MSMS Basilio Bona DAUIN PoliTo
|
|
- Bruno Stevens
- 5 years ago
- Views:
Transcription
1 MSMS Basilio Bona DAUIN PoliTo Problem 2 The planar system illustrated in Figure 1 consists of a bar B and a wheel W moving (no friction, no sliding) along the bar; the bar can rotate around an axis perpendicular to the plane at the pivot point A. The wheel can rotate around its center and maintains always a contact point with the bar. No elastic elements are present. Questions Answer to the following questions: 1. Defineapossible set ofgeneralized coordinatesq i (t). Trytolinkq 1 tothebarrotation and q 2 to the wheel motion. 2. Write the homogeneous transformation T 2 between R and R Compute the constraint between the angular motion of W and the linear motion of its center. 4. Find the total angular velocity of the wheel W. 5. Compute the kinetic co-energy of the system. 6. Compute the potential energy of the system if g = [ G ] T. 7. Compute the total virtual work due to force F and torque N. 8. Write the two second order differential equations obtained using the Lagrange approach. 9. Using the Lagrange equations and the data below, compute the force F required to keep the bar in equilibrium, assuming that the bar is horizontal, that τ =, and the that the wheel is at the left extremity of the bar. 1
2 MSMS Problem Data Description Symbol Value Unit Bar mass m b 2 kg Bar length L 2 m Bar height ε negligible.7 Bar inertia matrix Γ b ε kg m 2.7 Wheel mass m w 1 kg Wheel radius R.1 m.25 Wheel inertia matrix Γ w.25 kg m 2.5 Applied force Applied torque F N [ f(t) ] T N [ τ ] T N m Solution 1A: Generalized coordinates are angles The most immediate choice for q 1 is the angle between i and i 1 ; q 2 can be chosen as the rotation angle of the wheel, see Figure 1. The first component of the position vector of the wheel center along the axis i 1 is constrained by the relation R(q 2 +θ) = x 1 ; the negative sign is necessary since for positive q 2 the wheel center moves to the left (negative values of component along i 1 ); θ is a constant angle that depends on the orientation of the wheel when x 1 =. We assume for simplicity θ = ; this means that x 1 = when q 2 =. Solution 2: Homogeneous transformations c 1 s 1 L T 1 = +c 2 s 1 c 1 1 T1 2 = 1 c 2 s 2 Rq 2 s 2 c 2 R 1 1 = where c is a constant that depends on the position of the origin of R with respect to the left side of the bar. For simplicity we assume c =.
3 MSMS Problem Therefore c 12 s 12 Rq 2 c 1 Rs 1 + L 2 T 2 = s 12 c 12 Rq 2 s 1 +Rc We can compute the position vectors of the points A,B,W with respect to the reference frame R ; they will be used to answer the various questions: L L (1+c 2 2 1) Rq 2 c 1 Rs 1 + L r A = r B = L s r W = Rq 2 s 1 +Rc 1 from T 2 We have used the symbols r instead of p for convenience (see also Solution 7 paragraph) Solution 3 See discussion in Solution 1A paragraph. Solution 4: Angular velocity A simple geometrical consideration gives the total angular velocity of the wheel as ω W = q 1 + q 2 A formal approach consists in taking the rotation matrix R 2 from T 2 c 12 s 12 R 2 = s 12 c 12 1 derive it with respect to time s 12 c 12 Ṙ 2 = c 12 s 12 ( q 1 + q 2 ) and use the relation Ṙ 2 = S(ω W)R 2, or Ṙ 2 (R 2 )T = S(ω W ), that produces
