DOCSLIB.ORG

EP 222: Classical Mechanics Tutorial Sheet 1

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
EP 222: Classical Mechanics Tutorial Sheet 1 EP 222: Classical Mechanics Tutorial Sheet 1 This tutorial sheet contains problems on the Newton’s laws of motion and Lagrangian formalism. 1. Show that for a single particle with a constant mass the equation of motion implies the following differential equation for the kinetic energy: dT = F · v; dt while if the mass varies with time the corresponding equation is d(mT ) = F · p: dt 1 2 1 Soln: (a) We know T = 2 mv = 2 mv · v. Thus, if m is constant dT dv = m · v = ma · v = F · v dt dt (b) For a variable mass particle, let us consider d(mT ) d 1 dm dm = m2v · v = m v · v + m2v · a = v + ma · (mv) = F · p; dt dt 2 dt dt dp dm because for a variable mass particle dt = dt v + ma. 2. Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation X 1 X M 2R2 = M m r2 − m m r2 i i 2 i j ij i i;j Soln: R is defined as P m r R = i i i M X =) MR = miri i 2 2 X X =) M R = ( miri) · ( mjrj) i j 2 2 X X 2 2 X =) M R = mimjri · rj = mi ri + mimjri · rj i;j i i6=j 1 Using 2 2 2 2 rij = (ri − rj) = ri + rj − 2ri · rj 1 =) r · r = (r2 + r2 − r2 ); i j 2 i j ij we obtain X 1 X M 2R2 = m2r2 + m m (r2 + r2 − r2 ) i i 2 i j i j ij i i6=j X 1 X 1 X 1 X = m2r2 + m m r2 + m m r2 + − m m r2 i i 2 i j i 2 i j j 2 i j ij i i6=j i6=j i6=j X X 1 X = m m r2 + m2r2 − m m r2 i j i i i 2 i j ij i6=j i i6=j X X 1 X = ( m )( m r2) − m m r2 j i i 2 i j ij j i i6=j X 1 X = M( m r2) − m m r2 i i 2 i j ij i i6=j X 1 X = M( m r2) − m m r2 i i 2 i j ij i i;j In the second sum on the RHS, we can perform unrestricted sum over i and j, because for i = j, rij = 0. 3. Show that the Lagrange equations d @T @T − = Qj; dt @q_j @qj can also be written as @T_ @T − 2 = Qj: @q_j @qj These are sometimes called the Nielsen form of Lagrange equations. Soln: Assuming that T = T (qi;q_i; t); we have dT X @T X @T @T T_ = = q¨ + q_ + dt @q_ i @q i @t i i i i so that @T_ X @2T X @2T @T @2T = q¨ + q_ + + @q_ @q_ @q_ i @q_ @q i @q @q_ @t j i j i i j i j j 2 leading to @T_ @T X @2T X @2T @2T − = q¨ + q_ + @q_ @q @q_ @q_ i @q_ @q i @q_ @t j j i j i i j i j X @ @T @ @T @ @T = q¨ + q_ + @q_ @q_ i @q @q_ i @t @q_ i i j i j j d @T = dt @q_j Substituting this on the LHS of Lagrange equation, we obtain d @T @T − = Qj dt @q_j @qj @T_ @T @T =) − − = Qj @q_j @qj @qj @T_ @T =) − 2 = Qj @q_j @qj 4. If L is a Lagrangian for a system of n degrees of freedom satisfying the Lagrange equations, show by direct substitution that 0 dF (q ; : : : ; q ; t) L = L + 1 n dt also satisfies Lagrange’s equations where F is any arbitrary, but differentiable, func- tion of its arguments. Soln: We have dF X @F @F = q_ + : dt @q i @t i i Therefore 0 dF (q1; : : : ; qn; t) X @F @F L = L + = L + q_ + ; dt @q i @t i i so that @L0 @L @F = + ; @q_j @q_j @qj leading to d @L0 d @L d @F = + dt @q_j dt @q_j dt @qj d @L X @2F @2F = + q_ + dt @q_ @q @q i @t@q j i i j j d @L X @2F @2F = + q_ + dt @q_ @q @q i @q @t j i j i j 3 and 0 @L @L X @2F @2F = + q_ + : @q @q @q @q i @q @t j j i j i j Thus 0 0 d @L @L d @L X @2F @2F @L X @2F @2F − = + q_ + − − q_ − dt @q_ @q dt @q_ @q @q i @q @t @q @q @q i @q @t j j j i j i j j i j i j d @L0 @L0 d @L @L =) − = − = 0 dt @q_j @qj dt @q_j @qj QED. 5. Obtain the Lagrange equations of motion for a spherical pendulum, i.e., a point mass suspended by a rigid weightless rod. Soln: It is best to use spherical polar coordinates here. In the lectures we showed that the kinetic energy of a particle in spherical polar coordinates is 1 T = m r_2 + r2θ_2 + r2 sin2 θφ_2 : 2 If length