PHYS 705: Classical Mechanics. Introduction and Derivative of Moment of Inertia Tensor

Transcription

1 1 PHYS 705: Classical Mechanics Introduction and Derivative of Moment of Inertia Tensor

2 L and N depends on the Choice of Origin Consider a particle moving in a circle with constant v. If we pick the origin O to e at the center of the circle and calculate L wrt this origin, L is a fixed quantity. L O r v m L rmv In order for the particle to actually maintain this circular motion, there must e a centripetal force acting on it. O v F c m But, notice that this does not give rise to any torque since N rf F c (Note: r and are collinear) c 0 and LN 0

3 3 L and N depends on the Choice of Origin Now, consider the same situation ut with the origin O chosen to e elow the plane of rotation. O r v m (into page) And, there must e a torque acting on it. L rmv points as shown Most importantly, it is no longer a dl dt N constant, i.e, 0 N L O Fc r m N rfc 0 (out of page) (L changes in direction as given y N) L and N depends on the choice of the origin

4 4 Non-Collinear Relation etween L and When a rigid oject spins aout one of its axes of symmetry, the dynamical equations are simple: Trot 1 I L Iω where - the moment of inertia I is simply a scalar - L and point in the SAME direction ω nˆ However, this is not generally the case when a rigid ody is NOT rotating aout one of its Principle Axes of rotation.

5 5 Non-Collinear Relation etween L and Let consider the following dumells rotating aout some axis which is not an axis of symmetry. L 1 L ω v 1 m 1 r 1 - Here, the total angular momentum is, L m L L L 1 r v m r v Notice that L and do NOT point in the same direction. m v r - In fact, L is changing in time so that there must e a torque: L N expect an oject to rotate (In general, one just can t d dt -The rotational KE doesn t work out to e quite either. T rot aout a particular chosen axis in space w/o N.) 1 I

6 6 Derivation of L and T for a Rigid Body At any given instant of a rigid ody motion, we can consider the ody to e performing an infinitesimal rotation aout some fixed point O on the rigid ody Let say that in the fixed frame, O is moving with velocity v0 Now, let consider another point in the ody ω r O ody The point is located in the ody frame y r and its velocity in the fixed frame is given y, fixed The origin of the ody v axis is moving with 0 The ody frame is also rotating aout ω v f v0 ωr (r is fixed in the ody frame rigid ody) (note: we will also drop for ody frame variales from now on)

7 7 Derivation of T for a Rigid Body A particle (lael it y ) has T 1 m v 1 T m 0 0 v ωr v ωr 1 1 T m 0 m 0 m v v ωr ωr f v ω r 0 m v f v0 ωr Now, summing over all particles, we have Ttotal 1 m m m v v ω r ω r M MRcm

8 8 Derivation of T for a Rigid Body Now, let s pick the origin of the ody frame to e at the CM so that R cm 0 Then, we have the splitting of the total kinetic energy 1 1 T Mv m ωr T T total 0 CM rot T CM KE of total M as if all at CM KE aout the CM (the interesting part) T rot Writing out the cross product: ωr ωr ωr r ωr Recall ABCDACBDADBC

9 9 Derivation of T for a Rigid Body Writing out the components for explicitly, we have Inserting r 1 T m x x x x rot i i k k i i j j i j ij Defining the Moment of Inertia Tensor y x, x, x 1 T m x x x x 1 3 rot i j ij k k i j i j 1 Trot i m ij xk xk xi xj j T 1 I rot i ij j T (sum rule) I m x x x x ij ij k k i j

10 10 Derivation of T for a Rigid Body Notes: 1. Summary: I ij is defined relative to some origin O and a particular choice of axes through the origin. where the position of the th particle is measured relative to O at r I m x x x x ij ij k k i j r, r, r 1 3 T

11 11 Derivation of T for a Rigid Body I ij T rot. One could, in general, calculate for some other origin O which might not e the CM of the rigid ody and can still e given as But, if O is NOT at CM, then the total KE cannot e separated into: T 1 I rot i ij j T T T total CM rot (The Mv ωr terms in full expression T total 0 CM for might not e zero.) If O for the ody axes is fixed in the fixed frame, then and Ttotal 1 m v0 0 ω m r 1 T T I total rot i ij j v There are two natural choices for O: CM and a fixed point of rotation. v 1 m ω r

12 1 Derivation of T for a Rigid Body 3. There is a generalization of the Parallel Axis Theorem that allows one to relate the Momentum of Inertia Tensor wrt the CM to one relative to another parallelly translated origin. I J M a aa ij ij ij i j where I ij is moment of inertia tensor aout the CM of the rigid ody J ij a is moment of inertia tensor aout some other point O a, a, a 1 3 T is a vector pointing from O to CM O a CM NOTE: the two sets of axes must e PARALLEL

