal up sosial kepribadian pendidik 3
 
Soal up sosial kepribadian pendidik 2
Soal up sosial kepribadian pendidik 2Soal up sosial kepribadian pendidik 2
Soal up sosial kepribadian pendidik 2
 
Soal up sosial kepribadian pendidik 1
Soal up sosial kepribadian pendidik 1Soal up sosial kepribadian pendidik 1
Soal up sosial kepribadian pendidik 1
 
Soal up akmal
Soal up akmalSoal up akmal
Soal up akmal
 
Soal tkp serta kunci jawabannya
Soal tkp serta kunci jawabannyaSoal tkp serta kunci jawabannya
Soal tkp serta kunci jawabannya
 
Soal tes wawasan kebangsaan
Soal tes wawasan kebangsaanSoal tes wawasan kebangsaan
Soal tes wawasan kebangsaan
 
Soal sospri ukm ulang i 2017 1 (1)
Soal sospri ukm ulang i 2017 1 (1)Soal sospri ukm ulang i 2017 1 (1)
Soal sospri ukm ulang i 2017 1 (1)
 
Soal perkembangan kognitif peserta didik
Soal perkembangan kognitif peserta didikSoal perkembangan kognitif peserta didik
Soal perkembangan kognitif peserta didik
 
Soal latihan utn pedagogik plpg 2017
Soal latihan utn pedagogik plpg 2017Soal latihan utn pedagogik plpg 2017
Soal latihan utn pedagogik plpg 2017
 
Rekap soal kompetensi pedagogi
Rekap soal kompetensi pedagogiRekap soal kompetensi pedagogi
Rekap soal kompetensi pedagogi
 
Bank soal pedagogik terbaru 175 soal-v2
Bank soal pedagogik terbaru 175 soal-v2Bank soal pedagogik terbaru 175 soal-v2
Bank soal pedagogik terbaru 175 soal-v2
 
Bank soal ppg
Bank soal ppgBank soal ppg
Bank soal ppg
 
Soal cpns-paket-17
Soal cpns-paket-17Soal cpns-paket-17
Soal cpns-paket-17
 
Soal cpns-paket-14
Soal cpns-paket-14Soal cpns-paket-14
Soal cpns-paket-14
 
Soal cpns-paket-13
Soal cpns-paket-13Soal cpns-paket-13
Soal cpns-paket-13
 
Soal cpns-paket-12
Soal cpns-paket-12Soal cpns-paket-12
Soal cpns-paket-12
 

Recently uploaded

AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum Revolution
AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum RevolutionAI Health Agents: Longevity as a Service in the Web3 GenAI Quantum Revolution
AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum RevolutionMelanie Swan
 
Building AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxBuilding AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxUdaiappa Ramachandran
 
Cybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxCybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxGDSC PJATK
 
Reference Domain Ontologies and Large Medical Language Models.pptx
Reference Domain Ontologies and Large Medical Language Models.pptxReference Domain Ontologies and Large Medical Language Models.pptx
Reference Domain Ontologies and Large Medical Language Models.pptxChimezie Ogbuji
 
Linked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesLinked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesDavid Newbury
 
Digital magic. A small project for controlling smart light bulbs.
Digital magic. A small project for controlling smart light bulbs.Digital magic. A small project for controlling smart light bulbs.
Digital magic. A small project for controlling smart light bulbs.francesco barbera
 
RTL Design Methodologies_Object Automation Inc
RTL Design Methodologies_Object Automation IncRTL Design Methodologies_Object Automation Inc
RTL Design Methodologies_Object Automation IncObject Automation
 
DS Lesson 2 - Subsets, Supersets and Power Set.pdf
DS Lesson 2 - Subsets, Supersets and Power Set.pdfDS Lesson 2 - Subsets, Supersets and Power Set.pdf
DS Lesson 2 - Subsets, Supersets and Power Set.pdfROWELL MARQUINA
 