of the rod is l, then r = l, and r_ = 0, we are left with two generalized coordinates (θ; φ), with 1 T = m l2θ_2 + l2 sin2 θφ_2 ; 2 and using the point of suspension as the reference for potential energy, we have V = −mgl cos θ: 4 Thus 1 L = T − V = T = m l2θ_2 + l2 sin2 θφ_2 + mgl cos θ: 2 Now the two Lagrange equations are d @L @L − = 0 dt @θ_ @θ 1 =) ml2θ¨ − ml2 sin 2θφ_2 + mgl sin θ = 0 2 and d @L @L − = 0 dt @φ_ @φ d(ml2 sin2 θφ_) =) = 0: dt 6. Obtain the Lagrangian and equations of motion for a double pendulum, where the lengths of the pendula are l1 and l2 with corresponding masses m1 and m2, confined to move in a plane. Soln: As discussed in the lectures, this system has two generalized coordinates θ1 and θ2, the angles which the upper and the lower pendula make with respect to the vertical. With the motion of the pendula confined in a plane (say, xy plane), then the Cartesian coordinates of the two particles can be written as x1 = l1 sin θ1 y1 = −l1 cos θ1 and x2 = x1 + l2 sin θ2 = l1 sin θ1 + l2 sin θ2 y2 = y1 − l2 cos θ2 = −l1 cos θ1 − l2 cos θ2 5 So that 1 1 T = m (_x2 +y _2) + m (_x2 +y _2): 2 1 1 1 2 2 2 2 Now _ x_ 1 = l1 cos θ1θ1 _ y_1 = l1 sin θ1θ1 _ _ x_ 2 = l1 cos θ1θ1 + l2 cos θ2θ2 _ _ y_2 = l1 sin θ1θ1 + l2 sin θ2θ2 Easy to verify 2 2 2 _2 x_ 1 +y _1 = l1θ1; 2 2 2 _2 2 _2 _ _ x_ 2 +y _2 = l1θ1 + l2θ2 + 2l1l2θ1θ2 cos(θ1 − θ2): With this 1 1 T = (m + m )l2θ_2 + m l2θ_2 + m l l θ_ θ_ cos(θ − θ ); 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 2 and V = m1gy1 + m2gy2 = −(m1 + m2)gl1 cos θ1 − m2gl2 cos θ2; so that L = T − V 1 1 = (m + m )l2θ_2 + m l2θ_2 + m l l θ_ θ_ cos(θ − θ ) 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 2 + (m1 + m2)gl1 cos θ1 + m2gl2 cos θ2: θ1 equation of motion d @L @L − = 0; _ dt @θ1 @θ1 leads to d (m + m )l2θ_ + m l l θ_ cos(θ − θ ) + m l l θ_ θ_ sin(θ − θ ) + (m + m )gl sin θ = 0: dt 1 2 1 1 2 1 2 2 1 2 2 1 2 1 2 1 2 1 2 1 1 Upon taking the time derivative, we obtain the final form 2 ¨ ¨ _2 (m1 + m2)l1θ1 + m2l1l2 cos(θ1 − θ2)θ2 + m2l1l2 sin(θ1 − θ2)θ2 + (m1 + m2)gl1 sin θ1 = 0 For θ2 d @L @L − = 0; _ dt @θ2 @θ2 6 which yields d m l2θ_ + m l l θ_ cos(θ − θ ) − m l l θ_ θ_ sin(θ − θ ) + m gl sin θ = 0: dt 2 2 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 2 2 Upon taking the time derivative, we obtain the final form ¨ 2 ¨ _2 m2l1l2 cos(θ1 − θ2)θ1 + m2l2θ2 − m2l1l2 sin(θ1 − θ2)θ1 + m2gl2 sin θ2 = 0 7. If we want to obtain the equations of motion for a charged particle of mass m, moving in an electromagnetic field (E, B), the potential in the Lagrangian has to be velocity dependent U = qφ − qA · v, where q is the charge of the particle, and φ, and A, respectively, are the scalar and vector potentials of the electromagnetic field so that @A E = −∇φ − @t B = r × A: Show that using this Lagrangian, we obtain the correct equations of motion for the particle. Soln: Using Cartesian coordinates and the fact that v =x _^i +y _^j +z _k^, and A = ^ Ax^i + Ay^j + Azk, we have 1 L = m(_x2 +y _2 +z _2) − qφ + q(A x_ + A y_ + A z_): 2 x y z Lagrange equation for x component d @L @L − = 0 dt @x_ @x d @φ @A @A @A =) (mx_ + qA ) + q − qx_ x − qy_ y − qz_ z = 0 dt x @x @x @x @x dA @φ @A @A @A =) mx¨ = −q x − q + qx_ x + qy_ y + qz_ z dt @x @x @x @x But dA @A @A @A @A x = x x_ + x y_ + x z_ + x ; dt @x @y @z @t so that @φ @A @A @A @A @A @A @A mx¨ = q − − x − q x x_ − q x y_ − q x z_ + qx_ x + qy_ y + qz_ z @x @t @x @y @z @x @x @x @φ @A @A @A @A @A = q − − x + qy_ y + qz_ z − q x y_ − q x z_ @x @t @x @x @y @z Using the fact that @A E = −∇φ − @t B = r × A; 7 we obtain above mx¨ = qEx + q(v × B)x; which is the x component of the Lorentz force equation.