4 MSMS Problem s 12 c 12 c 12 s 12 Ṙ 2 (R 2 )T = c 12 s 12 s 12 c 12 ( q 1 + q 2 ) 1 s 12 c 12 +s 12 c 12 (s c2 12 ) = s c 2 12 s 12 c 12 s 12 c 12 ( q 1 + q 2 ) 1 = 1 ( q 1 + q 2 ) Recalling the meaning of S(ω) as ω 3 ω 2 S(ω) = ω 3 ω 1 ω 2 ω 1 one finds again the ω W given above. Solution 5: Kinetic co-energy In order to compute the kinetic co-energy C it is necessary to compute the norms of the linear velocities of the centers-of-mass and the angular velocities; in particular we have ṙ A = ω A = q 1 R(q 2 s 1 q 1 c 1 q 1 c 1 q 2 ) ṙ W = R(q 2 c 1 q 1 +s 1 q 1 +s 1 q 2 ) ω W = q 1 + q 2 Ls 2 1 q 1 ṙ B = L 2 1 q 1 The norm ṙ W 2 is ṗ W 2 = R 2( (q 2 s 1 q 1 c 1 q 1 c 1 q 2 ) 2 +(q 2 c 1 q 1 +s 1 q 1 +s 1 q 2 ) 2) = R 2( q ) q q 1 q 2 +q2 2 q 1 2 = R 2( ) ( q 1 + q 2 ) 2 +q2 2 q2 1 (1)
5 MSMS Problem Hence C A = 1 2 q2 1Γ b,z C W = 1 2 m wr 2 (( q 1 + q 2 ) 2 +q 2 2 q2 1 )+ 1 2 ( q 1 + q 2 ) 2 Γ w,z C total = C A +C W where Γ b,z =.7 and Γ w,z =.5. Solution 6: Potential energy Since no elastic elements are present in the system, the potential energy is only due to the gravitational field. The formula for potential energy gives P A = m b g T r A = P W = m w g T r [ ] Rq 2 c 1 Rs 1 + L 2 W = m w G Rq 2 s 1 +Rc 1 = m w GR(c 1 q 2 s 1 ) P total = P A +P W = P W Solution 7: Virtual work There are two external generalized forces applied to the system, F and N; therefore the total virtual work is given by δw total = N T δα A +F T δr B (2) where with α A one indicates the virtual angular motion associated to the bar rotation and with δr B the virtual displacement of the right side of the bar, where F is applied. If one introduces the pose vector p = [ r α ] T and the generalized force vector Φ = [ F N ] T, one can write δw total = δw A +δw B = δp T Φ = Φ T δp (3) It appears that eqn. (2) and eqn. (3) are the same, apart from different symbols used. We have now the following vectors ṗ A = δp A = q 1 δq 1
6 MSMS Problem and and the virtual work is L s 2 1 q 1 L L c 2 1 q 1 s 2 1δq 1 L ṗ B = c 2 1δq 1 δp B = q 1 δq 1 Φ A = τ f(t) Φ B = δw total = τδq 1 f(t) L 2 c 1δq 1 = F 1 δq 1 +F 2 δq 2 where, equating the corresponding symbols, we obtain F 1 = τ f(t) L 2 c 1 F 2 = Solution 8: Lagrange equations Equation 1: angular coordinate, torques therefore C q 1 = Γ b,z q 1 +(m w R 2 +Γ w,z )( q 1 + q 2 )+m w R 2 q 2 2 q 1 d C = Γ b,z q 1 +(m w R 2 +Γ w,z )( q 1 + q 2 )+m w R 2 q dt q 2 q m w R 2 q 2 q 1 q 2 1 C q 1 = P q 1 = m w GR( s 1 q 2 c 1 ) (m w R 2 +m w R 2 q2 2 +Γ b,z +Γ w,z ) q 1 +(m w R 2 +Γ w,z) q 2 +2m w R 2 q 2 q 1 q 2 = τ f(t) L }{{}}{{} c }{{} 2 1 Γ 1 Γ 2 Coriolis acc.
7 MSMS Problem Equation 2: angular coordinate, torques therefore C q 2 = (m w R 2 +Γ w,z )( q 1 + q 2 ) d C = (m w R 2 +Γ w,z )( q 1 + q 2 ) dt q 2 C q 2 = m w R 2 q 2 q 2 1 P q 2 = m w GRs 1 (m w R 2 +Γ w,z ) q }{{} 1 +(m w R 2 +Γ w,z) q }{{} 2 m w R 2 q 2 q 1 2 }{{} Γ 2 Γ 2 centrifugal acc. m w GRs }{{} 1 = weight force Summarizing, we have can write the second order differential equations as Γ 1 q 1 +Γ 2 q 2 +2m w R 2 q 2 q 1 q 2 = τ f(t) Lc 2 1 Γ 2 q 1 +Γ 2 q 2 m w R 2 q 2 q 2 1 = m wgrs 1 Notice that the system is linear with respect to q 1 and q 2, but contains also the nonlinear terms (2m w R 2 q 2 q 1 q 2 ), (m w R 2 q 2 q 2 1), and the two right-hand torques, with trigonometric functions. Solution 9: Equilibrium Considering the given conditions 1. the bar is in equilibrium (static). The static equilibrium is intended as mechanical engineers define it, i.e., q 1 = q 2 =, but generic nonzero accelerations. 