whether we invert the motion of the plate, or make the magnet revolve in opposite directions. When the magnet is suspended directly over the centre of the revolving plate, (and which is called its concentric position,) it receives no impulse; because the voltaic currents thus generated, lie in planes passing through the magnetic axis of N, and those of the same denomination meet at the centre. The counter currents that thus arise upon opposite sides of the magnetic pole, exactly neutralize each other. ART. IV.-Motion of a System of Bodies; by Prof. THEODORE STRONG. Continued from Vol. xxv, p. 289. Again, supposing T, T', &c. to denote the same things as before, we have Qz - Py=(~×Q-"; ×P)r=Tr, for ×Q= the force Q y when resolved at right angles to r, and ×P= the force P resolved r x y r r the forces XQ, XP, since they act in contrary directions; in the xd3y-yd2x same way Q'x'-P'y' =T'r', and so on; hence we have dta =Tr, x'd3y' — y'd2x' - dt 2 =T'r', &c.; .. it may be shown in the very same way as at p. 43, that if we multiply these equations by m, m', &c. respectively, we shall have, (by adding the products,) the equation Sm =SmTr, which is independent of the recipro 'xd3y — yd2x\ cal actions of the bodies m, m', &c.; by restoring the values of Tr, Tr', &c. we have the first of (18); and in a similar way may the second and third be found. Let h denote the distance of m from the origin of the coördinates, then if m is acted upon by any force mF in the direction of the straight line h, we shall have F for the force which acts on a unit of m in that direction; .. by resolving F in the directions of the axes of x and y, we have and Q respectively which arise from F, x h y XF,F for the parts of P h .. by considering these forces [xy yx only, we have Q¤—Py=(27 F=0; hence it is evident that (18) are independent of any forces which act upon m, m', &c. in the directions of straight lines drawn to the origin of the coördinates; the same thing is also evident from the expressions Tr, T'r', &c., which require F to be resolved in the direction of a straight line which is at right angles to the extremity of r in the plane x, y;.. resolving F into two, one of which is perpendicular to the plane x, y, and the other in the direction of r; the first of these will not affect T, and the second =×F, but as this acts in the direction of r, it will give nothing when resolved in a direction at right angles to r, indeed F will not affect T, since their directions are perpendicular to each other; hence Tr, T'r', &c. are independent of any forces which act on m, m', &c. in the directions of straight lines drawn to the origin of the coördinates; .. as before (18) are independent of such forces. Let X, Y, Z denote the coördinates of the centre of gravity of the system, then put =X+,x, y=Y+y, z=Z+,z, x'=X+,x', &c. ; by substituting these values of x, y, z, &c. in (18), (since by the nature of the centre of gravity Sm,x=0, Sm,y=0, Sm,z=0, Smd2,x=0, -Q), (19). Hence since (19) are independent of the coördinates of the centre of gravity, the motion of that centre is found in the same way as it would be if all the bodies of the system were united at the centre, and the motion of the system about the centre is found by (19) in the same way that it would be if the centre was at rest, and the same forces were applied, and in the same manner, as when the centre is in motion; that is, the motion of the system is resolved into two, viz. the motion of the centre of gravity, and the motion of the system about the centre, which are independent of each other. Again, it is evident that (1), (2), (3) will remain the same if the origin of the coördinates has a uniform rectilineal motion in space : .. (4), and (18) or (7), which are merely transformations of (1), (2), (3) will exist relative to the moveable origin; supposing the axes of x, y, z to be reckoned from the moveable origin, and each to move 1 parallel to itself. If the system is not subjected to the actions of any foreign forces, (11) will exist relative to the moveable origin, whether the changes of the motions of the bodies are finite in an instant or not, and A, A,,,A, will each be invariable during the motion of the system; .. (16) will each be invariable, hence the invariable plane always passes through the moveable origin and moves parallel to itself. It has been proved when the system is not subjected to the actions of any foreign forces, that the centre of gravity is either at rest or moves uniformly forward in a straight line; .. by fixing the origin at the centre of gravity, the invariable