Lecture Notes for PHY 405 Classical Mechanics From Thorton & Marion's Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary's University Department of Astronomy & Physics November 21, 2004 Problem Set #6 due Thursday December 1 at the start of class late homework will not be accepted text problems 10{8, 10{18, 11{2, 11{7, 11-13, 11{20, 11{31 Final Exam Tuesday December 6 10am-1pm in MM310 (note the room change!) Chapter 11: Dynamics of Rigid Bodies rigid body = a collection of particles whose relative positions are fixed. We will ignore the microscopic thermal vibrations occurring among a `rigid' body's atoms. 1 The inertia tensor Consider a rigid body that is composed of N particles. Give each particle an index α = 1 : : : N. The total mass is M = α mα. This body can be translating as well as rotating. P Place your moving/rotating origin on the body's CoM: Fig. 10-1. 1 the CoM is at R = m r0 M α α α 0 X where r α = R + rα = α's position wrt' fixed origin note that this implies mαrα = 0 α X 0 Particle α has velocity vf,α = dr α=dt relative to the fixed origin, and velocity vr,α = drα=dt in the reference frame that rotates about axis !~ . Then according to Eq. 10.17 (or page 6 of Chap. 10 notes): vf,α = V + vr,α + !~ × rα = α's velocity measured wrt' fixed origin and V = dR=dt = velocity of the moving origin relative to the fixed origin. What is vr,α for the the particles that make up this rigid body? 2 Thus vα = V + !~ × rα after dropping the f subscript 1 1 and T = m v2 = m (V + !~ × r )2 is the particle's KE α 2 α α 2 α α N so T = Tα the system's total KE is α=1 X1 1 = MV2 + m V · (!~ × r ) + m (!~ × r )2 2 α α 2 α α α α X X 1 1 = MV2 + V · !~ × m r + m (!~ × r )2 2 α α 2 α α α ! α X X where M = α mα = the system's total mass. P Recall that α mαrα = 0, so thusP T = Ttrans + Trot 1 where T = MV2 = KE due to system's translation trans 2 1 and T = m (!~ × r )2 = KE due to system's rotation rot 2 α α α X Now focus on Trot, and note that (A × B)2 = A2B2 − (A · B)2 see text page 28 for proof. Thus 1 T = m [!2r2 − (!~ · r )2] rot 2 α α α α X 3 In Cartesian coordinates, !~ = !xx^ + !yy^ + !z^z 3 2 2 2 2 2 so ! = !x + !y + !z ≡ !i i=1 3 X and rα = xα;ix^i i=1 X3 so !~ · rα = !ixα;i i=1 X Thus 3 3 3 3 1 T = m !2 x2 − ! x ! x rot 2 α 2 i α;k i α;i 0 j α;j13 α i=1 ! =1 ! i=1 ! j=1 X X Xk X X 4 @ A5 We can also write 3 !i = !jδi;j j=1 X 1 i = j where δ = i;j 0 i 6= j 3 3 2 so !i = !i !jδi;j = !i!jδi;j j=1 j=1 X X and also note that ai bi = aibi ≡ aibi i ! i ! i j i;j X X X X X 4 3 1 so T = m ! ! δ x2 − ! ! x x rot 2 α 2 i j i;j α;k i j α;i α;j3 α i;j =1 ! i;j X X Xk X 3 1 4 5 = m ! ! δ x2 − x x 2 α i j i;j α;k α;i α;j α i;j " k=1 # X X X3 1 = ! ! m δ x2 − x x 2 i j α i;j α;k α;i α;j i;j α " k=1 # X3 3 X X 1 ≡ I ! ! 2 i;j i j i=1 j=1 X X 3 2 where Ii;j ≡ mα δi;j xα;k − xα;ixα;j α " =1 # X Xk are the 9 elements of a 3 × 3 matrix called the inertial tensor fIg. 2 Note that the Ii;j has units of mass×length . Since xα;1 = xα, xα;2 = yα, xα;3 = zα, 2 2 α mα(yα + zα) − α mαxαyα − α mαxαzα fIg = − m y x m (x2 + z2) − m y z 8 P α α α α α Pα α α Pα α α α 9 − m z x − m z y m (x2 + y2 ) < Pα α α α P α α α α α Pα α α = : P P P ; Be sure to write your matrix elements Ii;j consistently; here, the i index increments down along a column while the j index increments across