Monitoring Java Application Security with JDK Tools and JFR Events.pdf
Monitoring Java Application Security with JDK Tools and JFR Events.pdfMonitoring Java Application Security with JDK Tools and JFR Events.pdf
Monitoring Java Application Security with JDK Tools and JFR Events.pdfAna-Maria Mihalceanu
 
Future Research Directions for Augmented Reality
Future Research Directions for Augmented RealityFuture Research Directions for Augmented Reality
Future Research Directions for Augmented RealityMark Billinghurst
 
CHIPS Alliance_Object Automation Inc_workshop
CHIPS Alliance_Object Automation Inc_workshopCHIPS Alliance_Object Automation Inc_workshop
CHIPS Alliance_Object Automation Inc_workshopObject Automation
 
20200723_insight_release_plan
20200723_insight_release_plan20200723_insight_release_plan
20200723_insight_release_planJamie (Taka) Wang
 
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCost
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCostKubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCost
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCostMatt Ray
 
Things you didn't know you can use in your Salesforce
Things you didn't know you can use in your SalesforceThings you didn't know you can use in your Salesforce
Things you didn't know you can use in your SalesforceMartin Humpolec
 
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.Unleashing the power of AI in UiPath Studio with UiPath Autopilot.
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.DianaGray10
 
Dragino Technology LoRaWANデバイス、ゲートウェイ ユースケース
Dragino Technology   LoRaWANデバイス、ゲートウェイ ユースケースDragino Technology   LoRaWANデバイス、ゲートウェイ ユースケース
Dragino Technology LoRaWANデバイス、ゲートウェイ ユースケースCRI Japan, Inc.
 
COMPUTER 10: Lesson 7 - File Storage and Online Collaboration
COMPUTER 10: Lesson 7 - File Storage and Online CollaborationCOMPUTER 10: Lesson 7 - File Storage and Online Collaboration
COMPUTER 10: Lesson 7 - File Storage and Online Collaborationbruanjhuli
 
AI-based audio transcription solutions (IDP)
AI-based audio transcription solutions (IDP)AI-based audio transcription solutions (IDP)
AI-based audio transcription solutions (IDP)KapilVaidya4
 
Checklist to troubleshoot CD moisture profiles.docx
Checklist to troubleshoot CD moisture profiles.docxChecklist to troubleshoot CD moisture profiles.docx
Checklist to troubleshoot CD moisture profiles.docxNoman khan
 
What Developers Need to Unlearn for High Performance NoSQL
What Developers Need to Unlearn for High Performance NoSQLWhat Developers Need to Unlearn for High Performance NoSQL
What Developers Need to Unlearn for High Performance NoSQLScyllaDB
 

Recently uploaded (20)

AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum Revolution
AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum RevolutionAI Health Agents: Longevity as a Service in the Web3 GenAI Quantum Revolution
AI Health Agents: Longevity as a Service in the Web3 GenAI Quantum Revolution
 
Building AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxBuilding AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptx
 
Cybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxCybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptx
 
Reference Domain Ontologies and Large Medical Language Models.pptx
Reference Domain Ontologies and Large Medical Language Models.pptxReference Domain Ontologies and Large Medical Language Models.pptx
Reference Domain Ontologies and Large Medical Language Models.pptx
 
Linked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesLinked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond Ontologies
 
Digital magic. A small project for controlling smart light bulbs.
Digital magic. A small project for controlling smart light bulbs.Digital magic. A small project for controlling smart light bulbs.
Digital magic. A small project for controlling smart light bulbs.
 
