3 d Rotations Rotating dumbbells in lab frame Moment of Inertial Tensor

Transcription

1 d Rotatons Rotatng dumells n la frame Moment of Inertal Tensor Revew of BCS and FCS sstems Component notaton for I β Moments and Products of Inerta I x, I, I z P x, P xz, P z Moment of Inerta for a cue Cue centered on dagonal L not parallel to ω Requred torque for rotaton Prncpal axes Axes where ω s parallel to L Egenvector equaton for PA Rotatng dumells Moment of nterta tensor L n BCS coordnates L vewed n FCS Torque of the dumells Torque computed n BCS Precesson of L n a space cone Cone precesson of L requres torque Smple wa vew of dumell torque Centrfugal force descrpton Axle forces I for a rectangular plate Reducng 1 Product of nerta I z I x +I z PH6_L11 1

2 We frst encounter what happens when the angular momentum s angled wth respect to the angular veloct. As a result, L processes n a (space) cone aout ω, and a torque s requred to keep an oject unforml rotatng wth a constant ω. Two frames are used to descre the moton: a BCS frame whch s fxed n the rotatng od and a FCS frame whch s fxed n space. In the BCS, L I ω where I s tensor (a matrx). The dagonal components of ths matrx are moments of nerta, and the off dagonal elements are products of nerta. We show how to compute the I tensor for a unform cue wth an orgn on one corner. We show that there are specal rotaton axes known as Prncpal Axes or PA. When the oject s rotated aout a PA, L s parallel to ω and no torque s requred for a unform rotaton. One fnds the PA solvng an egenvector equaton smlar to that used n normal mode prolems. We guess from smmetr one of the PA for the corner centered cue. We next llustrate man of the formal ponts wth a smple sstem consstng of a dumell rotatng at an angle wth respect to ts central axs. We frst compute the angular momentum and torque n the BCS. We next dscuss the moton of the angular momentum n the FCS. Fnall we gve a smple ntutve pcture of wh a rotatng dumell requres a torque. We next compute the moment of nerta tensor for a rectangular plate and llustrate the formal wa of fndng the PA from the egenvector equatons solved usng the secular equaton. We prove that two non degenerate PA (.e. PA wth unequal egenvalues ) are perpendcular. PH6_L11

3 m ω m 1 La frame moton of a rotatng dumell ˆk ĵ î We vew moton n the la. At ths nstant frst mass at r cos ˆ + snkˆ At later tmes the frst mass appears at x1 () t cosωt snωt 0 cos () t snωt cosωt z1() t 0 0 1sn x coscosωt x ωcossnωt 1 1 cossnωt ωcoscosωt 1 1 z1 sn z 1 0 Hence ntall: v (0) cos ˆ ω j and v (0) ωcos j 1 At t0 oth angular momenta are gven ˆ ˆj kˆ L r mv cos 0 sn 0 mωcos 0 L1 L m ω L L1+ L m sn cos,0,cos The L vector s n the plane of the dumell ( sn cos,0,cos ) or ω( ) and rotates along wth t. In the la ts components are gven at later tmes : Lx cosωt snωt 0 m ωsn L snωt cosω 0 0 L z 0 0 1m ω cos m ω ωt Lx sn cos m ω ωt L sn sn PH6_L11 ˆ Lz m ω cos

4 We can take the dervatves of L to fnd the torques n the la frame ω Torques n oth frames dcosωt τx m ωsn m ω sn snωt dt dsnωt τ m ωsn m ω sn cos ωt dt dm ωcos τ z 0 dt We can wrte τ components n the dumell frame usng the nverse rotaton matrx τ ' x cosωt snωt 0 snωt τ ' m ω sn snωt cosω 0 cosωt τ ' z ˆk î Multplng ths out shows a constant torque of τ ω sn ˆj 'n the dumell frame. ĵ m τ L ω ˆ ' k ˆ' PH6_L11 4 ˆ ' j The aove fgure shows that dumell, the angular momenta, and the torque rotate together wth the angular momentum vector formng a cone. We wll call ths the od cone The torque les tangent to the ase of the cone. But there must e an easer wa of gettng these results! It s much easer usng a concept of the