a row. Also note that Ii;j = Ij;i ) the inertia tensor is symmetric. The diagonal elements Ii;i = the moments of inertia while the off–diagonal elements Ii6=j =the products of inertia 5 For a continuous body, mα ! dm = ρ(r)dV 3 2 dIi;j ! δi;j xk − xixj dm =1 ! Xk = the contribution to Ii;j due to mass element dm = ρdV 3 2 thus Ii;j = dIi;j = ρ(r) δi;j xk − xixj dV V V =1 ! Z Z Xk where the integration proceeds over the system's volume V . Example 11.3 Calculate fIg for a cube of mass M and width b and uniform density ρ = M=b3. Place the origin at one corner of the cube. Fig. 11-3. 6 b b b 2 2 2 2 I1;1 = ρ dx dy dz(x + y + z − x ) Z0 Z0 Z0 b 1 = ρb dy(y2b + b3) 0 3 Z1 1 = ρb( b4 + b4) 3 3 2 2 = ρb5 = Mb2 3 3 It is straightforward to show that the other diagonal elements are I2;2 = I3;3 = I1;1. The off–diagonal elements are similarly identical: b b b I1;2 = −ρ dx dy dz · xy Z0 Z0 Z0 1 1 1 1 = −ρ b2 b2 b = − ρb5 = − Mb2 2 2 4 4 = Ii6=j Thus this cube has an inertia tensor 2=3 −1=4 −1=4 fIg = Mb2 −1=4 2=3 −1=4 8 9 < −1=4 −1=4 2=3 = : ; 7 Principal Axes of Inertia The axes x^i that diagonalizes a body's inertia tensor fIg is called the principal axes of inertia. In this case Ii;j = Iiδi;j 3 3 1 so T = I ! ! = rotational KE rot 2 i;j i j i=1 j=1 X3 X 1 = I !2 2 i i i=1 X where the three Ii are the body's principal moments of inertia. If this body is also rotating about one of its principal axis, say, the ^z, then !~ = !^z and 1 T = I !2 rot 2 z which is the familiar formula for the rotational KE in terms of inertia Iz found in the elementary textbooks. 1 2 )those formulae with Trot = 2I! are only valid when the rotation axis !~ lies along one of the body's three principal axes. The procedure for determining the orientation of a body's principal axes of inertia is given in Section 11.5. 8 Angular Momentum The system's total angular momentum relative to the moving/rotating origin Orot is L = mαrα × vα α X where rα and vα are α'a position and velocity relative to Orot. Recall that vα = !~ × rα for this rigid body (Eqn. 10.17), so L = mαrα × (!~ × rα) α X Chapter 1 and problem 1-22 show that A × (B × A) = A2B − (A · B)A 2 so L = mα[rα!~ − (rα · !~ )rα] α X and the ith component of L is 2 Li = mα[rα!i − (rα · !~ )xα;i] α X And again use 3 3 3 2 2 rα = xα;k rα · !~ = xα;j!j !i = !jδi;j =1 j=1 j=1 Xk X X 9 So 3 3 3 L = m x2 ! δ − x ! x i α 0 α;k j i;j α;j j α;i1 α =1 j=1 j=1 X Xk X X 3 @ 3 A 2 = !j mα δi;j xα;k − xα;ixα;j j=1 α k=1 ! X3 X X = Iij!j j=1 X where the Ii;j are the elements of the inertial tensor fIg. Note that 3 3 3 L = Lix^i = Iij!jx^i i=1 i=1 j=1 X X X = fIg · !~ Lets confirm this last step by explicitly doing the matrix multiplication: L1 I11 I12 I13 !1 L = L2 = fIg · !~ = I21 I22 I23 !2 0 1 0 1 0 1 L3 I31 I32 I33 !3 @ A @ A @ A I11!1 + I12!2 + I13!3 j I1j!j = I ! + I ! + I ! = I2 ! 0 21 1 22 2 23 3 1 0 Pj j j 1 I31!1 + I32!2 + I33!3 j I3j!j B P C @ A @ A which confirms L = fIg · !~ . P Evidently the inertia tensor relates the system's angular momentum L to its spin axis !~ . 10 Now examine 3 3 3 1 1 1 ! L = !~ · L = I ! ! 2 i i 2 2 ij i j i=1 i=1 j=1 X X X = Trot = the body's KE due to its rotation 1 1 ) T = !~ · L = !~ · fIg · !~ rot 2 2 Note that the RHS is a scalar (as is should) since fIg · !~ is a vector, and a vector·vector = scalar.