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

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

2 OUTLINE : CP1 REVISION LECTURE 3 : INTRODUCTION TO CLASSICAL MECHANICS 1. Angular velocity and angular acceleration 2. The Moment of Inertia 2.1 Example : MoI of a thin rectangular plate 2.2 Parallel axis theorem 2.3 Perpendicular axis theorem 3. Lagrangian Mechanics 3.1 The Lagrangian in various coordinate systems 3.2 Example : bead on rotating hoop 4. The Hamiltonian 4.1 Example: re-visit bead on rotating hoop 2

3 1. Angular velocity and angular acceleration Definition of angular velocity for rotation in a circle ṙ = ω r Angular acceleration: α = ω Definition of the Moment of Inertia of the system I of particles or body rotating about a common axis of symmetry (cf. p = mv for a linear system) J = Iω where I = i m i r 2 i Torque associated with the rotation τ = d dt J = Iα 3

4 2. The Moment of Inertia Calculation of moment of inertia of rigid body A rigid body may be considered as a collection of infinitesimal point particles whose relative distance does not change during motion. Mass = dm, where dm = ρdv and ρ is the volume density ( N ) I = i m i di 2 V d 2 ρ dv where d is the perpendicular distance to the axis of rotation. This integral gives the moment of inertia about the axis of rotation. 4

5 Rotation about a principal axis In general J = Ĩ ω, where Ĩ is the Moment of Inertia Tensor Jx Ixx Ixy Ixz J y = I yx I yy I yz ωx ω y J z I zx I zy I zz ω z 5 Whenever possible, one aligns the axes of the coordinate system in such a way that the mass of the body evenly distributes around the axes: we choose axes of symmetry. Jx J y J z = Ix I y I z ωx ω y ω z The diagonal terms are called the principal axes of the moment of inertia. Whenever we rotate about an axis of symmetry, for every point A there is a point B which cancels it, and J J z ẑ = I z ω z ẑ and where J is parallel to ω along the z axis

6 Moment of inertia & energy of rotation Particles rotating in circular motion about a common axis of rotation with angular velocity ω (where v i = ω r i ). Kinetic energy of mass mi : T i = 1 2 m ivi 2 Total KE = 1 2 i ( mi vi 2 ) vi = ω r i v i = ω r i sin φ i sin φ i = d i r i Trot = 1 2 ( N i m i d 2 i T rot = 1 2 I ω2 ) ω 2 where I is calculated about the axis of rotation 6

7 2.1 Example : MoI of a thin rectangular plate About the x axis I x = y 2 dm dm = ρ dx dy ; ρ = M ab Ix = y 2 ρ dx dy [ ] + = ρ y 3 b 2 [x] + a 2 3 b a 2 2 [ ] b = ρ a 3 + b [ ] = ρa b b 2 12 Hence I x = M b2 12 7

8 2.2 Parallel axis theorem I CM is the moment of inertia of body mass M about an axis passing through its centre of mass. I is the moment of inertia about a parallel axis a distance d from the first. ICM = r 2 dm r = d + r r 2 = d d r + r 2 About the parallel axis : I = r 2 dm = d 2 dm + 2 d r dm + r 2 dm }{{} = 0 (definition of CM) Hence I = ICM + Md 2 8

9 2.3 Perpendicular axis theorem Consider a rigid object that lies entirely within a plane. The perpendicular axis theorem links I z (MoI about an axis perpendicular to the plane) with I x, I y (MoI about two perpendicular axes lying within the plane). 9 Consider perpendicular axes x, y, z (which meet at origin O) ; the body lies in the xy plane Iz = d 2 dm = ( x 2 + y 2) dm = x 2 dm + y 2 dm I z = I x + I y This is the perpendicular axis theorem.