5 Recall that two coordnate sstems are used: FCS coordnates whch are wth respect to a fxed pont n an nertal coordnate sstem and BCS or od centered coordnate sstem whch rotates wth the rgd od wth an angular veloct ω. The tradtonal thng s to wrte the angular momentum L wth respect to the fxed sstem, ut wrte all vector components n the BCS. Moment of Inertal Tensor Thus L mr where unless ( ω r) specfed otherwse, all components are n BCS coordnates.usng the BAC - CAB L mω( rr) mr ( rω) We wsh to factor out ω and use component notaton wth mpled summaton: () () () () L mω ( rγ rγ ) mr rβ ωβ () () () () m ωδ β β ( rγ rγ ) mr rβ ωβ () () () ( ) m ( δ r r r r ) ω I where PH6_L11 5 β γ γ β β β () () ( ) () ( δ γ γ r ) I m r r r β β β ω I ecomes a matrx (or nd rank tensor) that multples the ω vector to gve the L vector rather than a smple scalar such as mass. Hence L wll not le parallel to ω whch creates nterestng new complcatons as we wll see. β

6 Moments and Products of 11 () () () () ( δ ) I m r r r r β β γ γ β Frst consder the dagonal elements or I, I, I () () () () x ( δ γ γ ) ( + + ) ( + ) ( + ) z ( ) I I m r r r r and m x z x x m z I I m x z I I m x + These are just the ordnar I aout the x,,z axes On to the off-dagonal elements where δ 0 () () ( ) Thus I m r r There are β β three "products of nerta. I P mx ; I P mx z 1 x 1 xz z z I P m β Inerta I P P I P I P Pxz Pz Iz x x xz x z xz z x + I m x x + z z z PH6_L z x xz Lets compute the moment of nerta tensor of a unform denst cue of dmenson aout a corner. We need to compute moments of nerta and products of nerta. x

7 x z I ρ dx d dz + z ρ Moment of Inerta for a cue x ( ) dx d dz z + ρ dx dz d z ρ + ρ 0 0 m ρ Snce m ρ, Ix m B smmetr Ix I Iz x Px ρ dz dx x d ρ ρ 4 4 Ix Px Pxz I Px I Pz Pxz Pz I z / 1/4 1/4 8 m I m 1/4 / 1/ /4 1/4 / 8 Around most axes through the corner L s not to ω. For example ω ω 0 0 ( ) m ω m ω L d d Applng + ω [ ] to L BCS dt FCS dt BCS τ L I ω + ω L, we see that a torque s FCS requred to rotate the cue aout (1, 0,0) axs even for unform rotaton ( ω 0). In BCS coords: ˆ ˆ ˆ ' j' k' 5 m ω mω m τ ( 0 ) P smmetr PH6_L11 7 x Px Pz Pxz 1 1 8

8 However there are some axes, called prncpal axes,where L ω. For the cue, one prncpal axs s the dagonal or ω ω ( ) Prncpal axes m ω m ω L We see that ω ω( ) acts as an egenvector of I n the sense Iω λω. The egenvalue λ s the moment of nterta aout the prncpal axs (or I ). A unform rotaton ( ω) aout a prncpal axs requres no external torque snce τ ω L 0 f ω L and ω 0. Because such rotatons requre no torque these are often preferred axes for rotaton to mnmze earng wear. We can reccle our egenvector technques learned for normal modes to fnd the prncpal axes. For example, the prncpal axes are. As a second example, we wll work out I for the case where the cue s centered on the orgn of / / / the I BCS ρ dx d dz + z x ρ ρ / / / / / / / / / / / / PH6_L11 8 z ( ) / / / / / z ρ + ρ 4 m ρ I Iz 8 6 x dx d dz z dx dz d / / / / / / / P ρ dz dx x + / d 0 P P m m I Snce I, all 6 ω ω axes are prncpal axes for the centered cue! z x xz

9 z ' Rotatng dumells n PA sstem ' x ' I m These are two pont masses a dstance apart. Snce there s onl separaton of the masses along the x axs there are no products of nerta. The two moments of nerta are m z ' ω x ' cos ' We now rotate the dumell at an angle wth wth respect to x. Lets compute L n the BCS frame ω sn L Iω m ω cos L m ω kˆ z ' ω x ' v ωcos ˆj' r ˆ' L mr v m ˆ' ω cos ˆj ' m ω kˆ cos ' We also easl vew ths sstem n the FCS ut wth BCS unt vectors The upper mass moves nto the plane of the paper and the lower mass moves out of the plane of the paper wth ndcated veloctes. The r vector ponts along the dumell, and the cross product of r and v les along the z axs whch s transverse to the dumell axs. I ve douled the L of one of the pont masses. Snce L ponts along z t rotates wth the od n a cone wth a 1/ angle of aout the vertcal when vewed n the FCS. L s statonar n the BCS sstem and has the value PH6_L11 9 that we computed from the I tensor.