2. the bar is horizontal, i.e., q 1 =, 3. τ =, 4. the wheel is at the left extremity of the bar, i.e., q 2 = L 2R The two Lagrange equations become Γ 1 q 1 +Γ 2 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 =
8 MSMS Problem When condition 1) is relaxed and the equilibrium is dynamic, i.e., velocities and accelerations are generic ones, the two equations become where Γ 1 q 1 +Γ 2 q 2 m w RL q 1 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q m wrl q 2 1 = Γ 1 = (m w R 2 +m w R 2L2 4 +Γ b,z +Γ w,z ) Γ 1 The two equations are still nonlinear due to the terms q 1 q 2 and q 1, 2 but if we assume a further condition, i.e., that the velocities are small, so that q 1 q 2 ε and q 1 2 ε, we can write Γ 1 q 1 +Γ 2 q 2 ε = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 ε = Setting ε =, we can write the resulting equations as Γ 1 q 1 +Γ 2 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 = or [ ] Γ 1 Γ 2 ][ q1 Γ 2 Γ 2 = q 2 [ ].5 f(t)l (4) Additional note: linearization and state variable equations The problem does not require to linearize the system or write the state variable equations, but we will do it, linearizing around the following equilibrium point q 1,eq =, q 2,eq = l R = C, q 1,eq, q 2,eq where l,c, q 1,eq and q 2,eq are given constant values. The main nonlinear parts are the terms 2m w R 2 q 2 q 1 q 2 and m w R 2 q 2 q 1 2 ; recalling that a generic nonlinear function f(x 1,x 2,x 3 ) = x 1 x 2 x 3 can be linearized around the values x 1, x 2, x 3 as f + f + f = x 2 x 3 + x 1 x 3 + x 1 x 2 x 1 x1, x 2, x 3 x 2 x1, x 2, x 3 x 3 x1, x 2, x 3 we can conclude that the linearized Lagrange equations are Γ 1 q 1 +Γ 2 q 2 +2m w R 2 ( q 1,eq q 2,eq +C q 2,eq +C q 1,eq ) = τ f(t) Lc 2 1 Γ 2 q 1 +Γ 2 q 2 +m w R 2 ( q 1,eq 2 +2C q 1,eq) = m w GRs 1
9 MSMS Problem Since c 1 = 1, s 1 = an τ = we have the final linearized form Γ 1 q 1 +Γ 2 q 2 +2m w R 2 ( q 1,eq q 2,eq +C q 2,eq +C q 1,eq ) = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 +m w R 2 ( q 2 1,eq +2C q 1,eq) = (5) When q 1,eq =, q 2,eq = l R = C, q 1,eq =, q 2,eq = the equations (5) reduce to [ ] Γ1 Γ 2 ][ q1 = Γ 2 Γ 2 q 2 [ ].5 f(t)l that are similar to (4), apart from the inertia values. In matrix term, we have Since Γ is invertible Γ 1 = 1 Γ 2 (Γ 1 Γ 2 ) Γ q(t) = bf(t) [ ] Γ2 Γ 2 = Γ 2 Γ Γ 1 Γ 2 Γ 1 Γ 2 1 Γ 1 Γ 2 Γ 1 Γ 2 (Γ 1 Γ 2 ) we can write where B = Γ 1 b is q(t) = Γ 1 bf(t) = Bf(t).5 Γ B = 1 Γ 2.5 Γ 1 Γ 2 Setting the states as x 1 = q 1, x 2 = q 2, x 3 = q 1, x 4 = q 2 the resulting linearized differential state equation is ẋ(t) = A 1 x(t)+bf(t) where 1 1 A 1 = 1
10 MSMS Problem A different choice of states, as x 1 = q 1, x 2 = q 1, x 3 = q 2, x 4 = q 2 will produce a different A 2 matrix 1 A 2 = 1 In any case the A matrix has four null eigenvalues, and the linearized system is unstable. Figure 1: Problem 2: choice of q i.