Load more

Recommended publications
  • Hamiltonian Mechanics
    Hamiltonian Mechanics Marcello Seri Bernoulli Institute University of Groningen m.seri (at) rug.nl Version 1.0 May 12, 2020 This work is licensed under a Creative Commons “Attribution-NonCommercial-ShareAlike 4.0 Inter- national” license. Contents 1 Classical mechanics, from Newton to Lagrange and back 3 1.1 Newtonian mechanics .................... 3 1.1.1 Motion in one degree of freedom ........... 5 1.2 Hamilton’s variational principle ................ 7 1.2.1 Dynamics of point particles: from Lagrange back to Newton 13 1.3 Euler-Lagrange equations on smooth manifolds ........ 18 1.4 First steps with conserved quantities ............. 24 1.4.1 Back to one degree of freedom ............ 24 1.4.2 The conservation of energy .............. 25 1.4.3 Back to one degree of freedom, again ......... 27 2 Conservation Laws 34 2.1 Noether theorem ....................... 34 2.1.1 Homogeneity of space: total momentum ....... 38 2.1.2 Isotropy of space: angular momentum ......... 40 2.1.3 Homogeneity of time: the energy ........... 44 2.1.4 Scale invariance: Kepler’s third law .......... 44 2.2 The spherical pendulum ................... 45 2.3 Intermezzo: small oscillations ................. 47 2.3.1 Decomposition into characteristic oscillations ..... 50 2.3.2 The double pendulum ................. 51 2.3.3 The linear triatomic molecule ............. 52 2.3.4 Zero modes ...................... 53 2.4 Motion in a central potential ................. 54 2.4.1 Kepler’s problem ................... 57 2.5 D’Alembert principle and systems with constraints ...... 62 3 Hamiltonian mechanics 67 3.1 The Legendre Transform in the euclidean plane ........ 68 3.2 The Legendre Transform ................... 69 3.3 Poisson brackets and first integrals of motion ........
    [Show full text]
  • Solution Midterm
    Midterm Physics 411 03.09.2018 Mechanics I 1. Consider the following statement: If at all times during a projectile's flight its speed is much les than the terminal speed vter, the effects of the air resistance aew usually very small. (a) Without reference to the explicit equations for the magnitude of vter, explain clearly why this is so. SOLUTION - Without air resistance, a free falling projectile would keep acceler- ating due to gravity, meaning its speed would only get bigger with time. Because there is air resistance, that doesn't happen: as the speed increases so does the air resistance, stalling the projectile. Terminal speed happens when when gravity is matched by the air resistance effects, thus once the projectile achieves vter its speed remains constant, ultimately working as an upper limit for speed. If the speed is much lower than this cap, air resistance effects won't be as relevant, as the drag should be much less than the weight. (b) By examining the explicit equations in the book (2.26) and (2.53) explain why the statement above is even more useful for the case of quadratic drag than for the lienar case. [Hint: Express the ratio f=mg of the drag to the weight in terms of the ratio v=vter.] SOLUTION - Writing the ratio of linear and quadratic drag to the weight, we have 2 jflinearj = bv; jfquadj = cv (1) 2 2 flinear bv v fquad cv v = = ; = = 2 (2) mg mg vter mg mg vter From the ratio, we have that a small value v wil result in a more dramatic change in the quadratic drag given its squared dependence on v, making the statement even more useful - for a v that is far from the terminal velocity the quadratic drag corrections would be even smaller.
