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 be at the center of the circle and calculate L wrt this origin, L is a fixed quantity. L v m O r L r m v In order for the particle to actually maintain this circular motion, there must be a centripetal force acting on it. v But, notice that this does not give rise m O to any torque since Fc N rFc 0 and L N 0 (Note: r and F c are collinear) 3 L and N depends on the Choice of Origin Now, consider the same situation but with the origin O chosen to be below the plane of rotation. L r m v points as shown v (into page) m Most importantly, it is no longer a dL r constant, i.e, N 0 O dt And, there must be a torque acting on it. Fc N rFc 0 (out of page) N m (L changes in direction as given by N) r L O L and N depends on the choice of the origin 4 Non-Collinear Relation between L and w When a rigid object spins about one of its axes of symmetry, the dynamical equations are simple: 1 T Iw 2 L Iω rot 2 ω w nˆ where - the moment of inertia I is simply a scalar - L and w point in the SAME direction However, this is not generally the case when a rigid body is NOT rotating about one of its Principle Axes of rotation. 5 Non-Collinear Relation between L and w Let consider the following dumbbells rotating about some axis which is not an axis of symmetry. v1 - Here, the total angular momentum is, m1 L L1 L2 ω L1 r L m1r 1 v 1 m2r 2 v 2 L 1 2 - Notice that L and w do NOT point in the same direction. r - In fact, L is changing in time so that there 2 must be a torque: m (In general, one just can’t 2 v dL expect an object to rotate 2 N about a particular chosen dt axis in space w/o N.) 1 -The rotational KE doesn’t work out to be quite T I w 2 either. rot 2 6 Derivation of L and T for a Rigid Body At any given instant of a rigid body motion, we can consider the body to be performing an infinitesimal rotation about some fixed point O on the rigid body Let say that in the “fixed” frame, O is moving with velocity v0 Now, let consider another point in the body… r body The point is located in the “body” frame by r ω O and its velocity in the “fixed” frame is given by, fixed The origin of the “body” vf v0 ω r axis is moving with v0 (r is fixed in the “body” frame – rigid body) The “body” frame is also rotating about ω (note: we will also drop ‘ for body frame variables from now on) 7 Derivation of T for a Rigid Body 1 2 A particle (label it by a) has T mv vf v0 ω r a 2 f 1 T mv ωr v ωr a 2 0 0 1 1 2 Tmmv2 v ωr m ωr a 20 0 2 v0 ω m r Now, summing over all particles, we have 12 1 2 Ttotal ma v0 v 0 ω mar a ma ωr a 2a a 2 a M MRcm 8 Derivation of T for a Rigid Body Now, let’s pick the origin of the “body” frame to be at the CM so that Rcm 0 Then, we have the splitting of the total kinetic energy 12 1 2 TMtotalv0 ma ωr a TT CM rot 2 2 a KE of total M KE about the CM (the TCM T as if all at CM interesting part) rot Writing out the cross product: 22 2 2 ωra ωrωr a aaa w r ωr Recall AB CD ACBD ADBC 9 Derivation of T for a Rigid Body a a a T Writing out the components for r a x 1 ,, x 2 x 3 explicitly, we have 1 aa a a Tmrota ww ii xx kk ww ii xx jj (sum rule) 2 Inserting wi w jij 1 T mww xxaa ww xx aa rot2 a ijijkk ijij 1 aa aa Trotw i ma ijkk xxxx ij w j 2 aa aa Defining the Moment of Inertia Tensor by Iij ma ijkk xxxx ij a 1 T w I w rot2 iij j 10 Derivation of T for a Rigid Body Notes: 1. Summary: Iij is defined relative to some origin O and a particular choice of axes through the origin. aa aa Iij ma ijkk xxxx ij a where the position of the ath particle is measured relative to O at a a a T ra r1,, r 2 r 3 11 Derivation of T for a Rigid Body 2. One could, in general, calculate I ij for some other origin O which might 1 not be the CM of the rigid body and T can still be given as T w I w rot rot2 iij j But, if O is NOT at CM, then the total KE cannot be separated into: (The M v 0 ω R CM terms in full expression Ttotal T CM T rot for T total might not be zero.) 2+. If O for the body axes is fixed in the fixed frame, then v 0 0 and 1 2 1 2 Ttotal ma v0 v0 ω ma r a maω r a 2 a a 2 a 1 TT w I w total rot2 iij j There are two natural choices for O: CM and a fixed point of rotation. 12 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. 2 Iij J ij M a ij a ij a where is moment of inertia tensor about the CM of the rigid body Iij Jij is moment of inertia tensor about some other point O T a a1,, a 2 a 3 is a vector pointing from O to CM a CM O NOTE: the two sets of axes must be PARALLEL 13 Derivation of L for a Rigid Body Now, we look at the general expression for the angular momentum of a rigid body… - Here, we again assume that the rigid body is O body r rotating at a constant angular velocity ω with ω one point stationary calling this O. - The origin of the “body” frame is assumed to fixed be fixed, i.e. v0 0 - Then, the velocity vector v f as measured in the “fixed” frame is: v f ω r (note: with v 0 0 , contribution for the velocity vector comes from the rotation of the body axes only [the explicit subscript f will be suppressed].) - Then, the total ang momentum about O as measured in the “fixed” frame is: Lram aa v r aa m ω r a (sum) 14 Derivation of L for a Rigid Body - Using the vector identity: A BC BAC CAB - We have: 2 L raa mωr a mr aaaa ω rrω (sum) a a a T -Again, writing out the components for r a x 1 ,, x 2 x 3 explicitly, we have Lmw rraa rr a a w ia ikk ijj (sum) - Inserting wi w jij Lmw rraa rr aaw mrraa rr aa w ia j kk ij ijj a kkij ij j - Recognizing the pre-factor as the momentum of inertia tensor, I m xxxxaa aa ij a ijkk ij Li I ijw j a 15 T and L for a Rigid Body Summary: 1 T w I w ( T is a scalar and the sum is over both i and j) rot2 iij j rot ( L is a vector and the sum is only over j) Li I ijw j aa aa ( is a tensor and the sum is only over a) Iij ma ijkk xxxx ij Iij a Notes: - The moment of inertia I is no longer a scalar as in the simpler case: L Iω - As we have seen in our earlier example, L and w are not necessary collinear ! - However, we can still formally write, 1 1 T ωL ωIω rot 2 2 16 Moment of Inertia Tensor We will see now that our old familiar scalar I is still in the equation. Let pick nˆ to align with the axis of rotation, then we can write ω w nˆ 1 Substituting w w n into T w I w , we have i i rot2 iij j w 2 T n I n rot2 iijj w 2 aa aa Trot n i ma ijkk x x x ij x n j 2 2 w aa aa ma xkk x n ijij n n ii x x jj n 2 1 2 w 2 T mr 2 r nˆ (still sum over a) rot 2 aa a 17 Moment of Inertia Tensor 2 w 2 T mr 2 r nˆ rot 2 aa a da Now, consider the picture to the right, nˆ O ra Notice that d a is the distance to the position r a with respect to the chosen axis of rotation nˆ and 2 2 2 ra d a r a nˆ So, we can write, 2 w 2 Trot mad a (The blue term is our usual definition 2 a for the scalar moment of inertia) 18 Moment of Inertia Tensor Comments: da nˆ 1.