10 Example : compound pendulum Rectangular rod length a width b mass m swinging about axis O, distance l from the CM, in plane of paper 10 Ix = mb2 12, I y = m ma2 12 Perpendicular axis theorem : I z I CM = m ( a2 +b 2 12 ) Parallel axis theorem : I = I CM + ml 2 Torque about O = r F : τ = m g l sin θ ˆk J = I ω = I θ ˆk Differentiate wrt t : τ = I θ ˆk Equate I θ = m g l sin θ

11 Compound pendulum continued I θ = m g l sin θ where I = m ( a2 +b 2 12 ) + m l2 Small angle approximation : θ + m g l I θ = 0 SHM with period T = 2π I a T = 2π 2 +b 2 +12l 2 12gl mgl 11

12 Example : solid ball rolling down slope [Energy of ball] = [Rotational KE in CM] + [KE of CM] + [PE] E = 1 2 Iω Mv 2 + Mgy Ball falls a distance h from rest at y = 0 : M g h = 1 2 Iω Mv 2 = 1 2 I ( ) v 2 R Mv 2 Solid sphere: I = 2 5 M R2 Mgh = 1 2 Mv 2 ( ) 10 v = 7 gh Compare with a solid cylinder I = 1 2 M R2 v = The ball gets to the bottom faster! 4 3 gh 12

13 Example : A rod receives an impulse A rectangular rod receives an impulse from a force distance x from its centre of mass. Describe the subsequent motion. 13

14 A rod receives an impulse, continued Impulse ( p = F t) at point x from its centre of mass Force applied to the CM : F = ma Moment of inertia wrt CM : I CM = 1 12 Mb2 Torque (couple) about O = ICM θ Hence ICM θ = Mb 2 12 θ = x F 2 2 Rotational motion θ = 12Fx Mb 2 Acceleration at A due to rotation a rot = b 2 θ Acceleration at A due to translation a trans = F m 14

15 15 3. Lagrangian Mechanics The Lagrangian : L = T U In 1D : Kinetic energy T = 1 2 mẋ 2 No dependence on x Potential energy U = U(x) No dependence on ẋ The Lagrangian in 1D : L = 1 2 mẋ 2 U(x) L ẋ = mẋ and L Differential wrt t : d dt x = U x gives force F ) = mẍ ( L ẋ Hence we get the ) Euler - Lagrange equation for x : d dt ( L ẋ = L x Now generalize : the Lagrangian becomes a function of 2n variables (n is the dimension of the configuration space). Variables are the positions and velocities L(q 1,, q n, q 1,, q n ) ( ) d L dt q k = L q k

16 16 Definitions Generalised coordinates : A set of parameters qk (t) that specifies the system configuration. q k may be a geometrical parameter, x, y, z, a set of angles etc Degrees of Freedom : The number of independent coordinates that is sufficient to describe the configuration of the system uniquely. Constraints : These are imposed when its components are not permitted to move freely in 3-D. Conjugate (generalized) momentum : pk = L q k Following on : E-L equation then reads ṗ k = L q k Cyclic (or ignorable) coordinate qk : If the Lagrangian L does not explicitly depend on q k then in this case L q k = 0 and p k = L q k = constant The momentum conjugate to a cyclic coordinate is a constant of motion

17 Example : simple pendulum Evaluate simple pendulum using Euler-Lagrange equation Single variable qk θ v = l θ T = 1 2 ml2 θ2 U = mgl cos θ L = 1 2 ml2 θ 2 + mgl cos θ ( ) L θ = ml2 θ d L dt = ml θ 2 θ L θ = mgl sin θ E-L ml 2 θ + mgl sin θ = 0 θ + g l sin θ = 0 17

18 18 ( d L dt ( θ d L dt ẋ ) ) Pendulum on a trolley Pendulum s pivot can now move freely in x direction Pivot coordinates : (x, 0) Pendulum coordinates : (x + l sin θ, l cos θ) ( ) 2 T = 1 2 m 1ẋ m d 2 dt (x + l sin θ) + ( ) m d 2 dt ( l cos θ) U = m 2 gl cos θ = L θ ẍ cos θ + θl + g sin θ = 0 L = 1 2 (m 1 + m 2 )ẋ m 2l 2 θ m 2 lẋ cos θ θ + m 2 gl cos θ = L x ẍ(m 1 +m 2 ) m 2 l θ 2 sin θ + θm 2 l cos θ = 0 Small angle approx and solve θ + (m 1 +m 2 ) g m 1 l θ = 0