    [Show full text]
  • Notes: Spherical Pendulum
    M3-4-5 A16 Notes for Geometric Mechanics: Oct{Nov 2011 Professor Darryl D Holm 25 October 2011 Imperial College London [email protected] http://www.ma.ic.ac.uk/~dholm/ Text for the course: Geometric Mechanics I: Dynamics and Symmetry, by Darryl D Holm World Scientific: Imperial College Press, Singapore, Second edition (2011). ISBN 978-1-84816-195-5 Where are we going in this course? 1. Euler{Poincar´eequation 4. Elastic spherical pendulum 2. Rigid body 3. Spherical pendulum 5. Differential forms e These notes: Spherical pendulum 1 Notes for Geometric Mechanics: Spherical pendulum DD Holm Nov 2011 2 What will we study about the spherical pendulum? 1. Newtonian, Lagrangian and Hamilto- 5. Reduced Poisson bracket in R3 nian formulations of motion equation. 6. Nambu Hamiltonian form 2. S1 symmetry and Noether's theorem 3. Lie symmetry reduction 7. Geometric solution interpretation 4. Reduced Poisson bracket in R3 8. Geometric phase 1 Spherical pendulum A spherical pendulum of unit length swings from a fixed point of support under the constant acceleration of gravity g (Figure 2.7). This motion is equivalent to a particle of unit mass moving on the surface of the unit sphere S2 under the influence of the gravitational (linear) potential V (z) with z = ^e3 · x. The only forces acting on the mass are the reaction from the sphere and gravity. This system is often treated by using spherical polar coordinates and the traditional methods of Newton, Lagrange and Hamilton. The present treatment of this problem is more geometrical and avoids polar coordinates.
    [Show full text]
  • Physics 6010, Fall 2016 Constraints and Lagrange Multipliers. Relevant Sections in Text: §1.3–1.6
    Constraints and Lagrange Multipliers. Physics 6010, Fall 2016 Constraints and Lagrange Multipliers. Relevant Sections in Text: x1.3{1.6 Constraints Often times we consider dynamical systems which are defined using some kind of restrictions on the motion. For example, the spherical pendulum can be defined as a particle moving in 3-d such that its distance from a given point is fixed. Thus the true configuration space is defined by giving a simpler (usually bigger) configuration space along with some constraints which restrict the motion to some subspace. Constraints provide a phenomenological way to account for a variety of interactions between systems. We now give a systematic treatment of this idea and show how to handle it using the Lagrangian formalism. For simplicity we will only consider holonomic constraints, which are restrictions which can be expressed in the form of the vanishing of some set of functions { the constraints { on the configuration space and time: Cα(q; t) = 0; α = 1; 2; : : : m: We assume these functions are smooth and independent so that if there are n coordinates qi, then at each time t the constraints restrict the motion to a nice n − m dimensional space. For example, the spherical pendulum has a single constraint on the three Cartesian configruation variables (x; y; z): C(x; y; z) = x2 + y2 + z2 − l2 = 0: This constraint restricts the configuration to a two dimensional sphere of radius l centered at the origin. To see another example of such constraints, see our previous discussion of the double pendulum and pendulum with moving point of support.