Dynamics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Dynamics Semester 1, / 18
Dynamics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Dynamics Semester 1, 2016-17 1 / 18 Dynamics Dynamics studies the relations between the 3D space generalized forces
More informationROBOTICS Laboratory Problem 02
ROBOTICS 2015-2016 Laboratory Problem 02 Basilio Bona DAUIN PoliTo Problem formulation The planar system illustrated in Figure 1 consists of a cart C sliding with or without friction along the horizontal
More informationGeneralized coordinates and constraints
Generalized coordinates and constraints Basilio Bona DAUIN Politecnico di Torino Semester 1, 2014-15 B. Bona (DAUIN) Generalized coordinates and constraints Semester 1, 2014-15 1 / 25 Coordinates A rigid
More informationMSMS Matlab Problem 02
MSMS 2014-2015 Matlab Problem 02 Basilio Bona DAUIN PoliTo Problem formulation The planar system illustrated in Figure 1 consists of a cart C sliding with friction along the horizontal rail; the cart supports
More informationNon-holonomic constraint example A unicycle
Non-holonomic constraint example A unicycle A unicycle (in gray) moves on a plane; its motion is given by three coordinates: position x, y and orientation θ. The instantaneous velocity v = [ ẋ ẏ ] is along
More informationNonholonomic Constraints Examples
Nonholonomic Constraints Examples Basilio Bona DAUIN Politecnico di Torino July 2009 B. Bona (DAUIN) Examples July 2009 1 / 34 Example 1 Given q T = [ x y ] T check that the constraint φ(q) = (2x + siny
More informationChapter 12. Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx
Chapter 1 Lecture Notes Chapter 1 Oscillatory Motion Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx When the mass is released, the spring will pull
More informationLecture 13 REVIEW. Physics 106 Spring What should we know? What should we know? Newton s Laws
Lecture 13 REVIEW Physics 106 Spring 2006 http://web.njit.edu/~sirenko/ What should we know? Vectors addition, subtraction, scalar and vector multiplication Trigonometric functions sinθ, cos θ, tan θ,
More informationKinematics. Basilio Bona. October DAUIN - Politecnico di Torino. Basilio Bona (DAUIN - Politecnico di Torino) Kinematics October / 15
Kinematics Basilio Bona DAUIN - Politecnico di Torino October 2013 Basilio Bona (DAUIN - Politecnico di Torino) Kinematics October 2013 1 / 15 Introduction The kinematic quantities used are: position r,
More informationKinematics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Kinematics Semester 1, / 15
Kinematics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2014-15 B. Bona (DAUIN) Kinematics Semester 1, 2014-15 1 / 15 Introduction The kinematic quantities used are: position r, linear velocity
More informationPLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)
PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when
More informationAP Physics QUIZ Chapters 10
Name: 1. Torque is the rotational analogue of (A) Kinetic Energy (B) Linear Momentum (C) Acceleration (D) Force (E) Mass A 5-kilogram sphere is connected to a 10-kilogram sphere by a rigid rod of negligible
More informationPhysics 8 Monday, October 28, 2013
Physics 8 Monday, October 28, 2013 Turn in HW8 today. I ll make them less difficult in the future! Rotation is a hard topic. And these were hard problems. HW9 (due Friday) is 7 conceptual + 8 calculation
More informationKinematics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Kinematics Semester 1, / 15
Kinematics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Kinematics Semester 1, 2016-17 1 / 15 Introduction The kinematic quantities used to represent a body frame are: position
More informationIn most robotic applications the goal is to find a multi-body dynamics description formulated
Chapter 3 Dynamics Mathematical models of a robot s dynamics provide a description of why things move when forces are generated in and applied on the system. They play an important role for both simulation
More informationArticulated body dynamics
Articulated rigid bodies Articulated body dynamics Beyond human models How would you represent a pose? Quadraped animals Wavy hair Animal fur Plants Maximal vs. reduced coordinates How are things connected?
More information5. Plane Kinetics of Rigid Bodies
5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.6 Work and energy relations 5.7 Impulse
More informationLecture 6 Physics 106 Spring 2006
Lecture 6 Physics 106 Spring 2006 Angular Momentum Rolling Angular Momentum: Definition: Angular Momentum for rotation System of particles: Torque: l = r m v sinφ l = I ω [kg m 2 /s] http://web.njit.edu/~sirenko/
More informationDIFFERENTIAL KINEMATICS. Geometric Jacobian. Analytical Jacobian. Kinematic singularities. Kinematic redundancy. Inverse differential kinematics
DIFFERENTIAL KINEMATICS relationship between joint velocities and end-effector velocities Geometric Jacobian Analytical Jacobian Kinematic singularities Kinematic redundancy Inverse differential kinematics
More information6. Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm.
1. During a certain period of time, the angular position of a swinging door is described by θ = 5.00 + 10.0t + 2.00t 2, where θ is in radians and t is in seconds. Determine the angular position, angular
More informationConstrained motion and generalized coordinates
Constrained motion and generalized coordinates based on FW-13 Often, the motion of particles is restricted by constraints, and we want to: work only with independent degrees of freedom (coordinates) k
More informationRigid body dynamics. Basilio Bona. DAUIN - Politecnico di Torino. October 2013
Rigid body dynamics Basilio Bona DAUIN - Politecnico di Torino October 2013 Basilio Bona (DAUIN - Politecnico di Torino) Rigid body dynamics October 2013 1 / 16 Multiple point-mass bodies Each mass is
More informationChapter 8: Momentum, Impulse, & Collisions. Newton s second law in terms of momentum:
linear momentum: Chapter 8: Momentum, Impulse, & Collisions Newton s second law in terms of momentum: impulse: Under what SPECIFIC condition is linear momentum conserved? (The answer does not involve collisions.)