RTL Design Methodologies_Object Automation Inc
RTL Design Methodologies_Object Automation IncRTL Design Methodologies_Object Automation Inc
RTL Design Methodologies_Object Automation Inc
 
DS Lesson 2 - Subsets, Supersets and Power Set.pdf
DS Lesson 2 - Subsets, Supersets and Power Set.pdfDS Lesson 2 - Subsets, Supersets and Power Set.pdf
DS Lesson 2 - Subsets, Supersets and Power Set.pdf
 
Monitoring Java Application Security with JDK Tools and JFR Events.pdf
Monitoring Java Application Security with JDK Tools and JFR Events.pdfMonitoring Java Application Security with JDK Tools and JFR Events.pdf
Monitoring Java Application Security with JDK Tools and JFR Events.pdf
 
Future Research Directions for Augmented Reality
Future Research Directions for Augmented RealityFuture Research Directions for Augmented Reality
Future Research Directions for Augmented Reality
 
CHIPS Alliance_Object Automation Inc_workshop
CHIPS Alliance_Object Automation Inc_workshopCHIPS Alliance_Object Automation Inc_workshop
CHIPS Alliance_Object Automation Inc_workshop
 
20200723_insight_release_plan
20200723_insight_release_plan20200723_insight_release_plan
20200723_insight_release_plan
 
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCost
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCostKubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCost
KubeConEU24-Monitoring Kubernetes and Cloud Spend with OpenCost
 
Things you didn't know you can use in your Salesforce
Things you didn't know you can use in your SalesforceThings you didn't know you can use in your Salesforce
Things you didn't know you can use in your Salesforce
 
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.Unleashing the power of AI in UiPath Studio with UiPath Autopilot.
Unleashing the power of AI in UiPath Studio with UiPath Autopilot.
 
Dragino Technology LoRaWANデバイス、ゲートウェイ ユースケース
Dragino Technology   LoRaWANデバイス、ゲートウェイ ユースケースDragino Technology   LoRaWANデバイス、ゲートウェイ ユースケース
Dragino Technology LoRaWANデバイス、ゲートウェイ ユースケース
 
COMPUTER 10: Lesson 7 - File Storage and Online Collaboration
COMPUTER 10: Lesson 7 - File Storage and Online CollaborationCOMPUTER 10: Lesson 7 - File Storage and Online Collaboration
COMPUTER 10: Lesson 7 - File Storage and Online Collaboration
 
AI-based audio transcription solutions (IDP)
AI-based audio transcription solutions (IDP)AI-based audio transcription solutions (IDP)
AI-based audio transcription solutions (IDP)
 
Checklist to troubleshoot CD moisture profiles.docx
Checklist to troubleshoot CD moisture profiles.docxChecklist to troubleshoot CD moisture profiles.docx
Checklist to troubleshoot CD moisture profiles.docx
 
What Developers Need to Unlearn for High Performance NoSQL
What Developers Need to Unlearn for High Performance NoSQLWhat Developers Need to Unlearn for High Performance NoSQL
What Developers Need to Unlearn for High Performance NoSQL
 

