Lecture 37: Principal Axes, Translations, and Eulerian Angles

Transcription

1 Lecture 37: Principal Axes, Translations, and Eulerian Angles

2 When Can We Find Principal Axes? We can always write down the cubic equation that one must solve to determine the principal moments But if we want to interpret these as physically meaningful quantities, the roots of that cubic have to be real Recall that in general, cubics can have two complex roots Fortunately, we re not in the general case here The inertia tensor is both real and symmetric in particular, it satisfies: * I = I ij Matrices that satisfy this restriction are called Hermitian For such matrices, the principal moments can always be found, and they are always real (see proof in text ij This mathematics will come up again in Quantum Mechanics Principal Moments Eigenvalues Principal Axes Eigenfunctions

3 Inertia Tensor Under Translation We ve already seen how the inertia tensor transforms under a rotation of the coordinate system Now we ll see how it behaves under translation that is, eeping the direction of the the coordinate axes fixed, but changing the origin In particular, we ll compare I for a coordinate system at the center of mass of an object with that for a different origin: x 3 Q X 3 r CM O x O is at the center of mass; Q can be anywhere X x X

4 In the X reference frame (the one with origin at Q, the position of any point in the rigid body can be written as: R = + rcm r which means the inertia tensor becomes: ( δ,,, I = m X X X ij ij i j Position in center-ofmass frame ( δ ( ( ( ij, CM,, i CM, i, j CM, j = m x + x x + x x + x ( (,,,, = m δ x + x x + x = ij CM CM x x x x x x x x, i, j CM, i, j, i CM, j CM, i CM, j ( δ ij,, i, j m x x x ( ( δ ij CM,, CM, x CM, x i, x j, x i CM, x j CM, x i CM, j + m x x + x Inertia tensor about the center of mass

5 So the inertia tensor about Q is: ( δ I = I + x x x m ij CM, ij ij CM, CM, i CM, j ( δ ij CM,, CM, i, j, i CM, j + m x x x x x x Each term in the second line has a factor that loos lie: mx For a general reference frame, this quantity equals the mass of the system times the position of the ith component of the center of mass But the x i and in the center of mass frame, so all these terms are zero Therefore we have the result (Steiner s Parallel Axis Theorem:,i ( δ I = I + M x x x ij CM, ij ij CM, CM, i CM, j

6 Eulerian Angles In analyzing the motion of a rigid body, it is most convenient to use a reference frame that is fixed to the body For example, a symmetry axis of the body could be the z axis in this body frame But that s usually not an inertial frame So we need to define a translation between some inertial fixed frame and the body frame We learned way bac in the first wee of class that three angles are needed to specify how one coordinate system is rotated with respect to another (in three dimensions These angles determine the elements of the rotation matrix λ

7 There are many ways the three angles can be specified One convenient choice is the Eulerian Angles, in which the 3D rotation is the result of three D rotations: = 3 3 φ. Rotate CCW by φ around 3 axis. Rotate CCW by θ around axis 3. Rotate CCW by ψ around axis 3 = θ x = (,, 3 (,, 3 (,, 3 ( x, x, x3 = x x x ψ Line of nodes

8 Note that in each rotation, one axis (from the previous rotation is held fixed. That means the rotation matrices have the form: λ φ cosφ sinφ = sinφ cosφ 0 ; λ 0 cosθ sinθ θ = sinθ cosθ cosψ sinψ 0 λ = sinψ cosψ 0 ψ 0 0 The complete rotation is given by: x = λ ψλθλφx λx i.e., the elements of λ are obtained by multiplying the three individual rotation matrices (see Equation.99 in the text for their values

9 Angular Velocity in the Body Frame It s going to be easiest to find the equations of motion for the object in the body frame In general, we can write: ω= φ+ θ + ψ To write the components of this vector in the body frame we need to do some geometry: ψ x 3 3 θ φ φ x ψ is along the x 3 direction: ψ ψ e = 3 x θ is in the x -x plane: ψ θ = θ cosψ e θ sinψ e φ has components in all three θ directions: φ = φ sinθ sinψe + φsinθ cosψe + φcosθ e 3

10 Adding up these vectors, we find that the components of ω in the body frame are: ω = φ sinθ sinψ + θ cosψ ω = φ sinθ cosψ θ sinψ ω = φ cosθ + ψ 3

11 Euler s Equations for Rigid Body Motion Now that we ve defined the geometry, we can set up the Lagrangian for rigid body motion To mae things easy, we ll first consider the case where no external forces act on the body, so: L= T In that case, we can put the origin of both the fixed (meaning inertial and body frames at the center of mass Furthermore, we can cleverly choose the body frame to coincide with the principal axes of the body, so that: T = Iiωi i Principal moment of inertia associated with the ith acex

12 Since the Eulerian angles fully specify the orientation of the rigid body in the inertial frame, we can tae them to be our generalized coordinates The equation of motion for ψ is: T d T = 0 ψ dt ψ which becomes (using the chain rule: T ω d T ω = ω ψ ω ψ i i i i dt i i T One piece that enters everywhere is, which is simply: ωi T = Iiωi ω i 0

13 The other relations we need are: ω = φ sinθ cosψ θ sinψ = ω ψ ω ψ ω3 ψ = φ sinθ sinψ θ cosψ = ω = 0 ω = 0 ψ ω = 0 ψ ω3 = ψ Putting these bac into the Lagrange equation gives: d Iωω + Iω( ω I3ω3 = dt I I ωω I ω = 0 ( 3 3 0

14 The choice of which axis to call x 3 was completely arbitrary That means we can find similar relations for the other components of the angular velocity ( ( ( I I ωω I ω = 3 3 I I ωω I ω = 3 3 I I ωω I ω = 3 3 These imply that the motion of any rigid object depends only on the three numbers I, I, and I 3 and not on any other details of the shape of the object We can define an equivalent ellipsoid for any object Euler s Equations for force-free motion