More informationRigid Body Kinetics :: Virtual Work
Rigid Body Kinetics :: Virtual Work Work-energy relation for an infinitesimal displacement: du = dt + dv (du :: total work done by all active forces) For interconnected systems, differential change in
More informationTorque and Rotation Lecture 7
Torque and Rotation Lecture 7 ˆ In this lecture we finally move beyond a simple particle in our mechanical analysis of motion. ˆ Now we consider the so-called rigid body. Essentially, a particle with extension
More informationPLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.
More informationRigid bodies - general theory
Rigid bodies - general theory Kinetic Energy: based on FW-26 Consider a system on N particles with all their relative separations fixed: it has 3 translational and 3 rotational degrees of freedom. Motion
More informationNon-Linear Response of Test Mass to External Forces and Arbitrary Motion of Suspension Point
LASER INTERFEROMETER GRAVITATIONAL WAVE OBSERVATORY -LIGO- CALIFORNIA INSTITUTE OF TECHNOLOGY MASSACHUSETTS INSTITUTE OF TECHNOLOGY Technical Note LIGO-T980005-01- D 10/28/97 Non-Linear Response of Test
More informationRigid Body Kinetics :: Force/Mass/Acc
Rigid Body Kinetics :: Force/Mass/Acc General Equations of Motion G is the mass center of the body Action Dynamic Response 1 Rigid Body Kinetics :: Force/Mass/Acc Fixed Axis Rotation All points in body
More informationLecture 4. Alexey Boyarsky. October 6, 2015
Lecture 4 Alexey Boyarsky October 6, 2015 1 Conservation laws and symmetries 1.1 Ignorable Coordinates During the motion of a mechanical system, the 2s quantities q i and q i, (i = 1, 2,..., s) which specify
More informationClassical Mechanics and Electrodynamics
Classical Mechanics and Electrodynamics Lecture notes FYS 3120 Jon Magne Leinaas Department of Physics, University of Oslo 2 Preface FYS 3120 is a course in classical theoretical physics, which covers
More informationPhysics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium
Physics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium Strike (Day 10) Prelectures, checkpoints, lectures continue with no change. Take-home quizzes this week. See Elaine Schulte s email. HW
More information06. Lagrangian Mechanics II
University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 06. Lagrangian Mechanics II Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
More informationChapter 9. Rotational Dynamics
Chapter 9 Rotational Dynamics In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of translation and rotation. 1) Torque Produces angular
More informationDefinition. is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau)
Torque Definition is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau) = r F = rfsin, r = distance from pivot to force, F is the applied force
More informationExam 3 Practice Solutions
Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at
More informationPhysics 141 Rotational Motion 2 Page 1. Rotational Motion 2
Physics 141 Rotational Motion 2 Page 1 Rotational Motion 2 Right handers, go over there, left handers over here. The rest of you, come with me.! Yogi Berra Torque Motion of a rigid body, like motion of
More informationGeneralized Coordinates, Lagrangians
Generalized Coordinates, Lagrangians Sourendu Gupta TIFR, Mumbai, India Classical Mechanics 2012 August 10, 2012 Generalized coordinates Consider again the motion of a simple pendulum. Since it is one
More informationIntroduction to centralized control
ROBOTICS 01PEEQW Basilio Bona DAUIN Politecnico di Torino Control Part 2 Introduction to centralized control Independent joint decentralized control may prove inadequate when the user requires high task
More informationClassical Mechanics and Electrodynamics
Classical Mechanics and Electrodynamics Lecture notes FYS 3120 Jon Magne Leinaas Department of Physics, University of Oslo December 2009 2 Preface These notes are prepared for the physics course FYS 3120,
More informationPhysics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 20: Rotational Motion. Slide 20-1
Physics 1501 Fall 2008 Mechanics, Thermodynamics, Waves, Fluids Lecture 20: Rotational Motion Slide 20-1 Recap: center of mass, linear momentum A composite system behaves as though its mass is concentrated
More informationProblem 1 Problem 2 Problem 3 Problem 4 Total
Name Section THE PENNSYLVANIA STATE UNIVERSITY Department of Engineering Science and Mechanics Engineering Mechanics 12 Final Exam May 5, 2003 8:00 9:50 am (110 minutes) Problem 1 Problem 2 Problem 3 Problem
More information15. Hamiltonian Mechanics
University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 15. Hamiltonian Mechanics Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
More information7. The gyroscope. 7.1 Introduction. 7.2 Theory. a) The gyroscope
K 7. The gyroscope 7.1 Introduction This experiment concerns a special type of motion of a gyroscope, called precession. From the angular frequency of the precession, the moment of inertia of the spinning
More informationMultibody simulation
Multibody simulation Dynamics of a multibody system (Euler-Lagrange formulation) Dimitar Dimitrov Örebro University June 16, 2012 Main points covered Euler-Lagrange formulation manipulator inertia matrix
More informationRotation review packet. Name:
Rotation review packet. Name:. A pulley of mass m 1 =M and radius R is mounted on frictionless bearings about a fixed axis through O. A block of equal mass m =M, suspended by a cord wrapped around the
More information1. Which of the following is the unit for angular displacement? A. Meters B. Seconds C. Radians D. Radian per second E. Inches
AP Physics B Practice Questions: Rotational Motion Multiple-Choice Questions 1. Which of the following is the unit for angular displacement? A. Meters B. Seconds C. Radians D. Radian per second E. Inches
More informationKinematics. Chapter Multi-Body Systems
Chapter 2 Kinematics This chapter first introduces multi-body systems in conceptual terms. It then describes the concept of a Euclidean frame in the material world, following the concept of a Euclidean
More informationArtificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik. Robot Dynamics. Dr.-Ing. John Nassour J.
Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Robot Dynamics Dr.-Ing. John Nassour 25.1.218 J.Nassour 1 Introduction Dynamics concerns the motion of bodies Includes Kinematics
More informationRigid Manipulator Control
Rigid Manipulator Control The control problem consists in the design of control algorithms for the robot motors, such that the TCP motion follows a specified task in the cartesian space Two types of task
More informationChapter 9 Rotational Dynamics
Chapter 9 ROTATIONAL DYNAMICS PREVIEW A force acting at a perpendicular distance from a rotation point, such as pushing a doorknob and causing the door to rotate on its hinges, produces a torque. If the
More informationz F 3 = = = m 1 F 1 m 2 F 2 m 3 - Linear Momentum dp dt F net = d P net = d p 1 dt d p n dt - Conservation of Linear Momentum Δ P = 0
F 1 m 2 F 2 x m 1 O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = = Ma Ma Ma com, x com, y com, z p = mv - Linear Momentum F net = dp dt F net = d P dt = d p 1 dt +...+ d p n dt Δ P = 0 - Conservation
More informationClassical Mechanics Comprehensive Exam Solution
Classical Mechanics Comprehensive Exam Solution January 31, 011, 1:00 pm 5:pm Solve the following six problems. In the following problems, e x, e y, and e z are unit vectors in the x, y, and z directions,
More informationMassachusetts Institute of Technology Department of Physics. Final Examination December 17, 2004
Massachusetts Institute of Technology Department of Physics Course: 8.09 Classical Mechanics Term: Fall 004 Final Examination December 17, 004 Instructions Do not start until you are told to do so. Solve
More informationChapter 14 Periodic Motion
Chapter 14 Periodic Motion 1 Describing Oscillation First, we want to describe the kinematical and dynamical quantities associated with Simple Harmonic Motion (SHM), for example, x, v x, a x, and F x.
More informationCase Study: The Pelican Prototype Robot
5 Case Study: The Pelican Prototype Robot The purpose of this chapter is twofold: first, to present in detail the model of the experimental robot arm of the Robotics lab. from the CICESE Research Center,
More informationAn Overview of Mechanics
An Overview of Mechanics Mechanics: The study of how bodies react to forces acting on them. Statics: The study of bodies in equilibrium. Dynamics: 1. Kinematics concerned with the geometric aspects of
More informationPhysics for Scientists and Engineers 4th Edition, 2017
A Correlation of Physics for Scientists and Engineers 4th Edition, 2017 To the AP Physics C: Mechanics Course Descriptions AP is a trademark registered and/or owned by the College Board, which was not
More informationα = p = m v L = I ω Review: Torque Physics 201, Lecture 21 Review: Rotational Dynamics a = Στ = I α
Physics 1, Lecture 1 Today s Topics q Static Equilibrium of Rigid Objects(Ch. 1.1-3) Review: Rotational and Translational Motion Conditions for Translational and Rotational Equilibrium Demos and Exercises
More informationKinetic Energy of Rolling
Kinetic Energy of Rolling A solid disk and a hoop (with the same mass and radius) are released from rest and roll down a ramp from a height h. Which one is moving faster at the bottom of the ramp? A. they
More informationRutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 19. Home Page. Title Page. Page 1 of 36.
Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 19 Page 1 of 36 12. Equilibrium and Elasticity How do objects behave under applied external forces? Under
More informationChapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I
More informationAdvanced Dynamics. - Lecture 4 Lagrange Equations. Paolo Tiso Spring Semester 2017 ETH Zürich
Advanced Dynamics - Lecture 4 Lagrange Equations Paolo Tiso Spring Semester 2017 ETH Zürich LECTURE OBJECTIVES 1. Derive the Lagrange equations of a system of particles; 2. Show that the equation of motion
More informationTable of Contents. Pg. # Momentum & Impulse (Bozemanscience Videos) Lab 2 Determination of Rotational Inertia 1 1/11/16
Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum 5 1/26/16 Rotational tatics
More informationPhysics 5A Final Review Solutions
Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone
More informationPhys101 Lectures 19, 20 Rotational Motion
Phys101 Lectures 19, 20 Rotational Motion Key points: Angular and Linear Quantities Rotational Dynamics; Torque and Moment of Inertia Rotational Kinetic Energy Ref: 10-1,2,3,4,5,6,8,9. Page 1 Angular Quantities
More informationRotational Kinematics
Rotational Kinematics Rotational Coordinates Ridged objects require six numbers to describe their position and orientation: 3 coordinates 3 axes of rotation Rotational Coordinates Use an angle θ to describe
More informationτ net l = r p L = Iω = d L dt = I α ΔL Angular momentum (one) Angular momentum (system, fixed axis) Newton s second law (system)
l = r p L = Iω = d L dt = I α ΔL = 0 Angular momentum (one) Angular momentum (system, fixed axis) Newton s second law (system) Conserva
More informationPHY131H1S - Class 20. Pre-class reading quiz on Chapter 12
PHY131H1S - Class 20 Today: Gravitational Torque Rotational Kinetic Energy Rolling without Slipping Equilibrium with Rotation Rotation Vectors Angular Momentum Pre-class reading quiz on Chapter 12 1 Last
More informationAssignment 9. to roll without slipping, how large must F be? Ans: F = R d mgsinθ.
Assignment 9 1. A heavy cylindrical container is being rolled up an incline as shown, by applying a force parallel to the incline. The static friction coefficient is µ s. The cylinder has radius R, mass
More informationChapter 8. Centripetal Force and The Law of Gravity
Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration
More informationVideo 2.1a Vijay Kumar and Ani Hsieh
Video 2.1a Vijay Kumar and Ani Hsieh Robo3x-1.3 1 Introduction to Lagrangian Mechanics Vijay Kumar and Ani Hsieh University of Pennsylvania Robo3x-1.3 2 Analytical Mechanics Aristotle Galileo Bernoulli
More informationIn-Class Problems 30-32: Moment of Inertia, Torque, and Pendulum: Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 TEAL Fall Term 004 In-Class Problems 30-3: Moment of Inertia, Torque, and Pendulum: Solutions Problem 30 Moment of Inertia of a
More informationfor changing independent variables. Most simply for a function f(x) the Legendre transformation f(x) B(s) takes the form B(s) = xs f(x) with s = df
Physics 106a, Caltech 1 November, 2018 Lecture 10: Hamiltonian Mechanics I The Hamiltonian In the Hamiltonian formulation of dynamics each second order ODE given by the Euler- Lagrange equation in terms
More informationAP Pd 3 Rotational Dynamics.notebook. May 08, 2014
1 Rotational Dynamics Why do objects spin? Objects can travel in different ways: Translation all points on the body travel in parallel paths Rotation all points on the body move around a fixed point An
More informationSolutions to Exam #1
SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 1 of 9 Solutions to Exam #1 1. Which of the following natural sciences most directly involves and applies physics? a) Botany (plant biology)
More informationFor a rigid body that is constrained to rotate about a fixed axis, the gravitational torque about the axis is
Experiment 14 The Physical Pendulum The period of oscillation of a physical pendulum is found to a high degree of accuracy by two methods: theory and experiment. The values are then compared. Theory For
More informationis acting on a body of mass m = 3.0 kg and changes its velocity from an initial
PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0-kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block
More informationPhysics 106a, Caltech 16 October, Lecture 5: Hamilton s Principle with Constraints. Examples
Physics 106a, Caltech 16 October, 2018 Lecture 5: Hamilton s Principle with Constraints We have been avoiding forces of constraint, because in many cases they are uninteresting, and the constraints can
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5
1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES
More informationLecture 14. Rotational dynamics Torque. Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.
Lecture 14 Rotational dynamics Torque Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes, 87 1 BC EXAM Tuesday March 6, 018 8:15 PM 9:45 PM Today s Topics:
More informationLinearize a non-linear system at an appropriately chosen point to derive an LTI system with A, B,C, D matrices
Dr. J. Tani, Prof. Dr. E. Frazzoli 151-0591-00 Control Systems I (HS 2018) Exercise Set 2 Topic: Modeling, Linearization Discussion: 5. 10. 2018 Learning objectives: The student can mousavis@ethz.ch, 4th
More informationQ1. Which of the following is the correct combination of dimensions for energy?