13 13 Derivation of L for a Rigid Body Now, we look at the general expression for the angular momentum of a rigid ody r O ody - Here, we again assume that the rigid ody is rotating at a constant angular velocity with ω ω one point stationary calling this O. fixed v f - The origin of the ody frame is assumed to e fixed, i.e. v0 0 - Then, the velocity vector as measured in the fixed frame is: v (note: with, contriution for the velocity vector comes from the 0 0 v f ωr rotation of the ody axes only [the explicit suscript f will e suppressed].) - Then, the total ang momentum aout O as measured in the fixed frame is: m m Lr v r ωr (sum)

14 14 Derivation of L for a Rigid Body - Using the vector identity: -We have: ABCBACCAB Lr m ωr m ωr r r ω I m x x x x ij ij k k i j r (sum) -Again, writing out the components for explicitly, we have x, x, x 1 3 Li m i rk rk ri rj j (sum) - Inserting i j ij Li m j rk rk ij ri rj m r j k rk ij ri r j j - Recognizing the pre-factor as the momentum of inertia tensor, L T I i ij j

15 15 T and L for a Rigid Body Summary: T L I m x x x x ij ij k k i j Notes: 1 I rot i ij j I i ij j T ( is a scalar and the sum is over oth i and j) - The moment of inertia I is no longer a scalar as in the simpler case: rot ( L is a vector and the sum is only over j) ( I ij is a tensor and the sum is only over ) L Iω - As we have seen in our earlier example, L and are not necessary collinear! - However, we can still formally write, 1 1 Trot ωl ωiω

16 16 Moment of Inertia Tensor We will see now that our old familiar scalar I is still in the equation. ˆn 1 Trot iiij i ni j Trot ni i ijnj Trot n i m ij xk xk xi xj nj m xk xk nn i jij nx i i xjn j 1 T ˆ rot m r r n (still sum over ) Let pick to align with the axis of rotation, then we can write Sustituting into, we have ω nˆ

17 17 Moment of Inertia Tensor T ˆ rot m r r n Now, consider the picture to the right, d Notice that is the distance to the O ˆn r d r position with respect to the chosen axis of rotation ˆn and r r n d ˆ So that, we can write, Trot m d (The lue term is our usual definition for the scalar moment of inertia)

18 18 Moment of Inertia Tensor Comments: 1. The projection is the Moment of Inertia aout the axis of rotation going through the origin O. The tensor itself has in it information for ALL I s in any direction. ni n nˆinˆ I ij i ij j 3. A particular representation of depends on the choice of origin and axes chosen in calculating it. ˆn O ˆn d r 4. (later) There will also e a generalization of the parallel axis thm: I I M a aa CM ij ij ij i j I ij ω a M CM

19 19 An Example: Moment of Inertia for a Cue x 3 O x x 1 A cue with uniform density, mass M, and side is placed as show with its corner located at the origin O The finite sum for is now an integral, I ij I r x x dv ij ij i j V As an exercise, let evaluate the components explicitly, I x x x x dxdx dx x x3dxdx 3 dx

20 0 An Example: Moment of Inertia for a Cue I x x dx dx dx xdxdx3 x3dx3dxdx M dx dx dx Sustituting, we then have, 3 Since the cue is symmetric, similar calculations will result for 33 Iii 3 M 0 I 11 3 M x dx1 3 x 3 5 O I, I x

21 1 An Example: Moment of Inertia for a Cue Now, consider off-diagonal terms, I x x dv ij i j V i j xixdxdx j i j dxk 00 0 x dx x dx dx 000 i i j j k (indices don t sum here) x 1 x 3 O x x j dx j dx k dxk 4 0 Iij 1 4 M i j

22 An Example: Moment of Inertia for a Cue Summarizing, we have, I ij M x 1 x 3 O x

23 3 Principal Axes & Principal Moments of Inertia I ij Notice that since is always a real symmetric matrix, one can always diagonalize it. I ij M diagonalize M (shown later) In the language of linear algera, this corresponds to rotating our system (with the same origin) to a set of orthogonal asis corresponding to the eigenvectors of I ij. These axes are called the Principal Axes. And, the diagonal elements will e the eigenvalues of I ij. They are the Principal Moment of Inertia.

