CAPUT V.

178

terminatae, ita ut arcus AB, AC et BC fiut quadrantes; axlbiisqne in A,B,Cterminatis refpoudeant momenta inertiae M«n, Mbb, M cc, quorum pri-mum fit maximum, fecundum medium, et tertium minimum. Qnodiljam alius axis quicunque per centrum inertiae tranfiens, qui fuperficieinIphaericant in pundto 8. trajiciat, confideretur, ejus refpetIt, momentuminertiae erit:

Mflfl cof AS 4 + M bb cof BS 2 + Mff cof CS *quod ob cof AS 2 q-ro/BS* 4 ~ to f’ CS 2 = , ? his modis exprimi poteft:Mna — M(aa — hb)cofBS 2 ‘ — M(aa — cc) cofCS Z velMbb + M (an — bb) cof A S 4 — M (bb — cc) cof CS 4 velMrr +M(ao—rr) ro/AS 4 -\-M(bb — rr)ro/BS 4 . -Hinc fi S fit in quadrante BC puta in D, erit momentum inertiae refpeflllaxis ID = M (bb cofKD 4 + cc cof CD 4 ) = Mbb — M (bb — cc) cof CD 4 =sMnr 4 - M (bb _ rr) ro/BD 4 ,feu momentum inertiae refpedlu axis ID erit:

Mbb-M(bb-cc)ftm z = Mcc + M(bb-tc)fiCV*.

Simili modo momentum iuertiae refpeftn axis 1F. efi

Maa — M (aa — cc) fi AE 2 = Mrc+M(aa_rr)_^CE 2 ;momentum autem inertiae refpedu axis IF fit

Maa — M(aa — bb)fit\¥ 2 z=:Mbb + M(aa-bb) fi BF a .

PROBLEMA 30.

458* Invenire omnes axes per centrum inertiae dudtos, quorum re-(pedlu momenta inertiae lint inter fe aequalia.

SOLUTIO.

Fig.48. Sint momenta inertiae refpedu axium principalium IA, IB, IC re/pe-&ive Maa, Mbb, M cc et aa > bb > cc : et quaerantur omnes axes per cen-trum inertiae l ducendi, quorum refpcdlu momenta inertiae fiut inter feaequalia, et quidem aequalia ei, quod refpondet axi IE, fumto E in quadranteAC, quoniam ab A ad C omnia momenta hujus corporis a maximo ad mi-nimum occurrunt. SitIS talis axis, et habebimus hanc aequationem:

Mon-M (aa-cc) fi AE 2 = Maa - M (aa - bb) ro/BS*— M (aa — cc) cof CS 4

feu (an — cc) fi AE 4 — ( aa — bb) cofR$ z + (aa — cc') cof CS 4ergo ob rq^BS 4 =zji AS 4 — cof CS 4 erit, ( aa — cc ) fi AE * = (aa — bb) f AS 4 -\-(bb — cc) cofC S*

Introducatur angulus CAS, et cum fit cofCS=fi AS cofCAS erit

(aa~cc)fi AE 4 = (aa - bb)fi AS 4 + (bb-cc) / AS 4 ro/CAS*

ergo