MECHANICS 733 Hence, if p be an undetermined multiplier, we have (k +p) a(a + 0/j. + yv) = , (kl +p)n - &(a + fifj. + yi>) = Q, (kl +p}v -y(a + flu + yv) = . But, if we consider a surface of the second order Jc +p k i +p k~l +p confocal with the ellipsoid the direction cosines of its normal at x, y, z are x _J/_ g " k+p kl + p kl+p Hence, if this surface pass through the point a, 0, 7, we have _W. where F is determined by the equation k +p k; +p Substitute this value of P in the preceding equations, and they become identical with those above given for determining the principal axes at o, 0, 7. Hence Binet s Theorem : The principal axes at any point of a body arc normals to the three surfaces of the second order which pass through that point and are confocal with the ellipsoid (a). ncipal 238. We will here tabulate the values of the moments iments O f inertia about principal axes through the centre of inertia, mer " in a few specially useful cases. 1. Plane uniform circular disk. Divide it into concentric rings, of radius r, of breadth Sr. Then the moment of inertia about the axis through the centre, and per pendicular to the plane, of the circle is where a is the radius, and p the mass of a square unit, of the disk. But the mass is ira 2 p, so that k = a?. This of course, applies to a circular cylinder. Obviously, in the disk fcJ-fcJ-JJ-Ja 8 . In fact the moment of inertia about an axis drawn perpendicular to any plane figure at any point is equal to the sum of the other two about rectangular axes which lie in the plane. The one is 2m(.c- + ?/ 2 ), and the others are Zmx 2 and S.mi/ respectively. 2. Uniform rod of length I, p mass per unit length. " * so that fc* = fr= = T y 2 . 3. Uniform rectangular plate, sides a and b, axis parallel to b. From this follows immediately 6. Ellipsoid, semiaxes a, b, c, and of uniform density : so that kl = T yi 2 , and fc* = T V& S . Hence, by the remark above, 4. Uniform sphere, radius a, p mass per unit volume. Here 2(ma; 2 ) = ^(my*) = 2(ms ! ) , and therefore the sum of any two is 7,2 _ 1-7,2 _ n.r,, 2 _ 2 , /" a .47._R_ 5 1 3 "* yo Thus But M = and thus A,-^ = kl = k = |a 2 . 5. Plane uniform elliptic disk, semiaxes a,b;p mass of unit area. Moment of inertia about a is From these we can, of course, reproduce the result for a sphere. 7. Rectangular parallelepiped, edges a, b, c : k = -h (b- + c 2 ), k = T V (c 2 + a 2 ), k - ,V ( 2 + 6 2 )- The determination of moments of inertia is, like that of centres of inertia, a purely mathematical matter, the full discussion of which would lead us away from the proper objects of this article. 239. The simplest cases that can present themselves so Rotation far as rotation is concerned (for the translational effects on about a rigid body are treated precisely as if it were a mere par- 1X ? ( tide, a process already sufficiently illustrated) are those in which there is one degree of freedom to rotate, i.e., when the body is rigidly attached to a fixed axis. Here the physical condition is simply that the rate of increase of moment of momentum is equal to the moment of the resultant couple about the axis of rotation. 240. Let us recur to Atwood s machine as a first example, Pulley of and suppose the string not to slip on the pulley, so that the A.twop<r pulley must turn. In this case we must observe that the mi two free parts of the string are now, as it were, separate strings, so that we have no right to assume their tensions to be equal. In fact if they were equal there would be no acceleration of the rotation of the pulley, nor of course of the common velocity of the two masses. We assume that the pulley is symmetrical, and the axis through its centre of inertia. Let a be the radius of the pulley, and <o its angular velocity, then aw is the linear velocity of either mass. Thus the linear acceleration of each of the masses is equal to a times the angular acceleration of the pulley. But the linear acceleration multiplied by the mass is the measure of the force producing it; while the angular acceleration multiplied by the moment of inertia is the measure of the moment of the couple producing it. Thus we have (M being the mass of the pulley, and k its radius of gyration) M& 2 x angular acceleration = (T - T)a , m x linear acceleration = m g - T , m x linear acceleration = T - mg . Eliminating T and T , and taking account of the above relation between the accelerations, we find at once m - m Linear Acceleration =
so that from which, by the last two of our equations, the separate values of T and T may be found. . If we compare this result with that obtained in 173, on the supposition that the pulley was perfectly smooth, we see that the only difference is in the addition of M 2 /a J to the sum of the two masses. Otherwise the nature of the motion remains unaffected. 241. Let us next take the case of a body of any form Corn- attached to a horizontal axis which does not pass through F 1 its centre of inertia. In such a case gravity is the force Jj^ ( " producing motion, and we have what is called a " com pound pendulum." Draw through the centre of inertia a line parallel to the axis ; let h be the distance between these lines, and the angle which their plane makes (at a given time) with the vertical. The moment producing angular acceleration is obviously - ?<//* sin . Divide by the moment of inertia about the axis, which by a previous proposition ( 235) is (where k is the radius of gyration about the line drawn through the centre of inertia), and we have for the angular acceleration
ah sin