24 4 Principal Axes & Principal Moments of Inertia I ij - If is NOT diagonalized, we will in general have L and not collinear and the Kinetic energy as a tensor product, i.e., I ij L I i ij j and T 1 I rot i ij j - If is diagonalized, then the situation simplifies and we get our regular T rot simpler results for L and : L1 I111, L I, L3 I and Trot Iiii - The physical significance is that when there is no torque acting on the system, a rigid ody when tossed into space will spin around one of these principal axes. It can e shown that rotation aout the principal axes with the largest and the smallest principal moments will e stale while rotation aout the middle one will not. (HW pro)

25 5 Principal Axes & Principal Moments Now, ack to our example for a cue with its corner at the origin, we had I ij M x 1 x 3 O x The characteristic polynomial (without the common pre-factor) is: det Solving for gives: 11,11,

26 6 Principal Axes & Principal Moments e The eigenvectors can then e evaluated y solving, 8 3 3e1 e e e e 3 e 3 For the solved eigenvalues, we have: e 1,1,1 1 1 e, e, e 1 3 In this asis, the Moment of Inertia tensor is diagonalized: I ij M e 1,1,0,3 e3 1, 0,1 (degenerate eigenspace)

27 7 Principal Axes & Principal Moments -The eigenvector (Principle Axis) corresponding to is shown: 1 - The other two eigenvectors can e e 1 1,1,1 x 3 e 1 e 1 any vectors in the plane to. x -Note: S e e e 1. with the as 1 3 e 's x 1 columns is a similarity transform which diagonizes, I ij M 1 M I diag I S IS

28 8 Principal Axes & Principal Moments I ij. For a given origin O, contains ALL the moments of inertia through ANY axes passing through O and for a given axis with unit vector I ˆ ˆ ˆ nin ni n n i ij j 3. Among all these possiilities, there are three Principle Axes given y the eigenvectors of I ij and their associated Principle Moments of Inertia. ˆn I ij 4. One can choose to represent using these Principle Axes as the coordinate asis and the resulting Moment of Inertia tensor will e diagonalized. I ij 5. Otherwise, will in general have non-zero off-diagonal elements ut all of them with the same origin are related y a similarity transformation.

29 9 Principal Axes & Principal Moments 6. The Principal Axes are special in the sense that: - if the RB is spinning along one of the Principal Axis, then i ii i - (later) it can e shown that those are the preferred directions for the oject to spin around when there is no external torque! L (no sum) I

30 30 Parallel Axis Theorem for Inertia Tensor I ij So, we know that all with the same origin are related y a rotation. Now, how are they related if they are calculated using different origin? Q x 1 x 3 x 3 r a CM I ij r ody x 1 x x J ij a is moment of inertia tensor aout the CM of the rigid is moment of inertia tensor aout Q a, a, a 1 3 T is vector pointing from Q to CM NOTE: the two sets of axes must e PARALLEL Let calculate the inertia tensor in the Q frame J m x x x x ij ij k k i j (sum)

31 31 Parallel Axis Theorem for Inertia Tensor Q x 1 x 3 x 3 r a x 1 r CM x x r a r - Note, we have, or J ij - Plugging it in our expression for, J m x x x x ij ij k k i j x a x i i i J m a x a x a x a x ij ij k k k k i i j j aa k k x xk xk ai j a x ajx i xi xj m a a ij k k i j m ijxk xk xi xj m ijakak aa i j m ijax k k ax i j ax j i

32 3 Parallel Axis Theorem for Inertia Tensor x 3 x 3 x i - Now, since all the are evaluated with the origin at CM, Q x 1 r a x 1 r CM x x ij we have, mx k J I M a a aa m - We also have eing the total mass, J ij a - So the ig complicated equation reduces simply to, ij ij k k i j 0 M m x x x x m a a aa ij k k i j ij k k i ij k m xk ai m x j aj m x i j wrt Q wrt CM

33 33 Parallel Axis Theorem for Inertia Tensor - Example: Cue again Now, we would like to I ij calculate with origin at CM, - From geometry, the displacement from the corner to the Q a CM CM is T 3 4 a,, and a - Also, recall that we have the inertia tensor wrt to Q (corner) as J ij M

34 34 Parallel Axis Theorem for Inertia Tensor - Applying the Parallel Axis Theorem: Diagonal elements: J I M a a aa ij ij ij k k i j Q a CM 8M 3 Iii Jii M a aa i i M M M M 1 6 Off-diagonal elements: J ij M M Iij Jij M aa i j M 1 4 3M M 0 1 4

35 35 Parallel Axis Theorem for Inertia Tensor - Putting everything together, we have: I ij M Q a CM Notes: I ij 1. Since is asically proportional to the identity matrix, it will e the same y ANY similarity transformation (any rotation aout CM). M M I S I S S S I ij ' ik kl lj ik kl lj ij ij. ANY set of axes through the CM will give the same for this cue. 3. Basically, wrt its CM, the cue looks like a sphere. I ij