    [Show full text]
  • Hamiltonian Dynamics
    Chapter 5. Hamiltonian Dynamics (Most of the material presented in this chapter is taken from Thornton and Marion, Chap. 7) 5.1 The Canonical Equations of Motion As we saw in section 4.7.4, the generalized momentum is defined by !L p j = . (5.1) !q! j We can rewrite the Lagrange equations of motion !L d !L " = 0 (5.2) !q dt !q! j j as follow !L p! j = . (5.3) !q j We can also express our earlier equation for the Hamiltonian (i.e., equation (4.89)) using equation (5.1) as H = p q! ! L. (5.4) j j Equation (5.4) is expressed as a function of the generalized momenta and velocities, and the Lagrangian, which is, in turn, a function of the generalized coordinates and velocities, and possibly time. There is a certain amount of redundancy in this since the generalized momenta are obviously related to the other two variables through equation (5.1). We can, in fact, invert this equation to express the generalized velocities as q! = q! q , p ,t . (5.5) j j ( k k ) This amounts to making a change of variables from q ,q! ,t to q , p ,t ; we, therefore, ( j j ) ( j j ) express the Hamiltonian as H (qk , pk ,t) = p jq! j ! L(qk ,q!k ,t) , (5.6) 91 where it is understood that the generalized velocities are to be expressed as a function of the generalized coordinates and momenta through equation (5.5). The Hamiltonian is, therefore, always considered a function of the (qk , pk ,t) set, whereas the Lagrangian is a function of the (qk ,q!k ,t) set: H = H (qk , pk ,t) (5.7) L = L(qk ,q!k ,t) Let’s now consider the following total differential " !H !H % !H (5.8) dH = $ dqk + dpk ' + dt, # !qk !pk & !t but from equation (5.6) we can also write # "L "L & "L dH = q! dp + p dq! ! dq ! dq! ! dt.
    [Show full text]
  • Online Stereo 3D Simulation in Studying the Spherical Pendulum in Conservative Force Field
    European J of Physics Education Vol.4 Issue 3 2013 Zabunov Online Stereo 3D Simulation in Studying the Spherical Pendulum in Conservative Force Field Svetoslav S. Zabunov Sofia University, Bulgaria [email protected] Abstract The current paper aims at presenting a modern e-learning method and tool that is utilized in teaching physics in the universities. An online stereo 3D simulation is used for e-learning mechanics and specifically the teaching of spherical pendulum as part of the General Physics course for students in the universities. This approach was realized on bases of interactive simulations on a personal computer, a part of the free online e-learning system at http://ialms.net/sim/. This system was practically applied with students at Sofia University, Bulgaria, among others. The shown simulation demonstrates the capabilities of the Web for online representations and visualizations of simulated physics processes that are hard to observe in laboratory conditions with all the accompanying parameters, vectors, quantities and trajectories. The discussed simulation allows the study of spherical pendulum both in conservative and non-conservative force fields. The conservative force field is created by the earth gravity force, whose magnitude may be varied in the simulation from positive to negative values, while its direction is always vertical. The simulation also supports various non-conservative forces that may be applied to the pendulum. The current article concentrates on the case of conservative forces acting. Keywords: Simulation of spherical pendulum, 3D-stereo simulation, E-learning physics through stereo 3D visualization. Introduction This article describes the capabilities of the modern web technologies through online stereo 3D simulations of physics phenomena.
    [Show full text]
  • Research Article 3 DOF Spherical Pendulum Oscillations with a Uniform Slewing Pivot Center and a Small Angle Assumption
    Hindawi Publishing Corporation Shock and Vibration Volume 2014, Article ID 203709, 32 pages http://dx.doi.org/10.1155/2014/203709 Research Article 3 DOF Spherical Pendulum Oscillations with a Uniform Slewing Pivot Center and a Small Angle Assumption Alexander V. Perig,1 Alexander N. Stadnik,2 Alexander I. Deriglazov,2 and Sergey V. Podlesny2 1 Manufacturing Processes and Automation Engineering Department, Engineering Automation Faculty, Donbass State Engineering Academy, Shkadinova 72, Donetsk Region, Kramatorsk 84313, Ukraine 2 Department of Technical Mechanics, Engineering Automation Faculty, Donbass State Engineering Academy, Shkadinova72,DonetskRegion,Kramatorsk84313,Ukraine Correspondence should be addressed to Alexander V. Perig; [email protected] Received 11 February 2014; Revised 7 May 2014; Accepted 21 May 2014; Published 4 August 2014 Academic Editor: Mickael¨ Lallart Copyright © 2014 Alexander V. Perig et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The present paper addresses the derivation of a 3 DOF mathematical model of a spherical pendulum attached to a crane boom tip for uniform slewing motion of the crane. The governing nonlinear DAE-based system for crane boom uniform slewing has been proposed, numerically solved, and experimentally verified. The proposed nonlinear and linearized models have been derived with an introduction of Cartesian coordinates. The linearizedodel m with small angle assumption has an analytical solution. The relative and absolute payload trajectories have been derived. The amplitudes of load oscillations, which depend on computed initial conditions, have been estimated. The dependence of natural frequencies on the transport inertia forces and gravity forces has been computed.