Sol85

  • 1. Solution Week 85 (4/26/04) Tower of cylinders Both cylinders in a given row move in the same manner, so we may simply treat them as one cylinder with mass m = 2M. Let the forces that the boards exert on the cylinders be labelled as shown. “F” is the force from the plank below a given cylinder, and “G” is the force from the plank above it. F G F G n an αn n n+1 n+1 Note that by Newton’s third law, we have Fn+1 = Gn, because the planks are massless. Our strategy will be to solve for the linear and angular accelerations of each cylinder in terms of the accelerations of the cylinder below it. Since we want to solve for two quantities, we will need to produce two equations relating the accelerations of two successive cylinders. One equation will come from a combination of F = ma, τ = Iα, and Newton’s third law. The other will come from the nonslipping condition. With the positive directions for a and α defined as in the figure, F = ma on the nth cylinder gives Fn − Gn = man, (1) and τ = Iα on the nth cylinder gives (Fn + Gn)R = 1 2 mR2 αn =⇒ Fn + Gn = 1 2 mRαn (2) Solving the previous two equations for Fn and Gn gives Fn = 1 2 man + 1 2 mRαn , Gn = 1 2 −man + 1 2 mRαn . (3) But we know that Fn+1 = Gn. Therefore, an+1 + 1 2 Rαn+1 = −an + 1 2 Rαn. (4) We will now use the fact that the cylinders don’t slip with respect to the boards. The acceleration of the board above the nth cylinder is an − Rαn. But the accel- eration of this same board, viewed as the board below the (n + 1)st cylinder, is an+1 + Rαn+1. Therefore, an+1 + Rαn+1 = an − Rαn. (5) 1
  • 2. Eqs. (4) and (5) are a system of two equations in the two unknowns, an+1 and αn+1, in terms of an and αn. Solving for an+1 and αn+1 gives an+1 = −3an + 2Rαn, Rαn+1 = 4an − 3Rαn. (6) We can write this in matrix form as an+1 Rαn+1 = −3 2 4 −3 an Rαn . (7) We therefore have an Rαn = −3 2 4 −3 n−1 a1 Rα1 . (8) Consider now the eigenvectors and eigenvalues of the above matrix. The eigenvectors are found via −3 − λ 2 4 −3 − λ = 0 =⇒ λ± = −3 ± 2 √ 2. (9) The eigenvectors are then V+ = 1√ 2 , for λ+ = −3 + 2 √ 2, V− = 1 − √ 2 , for λ− = −3 − 2 √ 2. (10) Note that |λ−| > 1, so λn − → ∞ as n → ∞. This means that if the initial (a1, Rα1) vector has any component in the V− direction, then the (an, Rαn) vectors will head to infinity. This violates conservation of energy. Therefore, the (a1, Rα1) vector must be proportional to V+.1 That is, Rα1 = √ 2a1. Combining this with the fact that the given acceleration, a, of the bottom board equals a1 + Rα1, we obtain a = a1 + √ 2a1 =⇒ a1 = a √ 2 + 1 = ( √ 2 − 1)a. (11) Remark: Let us consider the general case where the cylinders have a moment of inertia of the form I = βMR2 . Using the above arguments, you can show that eq. (7) becomes an+1 Rαn+1 = 1 1 − β −(1 + β) 2β 2 −(1 + β) an Rαn . (12) And you can show that the eigenvectors and eigenvalues are V+ = √ β 1 , for λ+ = √ β − 1 √ β + 1 , V− = √ β −1 , for λ− = √ β + 1 √ β − 1 . (13) 1 This then means that the (an, Rαn) vectors head to zero as n → ∞, because |λ+| < 1. Also, note that the accelerations change sign from one level to the next, because λ+ is negative. 2
  • 3. As above, we cannot have the exponentially growing solution, so we must have only the V+ solution. We therefore have Rα1 = a1/ √ β. Combining this with the fact that the given acceleration, a, of the bottom board equals a1 + Rα1, we obtain a = a1 + a1 √ β =⇒ a1 = √ β 1 + √ β a. (14) You can verify that all of these results agree with the β = 1/2 results obtained above. Let’s now consider a few special cases of the λ+ = √ β − 1 √ β + 1 (15) eigenvalue, which gives the ratio of the accelerations in any level to the ones in the next level down. • If β = 0 (all the mass of a cylinder is located at the center), then we have λ+ = −1. In other words, the accelerations have the same magnitudes but different signs from one level to the next. The cylinders simply spin in place while their centers remain fixed. The centers are indeed fixed, because a1 = 0, from eq. (14). • If β = 1 (all the mass of a cylinder is located on the rim), then we have λ+ = 0. In other words, there is no motion above the first level. The lowest cylinder basically rolls on the bottom side of the (stationary) plank right above it. Its acceleration is a1 = a/2, from eq. (14). • If β → ∞ (the cylinders have long massive extensions that extend far out beyond the rim), then we have λ+ = 1. In other words, all the levels have equal accelerations. This fact, combined with the Rα1 = a1/ √ β ≈ 0 result, shows that there is no rotational motion at any level, and the whole system simply moves to the right as a rigid object with acceleration a1 = a, from eq. (14). 3