Tuesday, June 15, 2010 Page: 1 Q1. Which of the following is the correct combination of dimensions for energy? A) ML 2 /T 2 B) LT 2 /M C) MLT D) M 2 L 3 T E) ML/T 2 Q2. Two cars are initially 150 kilometers
More informationProblem Set x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology. 1. Moment of Inertia: Disc and Washer
8.01x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology Problem Set 10 1. Moment of Inertia: Disc and Washer (a) A thin uniform disc of mass M and radius R is mounted on an axis passing
More informationPHYSICS 149: Lecture 21
PHYSICS 149: Lecture 21 Chapter 8: Torque and Angular Momentum 8.2 Torque 8.4 Equilibrium Revisited 8.8 Angular Momentum Lecture 21 Purdue University, Physics 149 1 Midterm Exam 2 Wednesday, April 6, 6:30
More informationRotational Kinetic Energy
Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.4 Rotational Work and Energy Work to accelerate a mass rotating it by angle φ F W = F(cosθ)x x = s = rφ = Frφ Fr = τ (torque) = τφ r φ s F to s θ = 0 DEFINITION
More informationPHYSICS 220. Lecture 15. Textbook Sections Lecture 15 Purdue University, Physics 220 1
PHYSICS 220 Lecture 15 Angular Momentum Textbook Sections 9.3 9.6 Lecture 15 Purdue University, Physics 220 1 Last Lecture Overview Torque = Force that causes rotation τ = F r sin θ Work done by torque
More informationq 1 F m d p q 2 Figure 1: An automated crane with the relevant kinematic and dynamic definitions.
Robotics II March 7, 018 Exercise 1 An automated crane can be seen as a mechanical system with two degrees of freedom that moves along a horizontal rail subject to the actuation force F, and that transports
More informationVibrations Qualifying Exam Study Material
Vibrations Qualifying Exam Study Material The candidate is expected to have a thorough understanding of engineering vibrations topics. These topics are listed below for clarification. Not all instructors
More information8.012 Physics I: Classical Mechanics Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 8.012 Physics I: Classical Mechanics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE
More informationAssignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class
Assignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class Homeworks VIII and IX both center on Lagrangian mechanics and involve many of the same skills. Therefore,
More informationChapter 15 Periodic Motion
Chapter 15 Periodic Motion Slide 1-1 Chapter 15 Periodic Motion Concepts Slide 1-2 Section 15.1: Periodic motion and energy Section Goals You will learn to Define the concepts of periodic motion, vibration,
More informationPre-AP Physics Review Problems
Pre-AP Physics Review Problems SECTION ONE: MULTIPLE-CHOICE QUESTIONS (50x2=100 points) 1. The graph above shows the velocity versus time for an object moving in a straight line. At what time after t =
More informationof the four-bar linkage shown in Figure 1 is T 12
ME 5 - Machine Design I Fall Semester 0 Name of Student Lab Section Number FINL EM. OPEN BOOK ND CLOSED NOTES Wednesday, December th, 0 Use the blank paper provided for your solutions write on one side
More informationPhysics 201, Lecture 21
Physics 201, Lecture 21 Today s Topics q Static Equilibrium of Rigid Objects(Ch. 12.1-3) Review: Rotational and Translational Motion Conditions for Translational and Rotational Equilibrium Demos and Exercises
More informationDynamics of Rotational Motion
Chapter 10 Dynamics of Rotational Motion To understand the concept of torque. To relate angular acceleration and torque. To work and power in rotational motion. To understand angular momentum. To understand
More informationCHAPTER 10 ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS WEN-BIN JIAN ( 簡紋濱 ) DEPARTMENT OF ELECTROPHYSICS NATIONAL CHIAO TUNG UNIVERSITY
CHAPTER 10 ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS WEN-BIN JIAN ( 簡紋濱 ) DEPARTMENT OF ELECTROPHYSICS NATIONAL CHIAO TUNG UNIVERSITY OUTLINE 1. Angular Position, Velocity, and Acceleration 2. Rotational
More informationAnnouncements. 1. Do not bring the yellow equation sheets to the miderm. Idential sheets will be attached to the problems.
Announcements 1. Do not bring the yellow equation sheets to the miderm. Idential sheets will be attached to the problems. 2. Some PRS transmitters are missing. Please, bring them back! 1 Kinematics Displacement
More informationMechanics Departmental Exam Last updated November 2013
Mechanics Departmental Eam Last updated November 213 1. Two satellites are moving about each other in circular orbits under the influence of their mutual gravitational attractions. The satellites have
More information