19 3.1 The Lagrangian in various coordinate systems Cartesian coordinates L = 1 2 m(ẋ 2 + ẏ 2 + ż 2 ) U(x, y, z) Cylindrical coordinates L = 1 2 m(ṙ 2 + r 2 φ2 + ż 2 ) U(r, φ, z) Spherical coordinates L = 1 2 m(ṙ 2 + r 2 θ2 + (r sin θ) 2 φ2 ) U(r, θ, φ) 19

20 3.2 Example : bead on rotating hoop A vertical circular hoop of radius R rotates about a vertical axis at a constant angular velocity ω. A bead of mass m can slide on the hoop without friction. Describe the motion of the bead. Use spherical coordinates : T = 1 2 m (Ṙ2 + R 2 θ 2 + (R sin θ) 2 φ 2 ) But Ṙ = 0, φ = ω = constant T = 1 2 m(r2 θ 2 + (R sin θ) 2 ω 2 ) U = mgr cos θ L = T U L = 1 2 m(r2 θ2 + (R sin θ) 2 ω 2 ) + mgr cos θ One single generalized coordinate : θ 20

21 Bead on rotating hoop, continued L = 1 2 m(r2 θ2 + R 2 sin 2 θ ω 2 ) + mgr cos θ ( ) E-L equation: d L dt θ = L θ ( ) ( ) d L dt θ = d dt m R 2 θ = m R 2 θ L θ = m R2 sin θ cos θ ω 2 mgr sin θ θ = sin θ cos θ ω 2 g R sin θ θ + ( ω 2 0 ω2 cos θ ) sin θ = 0 where ω 2 0 = g R If ω = 0, θ + ω 2 0 sin θ = 0 SHM, back to pendulum formula 21

22 4. The Hamiltonian Conjugate momentum : pk = L q k Define Hamiltonian H = k p k q k L Can show (see HT lectures) dh dt, from E-L ṗ k = L q k = L t If L does not depend explicitly on time, H is a constant of motion Take kinetic energy T = 1 2 m ( ẋ 2 + ẏ 2 + ż 2) L = 1 2 m ( ẋ 2 + ẏ 2 + ż 2) U(x, y, z) H = k p k q k L = m (ẋ.ẋ + ẏ.ẏ + ż.ż) (T U) = 2T (T U) = T + U = E total energy If L does not depend explicitly on time dh dt = 0 energy is a constant of the motion 22

23 4.1 Example: re-visit bead on rotating hoop First take the case of a free (undriven) system L = 1 2 m (R2 θ2 + R 2 sin 2 θ φ 2 ) + mgr cos θ H = k p k q k L ; p k = L q k 23 pθ = L θ = m R2 θ ; pφ = mr 2 sin 2 θ φ H = m R 2 θ 2 + mr 2 sin 2 θ φ 2 L = 1 2 m ( R 2 θ2 + R 2 sin 2 θ φ 2 ) mgr cos θ H = T + U = E L does not depend explicitly on t, H, E conserved Hamiltonian gives the total energy Note: L φ = m(r sin θ)2 θ : which is angular momentum about O. φ is CYCLIC L φ = 0 A.M conserved.

24 Example continued The system is now DRIVEN - hoop rotating at constant angular speed ω L = 1 2 m (R2 θ 2 + R 2 ω 2 sin 2 θ) + mgr cos θ H = k p k q k L ; p k = L q k pθ = L θ = m R2 θ ; a single coordinate θ H = m R 2 θ 2 L (Note p φ = L = 0) φ ) = 1 2 (R m 2 θ 2 R 2 ω 2 sin 2 θ mgr cos θ dh dt = L t E = 1 2 m (R2 θ2 + R 2 ω 2 sin 2 θ) mgr cos θ Hence E = H + m(r 2 ω 2 sin 2 θ) E = T + U H H is a constant of the motion, E is not const. In this case the hoop has been forced to rotate at an angular velocity ω. External energy is being supplied to the system. 24