10 z ' ω Torque of the dumells x ' We egn computng τ usng BCS. For unform moton ω 0 and τ ω L ˆ' ˆj' kˆ ' τ ω L m ω sn 0 cos 0 0 cos τ m ω ˆj sn cos ' L m ω cos kˆ ' Hence at the nstant depcted, τ ponts along the axs or out of the plane of the paper. We can also vew the L vector n the FCS. Snce L ponts along the z axs, t rotates around ω n a cone of 1/ angle aout the vertcal axs. The component of L along ω remans constant whle the component n the horzontal plane rotates n a crcle. L m ω cos ω L ω t L m ω cos sn The change L s agan out of the plane of the paper when vewed from sde. For small t, L L ω t and thus L τ L ω m ω cos sn t whch s same answer as efore. τ rotates wth the dumells as tangent to the rm of the cone. well and s alwas PH6_L11 10

11 ω cos A fnal wa of vewng dumell torque sn m ( sn ) cos mr m ( sn ) ω ω ω Ths s perhaps the smplest wa of vsualzng the torque requred to unforml rotate the dumell. Because the two masses are travellng n a crcle of radus sn, the undergo centrpetal acceleraton and a centrpetal force of F m ω sn acts on each mass. Each force s a transverse dstance of cos from the pvot pont. There s a torque drected out of the plane of The torque aout the center pont s τr F cos m sn ω ( ) ( ) m ω the rght hand rule that the torque s drected out of the paper. sn cos It s eas to see from How the requred torque s conveed to the dumell sstem? Imagne weldng the angled axs to a vertcal axle whch s held n place a rng of all earngs. We can crudel thnk of the earngs counterng the torque created the centrfugal force as the dumell tres to algn tself n the horzontal plane va centrfugal force. The purple earngs suppl ths torque. Note that the drecton of the rotaton s rrelevant to the earng force. ω PH6_L11 11

12 I for a rectangular plate We show how to fnd the moment of nerta and prncpal axs for a rectangular plate. A plate s a nce wa of llustratng the technques snce t essentall onl nvolves matrces and the secular equaton s just a (solvale) quadratc. We place the plate n the z0 plane. Snce the plate has no extent n the z axs, all of the products of nerta nvolvng z vansh. + z x xz I m x x + z z xz z x + Ix Px 0 I Px I Ix I + We also note that Iz Ix + I for a plate n the x- plane. Lets see wh I m + z m ( ) ( z ) ( ) x + mx I m x I m x + I + I z x a x a Ix σ dx d σ a 0 0 For the plate n the z0 plane, we need to compute ntegrals rather than 6 for the general extended oject. σ a m ma. Smlarl I ( a + ) PH6_L a a x Px σ xdx d σ a 0 0 σ a ma 4 4 a m I a a

13 Prncpal axes for a 4 a 0 m I a 4a ( a ) + Snce there s no mxng etween the z axes and another axs, one prncpal axs (PA) s the z-axs. Lets make sure. The crtera s I ω λω where λ s the moment of nerta aout the PA. 4 a 0 0 m a 4a ( a ) m4( a + ) The other PA are transverse to the ( 0,0,1) PA and therefore le n ths x- plane. rectangular plate The technque for fndng the other PA or egenvectors egns solvng the secular equaton for the egenvalues λ. Lets revew ths procedure I ω λω can e wrtten as I- 1 ω0. The onl wa to get a soluton other than ω0 s to make I-λ1 sngular ( λ ) ( ) or I-λ 1 0. We onl need consder the x- components snce onl these mx. The secular equaton reads: m4 -ma λ We could solve ths -ma ma 4 λ 1 1 for λ n general snce t s a quadratc equaton. Rather than ths lets work some specfc cases. Lets egn wth the case of a. PH6_L11 1