plane is either at rest or moves parallel to itself. Let the origin be at the centre of gravity, and suppose that the system is not subjected to the actions of any dt =A, Sm = foreign forces, then by (11) Sm [ydz - zdy ,A, Sm =A, (20). Now we have m'((x'—x)dy' — dt (y' —y)dx')=m'(x'dy' — y'dx')+m'(ydx' — xdy'),m"((x”—x)dy" — (y′′ —y)dx")=m"(x"dy" —y′′dx")+m"(ydx" —xdy"), and so on for all the bodies except m: hence Sm'((x' — x) dy' — (y' — y) dx') = Sm' (x'dy' — y'dx') + Sm' (ydx' — xdy'), but Sm'ydx' = ySm'dx', Sm'xdy'=xSmdy', and by the nature of the centre of gravity Sm'dx' (x' — x)dy' — (y' —y)dx' =-mdx, Sm'dy' = mdy; hence mSm' dt (x — x')dy — (y — y')dx dt — y')dx) = =m'A, and so on for all the bodies; .. mSm'((≈' — x)dy' – (y' — y)dx') + m'Sm ((x dt x')dy − ( y −y')dx) + &c. = mA + m'A + &c. or (x — — (y Smm' ((x-x).(dy' — dy) — (y' — y).(dx' — dx)) — ASm, (21), in the Let the plane x, y be the invariable plane, then A is a maximum, and A,,,A are each =0; .. the first member of (21) is a maximum, and those of (22) are each =0. Let a plane be drawn through any body of the system parallel to the plane x, y, then the first member of (21) is a maximum relative to the parallel plane; for its value is the same for the parallel plane as for that of x, y, as is evident by supposing the system to be reduced orthographically to the planes. Now since the plane x, y is either at rest or moves parallel to itself, its parallel plane will always be parallel to itself as the system moves: it is also manifest that the first members of (22) will each =0, relative to any plane drawn through the same body (as before,) perpendicular to the parallel plane; Mec. Cel. Vol. I, p. 63. Again, let, (for brevity,) any body of the system be indefinitely denoted by m, z, y, z, being its rectangular coördinates; also suppose (as in (d), (e), &c.) that x', y', z', are the rectangular coördinates of m, when referred to any other system of rectangular axes, which have the same origin as the axes of x, y, z; then denoting by a, b, &c. the same things as in (d,) (e,) &c., we have by (d),x=ax'+by' +cz', y=a'x'+b'y' + c'z', z = a'x' + b'y' + c'z', (a'): supposing dx the quantities in (a') to be functions of the time, we have dt x'da+y'db+z'dc, adx'+bdy'+cdz' dy x'da' + y'db' + z'dc' dt + dt = dt + + a'dx'+b'dy+c'dz' dz x'da"+y'db"+z'dc", a"dx'+b"dy'+c"dz' dt (b'). Put cdb+c'db'+c'db"=pdt, adc+a'dc' + a"de"= qdt, bda+ b'da'+b"da"= rdt; then by (f) cdb + c'db'+c"db" — — bdc — b'dc' -b"de"= pdt, ade+a'dc'+a"de"=-cda-c'da' - c"da" = qdt, bda +b'da'+b"da" =- adb - a'db' - a"db" rdt, (d'); by substituting the values of a, b, &c. from (o) in (d'), we have sin. sin. ed↓cos. çdê=pdt, cos. q sin. ed↓+sin. qdê==qdt, dp — cos. d↓=rdt, (e'). == dr' Put qz'-ry+L, rx'—pz'+ = dz' dt dy' dt =M, py' -qx+7=N, (ƒ'); then multiply (b') by a, a', a", respectively, add the products, and dr a, b, c severally, add the products, and we have t=a+b+cN, dy and in like manner=a'L+b'M+c'N, da′′L+b′′M+c′′N, (h'). +M+N2; hence by (f'), dx2+dy2+ dz By adding the squares of (h'), we have by (ƒ) 2 +z'3)q2 + (z'2 +y'3 )r2—2y'z'.qr-2z'z'.pr-2x'y'.pq+2(y'dz'z'dy'\ dt Py), d.Sm 2dx - xdz =dt. Sm(Qx dt = dt. = dt. Sm(Pz—Rx), d.Sm (ydz―zdy) Sm(RyQ), (23); put S(y'2 + z'2)m = A, S(x'2 + z'2)m = B, S(x'2+y'3)m=C, Sy'z'm=D, Sx'z'm=E, Sx'y'm=F, (k'), also put Ap Er Fq=p', Bq - Dr - Fp= q', Cr - Dq - Ep=r', - · 'z'dx' —x'dz' Sm --- [x'dy' — y'dx' ) =A, Sm dt =,B, Sm (ydz'z'dy') dt = C, (7). By substituting the values of x and y from (a) and those dx dy dt' dt' of from (h), in the first of (23), we shall have (since a, b, &c., are the same for all the bodies ;) d. [(bc'-cb').Sm(Ny' — Mz') +(ac-ac'). Sm(Lz' — Nx') + (ab' — ba'). Sm(Mx' — Ly')] = dt. Sm(Qx-Py,) (24). Put dt.Sm(Qx-Py)=dN", dt.Sm(Pz-Rx)=dN", dt.Sm(Ry— Qz)=dN', (m'); by substituting the values of L, M, N, from (ƒ'), we have Sm(Ny - Mz')=p'+,C, Sm(Lz' — Nx')=q'+,B, Sm(Mx' — Ly')=r'+‚A, (n'), by substituting these values, and those of bc' cb', &c., from (g), and using dN", (24) becomes d.[a"(p'+,C)+ b'(q'+,B)+c" (r'+,A)]=dN"', in like manner the second and third of (23) will give, d. [a'( p'+ ‚C) + b'(q' +‚B) + c'(r'+‚A)]=dN", d. [a(p'+,C)+b(q'+‚B)+c(r'+,A)] = dN', (25); the two last of these are easily derived from the first by making some very obvious changes in (24). By taking the differentials indicated in (25), then multiplying the resulting equations by a", a', a respectively, and adding the products, we shall have by (ƒ) and (d'), after dividing by dt, Put the right hand members of (26) equal to N,, N,,, N,,, respec |