    [Show full text]
  • Phys 7221 Homework #2
    Phys 7221 Homework #2 Gabriela Gonz´alez September 18, 2006 1. Derivation 1-9: Gauge transformations for electromagnetic potential If two Lagrangians differ by a total derivative of a function of coordinates and time, they lead to the same equation of motion: dF (qi, t) ∂F X ∂F L0(q , q˙ , t) = L(q , q˙ , t) + = L(q , q˙ , t) + + q˙ i i i i dt i i ∂t ∂q j j j as proven in the following lines: ∂L0 ∂L0 ∂F = + ∂q˙i ∂q˙i ∂qi ∂L0 ∂L ∂ dF = + ∂qi ∂qi ∂qi dt d ∂L0 ∂L0 d ∂L ∂L d ∂F ∂ dF − = − + − dt ∂q˙i ∂qi dt ∂q˙i ∂qi dt ∂qi ∂qi dt d ∂F ∂ X ∂ ∂F = + q˙ dt ∂q ∂t j ∂q ∂q i j j i ∂ ∂F X ∂F = + q˙ ∂q ∂t j ∂q i j j ∂ dF = ∂qi dt and thus the equations of motion are the same, derived from either Lagrangian: d ∂L0 ∂L0 d ∂L ∂L − = − dt ∂q˙i ∂qi dt ∂q˙i ∂qi 1 (What goes wrong if F = F (q, q,˙ t)?) A particle moving in an electromagnetic field, with scalar potential φ and vector potential A, has a generalized potential function U = φ − A · v and a Lagrangian 1 L = T − U = mv2 − q (φ − A · v) 2 Consider a gauge transformation for the scalar and vector potentials of the form ∂ψ(r, t) φ0 = φ − ∂t A0 = A + ∇ψ(r, t) The change in potential energy under this transformation is U 0 = φ0 − A0 · v ∂ψ = U − q + ∇ψ · v ∂t ∂ψ dr = U − q + ∇ψ · ∂t dt ∂ψ X ∂ψ dxi = U − q + ∂t ∂xi dt dψ = U − q dt and thus, the new Lagrangian differs from the original one by a total derivative: dψ d(qψ) L0 = T − U 0 = T − U + q = L + dt dt and thus we know the equations of motion are the same.
    [Show full text]
  • Animating Spherical Pendulum Using Jacobi's Elliptic Functions
    Animating Spherical Pendulum Using Jacobi's Elliptic Functions A.C. Baroudy [email protected] Abstract Some years ago I have written a Maple document ( already on Maple's online home page) on the subject of animating a simple pendulum for large angles of oscillation. This gave me the chance to test Maple command JacobiSN(time, k). I was very much pleased to see Maple do a wonderful job in getting these Jacobi's elliptic functions without a glitch. Today I am back to these same functions for a similar purpose though much more sophisticated than the previous one. The idea is: 1- to get the differential equations of motion for the Spherical Pendulum (SP), 2- to solve them, 3- to use Maple for finding the inverse of these Elliptic Integrals i.e. finding the displacement z as function of time, 4- to get a set of coordinates [x, y, z] for the positions of the bob at different times for plotting, 5- finally to work out the necessary steps for the purpose of animation. It turns out that even with only 3 oscillations where each is defined with only 20 positions of the bob for a total of 60 points on the graph, the animation is so overwhelming that Maple reports: " the length of the output exceeds 1 million". Not withstanding this warning, Maple did a perfect job by getting the animation to my satisfaction. Note that with only 60 positions of the bob, the present article length is equal to 11000 KB! To be able to upload it, I have to save it without running the last command related to the animation.
