ROTATIONAL MOTION. dv d F m m V v dt dt. i i i cm i

Transcription

1 ROTATIONAL MOTION Consder a collecton of partcles, m, located at R relatve to an nertal coordnate system. As before wrte: where R cm locates the center of mass. R Rcm r Wrte Newton s second law for the th partcle: Splt F nto nternal and external forces: dv d F m m V v cm F F F ext j j Snce we are lookng at rotatonal moton we are nterested n what causes rotaton. Remember the experment wth the teeter totter. We decded to defne torque (the quantty responsble for changes n rotaton) as: Thus take the cross product of RF R wth both sdes of the force equaton: d R r F F R r m V v Consder the rght hand sde. Note that: cm ext j cm cm j

2 d d d Rcm r Vcm v R r V v V v V v R r V v cm cm cm cm cm cm Thus the rght hand sde s: d d m Rcm r Vcm v m Rcm Vcm Rcm v r Vcm r v Now consder the left hand sde: Next add these N equatons: R F r F R F r F cm ext ext cm j j j j R F r F R F r F N N N N cm ext ext cm j j 1 1 j j,jr,jr d d d d M R V R m v m r V m r v cm cm cm cm d d R F r F 0 r F M R V r m v cm ext ext j cm cm j,j Now consder It wll contan terms n pars r Fj j r F r F r r F If the forces are central (drected along the lne of centers) each of these wll be zero. Electrcty and gravty are both central, hence ths s usually true. In ths case the sum s zero and we get: d d Rcm Fext ext M Rcm Vcm r mv

3 But d M R V MV V R A R MA R F cm m cm cm cm cm cm cm cm ext Hence the frst terms on each sde cancel and we get: d r m v ext where r F ext ext s the torque on the system measured relatve to the center of mass. Now defne: J rmv rp as the angular momentum of the th partcle. Then our result s: d dj ext J Ths s the rotatonal form of Newton s second law. Note that ths equaton does not requre the center of mass to be unaccelerated. Ths s very useful. We can now use ether an nertal coordnate system, or a non-rotatng one wth orgn at the center of mass. (Only for the torque equaton of course.) Now let s look at the angular momentum a lttle more closely. Assume a rdgd body rotatng about an axs through the center of mass as shown. What s the velocty of m relatve to the center of mass? It s due exclusvely to the rotaton,. It has magntude l and s drected nto the board. Thus:

4 v r Hence J m r r Thus J m r rr where we have used the trple product rule ABCCABBAC In Cartesan coordnates: J m xx yy zzxxˆ yyˆ zzˆ x y z xxˆ yyˆ zzˆ xm x xˆ xyy ˆ xzz ˆ r xˆ ym xyxˆ y yˆ yzzˆ r yˆ z m zxxˆ zyyˆ zzˆ r zˆ I Ixx Ixy Ixz x Iyx Iyy Iyz y Izx Izy I zz z Ixxx Ixyy Ixzz Iyxx Iyyy yzz Izxx IzyIzzz But we have found

5 Hence J xˆ xmr x ymxy zmxz ŷy r y xmxy yyzx ẑz r z xmxz zmyz Ixx m y z Ixy mxy Iyz mxz yy yx xy yz I m x z I I I m y z zz zy yz zx xz I m x y I I I I I s called the moment of nerta tensor. It s real sgnfcance s that J and are generally NOT parallel. However for symmetrc objects the components perpendcular to wll cancel and J and are parallel. We can see ths also from I. Suppose ẑ and the object s symmetrc about z. Then: Iyz Ixz 0 and we have Ixx Ixy J Ixy Iyy J Izz 0 0 I I zz zz z

6 wth zz I m x y m where l s the dstance from the z axs. Note that even for non-symmetrc objects we get (for ẑ ): J z I zz It s just that now J x and J y are not zero. As an example of the general case consder an out of balance wheel consstng of a tre of mass M on a rm of radus R (assume tre plus rm have mass M) wth an adonal mass m at one pont on the rm. Suppose the wheel s rotatng wth angular velocty as shown: We choose an nertal coordnate system as shown. We can calculate I. We fnd: M M Ixx m R sn s dr sn Ms s M m R m sn yy M I s Mm R mcos I M m R zz MR Ixy mr sn cos sn cos 0 mr sn

7 Ms Ixz mrscos cosd Iyz 0 mrssn mrs cos M mr s Mm R msn sn mrscos mr M I sn s Mm R mcos mrssn mrs cos mrs sn M m R 0 0 mrscos J I mrssn M m R Now θ = ωt. Thus mrs sn t dj mrs cos t 0 R cos txˆ R sn ty szˆ Fxxˆ Fyyˆ Fzzˆ xˆf zr sn t sfy yˆ FzR cos t sfx zˆf yr cos t FxR sn t But dj Thus RF sn t sf mrs sn t z y

8 x z FsFRcostmRs cos t FyR cos t FxR sn t 0 Fy Fx tan t From lnear momentum we know that F z = 0 (no change n P z ). Thus y F mr sn t x F mr cos t 1/ v v x y F F F mr mr m R R just as expected! Obvously ths was the hard way to do t, but shows how everythng fts together!