    [Show full text]
  • Chapter 1 HW Solution
    ME 516 Spring 2005 Chapter 1 HW Solution Problem 4. Using fixed Cartesian coordinates, the acceleration of the particle is given by a =x ¨i +y ¨j (1) The approach is to integratex ¨ to getx ˙ then x; then gety ˙,y ¨. Use (1.3.15) to get radius of curvature ρ. Sincex ¨(t) = 1 + 0.2t, integration plus initial conditions yields x˙(t) = t + 0.1t2 (2) t2 0.1t3 x(t) = + − 0.6 (3) 2 3 Then since the constraint equation is y = −x2 + x, we have (using the chain rule) dy dx y˙(t) = = (1 − 2x)x ˙ (4) dx dt d d dx d dx˙ y¨(t) = [(1 − 2x)x ˙] = [·] + [·] = −2x ˙ 2 + (1 − 2x)¨x (5) dt dx dt dx˙ dt Evaluate bothx ¨(t) andy ¨(t) at t = 0.5, and you get a =x ¨i +y ¨j = 1.1i + 1.5845j (6) Radius of curvature is given by (note y0 = dy/dx and y00 = d2y/dx2) 3/2 1 + y02 ρ = = 5.209 (7) |y00| Problem 8. (a) Here we have hθ z = (8) 10π which gives position vector hθ r = Re + k (9) R 10π Note that R is constant, and one can either differentiate r with respect to time or use text equations [1.3.47] and [1.3.48] to find velocity and acceleration. The results are hθ˙ v = Rθ˙e + k (10) θ 10π hθ¨ a = −Rθ˙2e + Rθ¨e + k (11) r θ 10π (b) The normal unit vector is opposite the radial unit vector,while the tangent unit vector is upward along the vehicle path, so en = −er (12) et = cos φeθ + sin φk (13) 1 ME 516 Spring 2005 where h φ = tan−1 (14) 10πR Radius of curvature ρ is found using the fact that along a curved path the normal acceleration is related to radius of curvature and velocity.
    [Show full text]
  • Eindhoven University of Technology BACHELOR the Spherical
    Eindhoven University of Technology BACHELOR The spherical pendulum different solutions on a 2-sphere Dohmen, Jesse L.P. Award date: 2017 Link to publication Disclaimer This document contains a student thesis (bachelor's or master's), as authored by a student at Eindhoven University of Technology. Student theses are made available in the TU/e repository upon obtaining the required degree. The grade received is not published on the document as presented in the repository. The required complexity or quality of research of student theses may vary by program, and the required minimum study period may vary in duration. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. • Users may download and print one copy of any publication from the public portal for the purpose of private study or research. • You may not further distribute the material or use it for any profit-making activity or commercial gain The Spherical Pendulum: Different Solutions on a 2-Sphere Bachelor Final Project Applied Mathematics Eindhoven University of Technology Supervisor: J.C. van der Meer by Jesse Dohmen 0899975 August 13, 2017 1 1 Introduction A well-known problem from mathematical physics is the simple (mathematical) pendulum, in which the movement of a pendulum through a plane is described. The trajectory of the pendulum is along a circle. By generalising this movement to a 3-dimensional space, we obtain the spherical pendulum.
    [Show full text]
  • Plane and Spherical Pendulums
    Plane and Spherical Pendulums Spherical Pendulum All spherical pendulums are similar to each other. Upon appropriate non-dimensionalization of time and action, they are described by one and the same dimensionless Lagrangian θ_2 '_ 2 L = + sin2 θ + cos θ ; (1) 2 2 where θ and ' are the polar and azimuthal angles, respectively. Consistent with the axial rotational symmetry of the problem, the variable ' is cyclic and corresponding equation of motion yields the law of conservation of the z-component of the (non-dimesionalized) angular momentum, lz: 2 '_ sin θ = lz : (2) This provides us with a simple relation between '(t) and θ(t): l Z dt '_ = z , '(t) = l : (3) sin2 θ z sin2 θ(t) This relation (and many other relations as well, as we will see in what follows) reveals a deep qualitative analogy between the problem of spherical pendulum and the problem of a two-dimensional particle performing a bound motion in a central potential. In both cases, we have a cyclic (azimuthal) angle associated with the rotational symmetry; the behavior of pendulums's polar angle θ is qualitatively equivalent to that of the polar radius of the 2D particle trapped in a central potential.1 The Lagrange equation for θ is θ¨ =' _ 2 sin θ cos θ − sin θ : (4) Relation (3) allows us to exclude' _ from (4), thus getting a closed equation of motion for θ(t). It is both convenient and instructive to directly arrive at the first integral of the resulting closed equation of motion from the conservation of energy, in a direct analogy with our treatment of a 2D particle in a central potential.
    [Show full text]