PHYSCS LOCUS 17 summation in mi ri becomes an integration. We imagine the body to be subdivided into infinitesimal elements, each of mass dm, as shown in figure 7.17. Let r be the distance from such an element to the axis of rotation. Then the moment of inertia is obtained from...(7.19 c) where the integral is taken over the whole body. The procedure by which the summation of a discrete distribution is replaced by the integral for a continuous distribution is the same as that discussed for the centre of mass. Moment of inertia of the element is = dm r ². Moment of inertia of the whole body = sum of moments of inertia of all elements =. For bodies of irregular shape the integrals may be difficult to evaluate. For bodies of simple geometrical shape the integrals are relatively easy when the axis of symmetry is chosen as the axis of rotation. (a) A small ball of mass m at a distance r from the axis: (b) Two small balls, each of mass m, with perpendicular bisector of the line joining the two balls as the axis of rotation: = mi ri = + = (c) Three balls, each of mass m, at the vertices of an equilateral triangle of side length l with the line passing perpendicularly through the centroid of the triangle as the axis of rotation: = i = mi ri = + + = = m = m l 4 4 l sec
PHYSCS LOCUS 18 (d) An uniform rod of mass m and length l with the line passing through its centre and perpendicular to its length as the axis of rotation: = Using equation 7.19, moment of inertia of the element is dm x m dx x l Therefore, moment of inertia of the rod is l l m m x dx x dx l l l l l m l m l l l l l l m l l m l ml l 8 8 l 8 1 (e) An uniform rod of length l and mass m with the line passing through one of its ends and perpendicular to its length as the axis of rotation: =
PHYSCS LOCUS 19 sum of moments of inertia of all elements about the same axis l m dm x x dx l l m l l 0 0 (f) An uniform ring of mass m and radius r with the line passing through its centre and perpendicular to its plane as the axis of rotation: sum of moments of inertia of all the elements of the ring about the chosen axis dm r r dm all elements are equidistant from the axis dm m The above result could also be established in the following way: Consider the point mass m at a distance r from the axis of rotation, as shown in figure 7.4 (a). n this case the moment of inertia of the mass m about the chosen axis is Now, if we redistribute this mass and make a ring of radius r about the same axis, as shown in figure 7.4 (b), the entire mass still remains at the same distance from the axis and hence the moment of inertia is only. Now, if we stretch the ring along the length of the axis and make a hollow cylinder of radius r, as shown in figure 7.4 (c), the entire mass is still at the same distance from the chosen axis and hence the moment of inertia is still only.
PHYSCS LOCUS 0 Same approach could be use to find the moments of inertia of uniform rectangular plates using the results for uniform rods, as shown in figure 7.5(a) and 7.5(b). We can say that when the mass of a system is redistributed in such a way that each element shifts parallel to the axis of rotation (so that its distance from the axis remains the same), the moment of inertia of the system remains unaffected by the redistribution. (g) An uniform disc of radius r and mass m with the line passing through its centre and perpendicular to its plane as the axis of rotation: The top view of the disc is shown in figure 7.6 (b). To calculate the moment of inertia of the disc about the chosen axis, it is divided into a large number of elements. Each element being a ring with the same centre as that of the disc. Such an element is shown in figure 7.6 (c). Mass of the element is dm, radius is x and thickness is dx. The moment of inertia of the element is dm x Using for an uniform ring m r x dx x mass of the element = mass per unit area area of the element m x r dx The moment of inertia of the disc, sum of moments of inertia of all elements
PHYSCS LOCUS 1 r r m x dx 0 4 m r r 4 (h) A disc of uniform mass density having mass M and radius R with a concentric circular hole of radius r with the line passing through the centre of the disc and perpendicular to the plane of the disc as the axis of rotation: To find the moment of inertia of the given body about its axis of symmetry we follow the same approach as we did in the last case. Divide the body in several elemental rings, calculate the moment of inertia of each ring and then add the moments of inertia of all elemental rings to get the moment of inertia of the whole body. Such an element, having mass dm, is shown in figure 6.7(c), r being its radius and it has a thickness dr, then its moment of inertia, dm r R M R1 r dr r z R M r dr R1 Therefore, moment of inertia of the given body is R 4 4 M M R R1 r dr 1 1 4 R R R R R 1 ( ) t could also be calculated using the method
PHYSCS LOCUS which is exactly what we have done in centre of mass. (i) A hollow sphere of mass M and radius R with the line coinciding with its diameter as the axis of rotation: To find the moment of inertia of the given hollow sphere, let us divide it into several elemental rings having the same axis as that of the hollow sphere, as shown in figure.8. Let the elemental ring shown in figure is at an angular position from the reference line. f dm be the mass of the ring, then, dm mass per unit area area of the ring 4 M R r Rd M Rcos 4 R Rd d M cos d The moment of inertia of the elemental ring, about the chosen axis is dm r M cos d R cos MR cos d Therefore, the moment of inertia of the given hollow sphere is sum of moments of inertia of all elemental rings MR cos d MR 4 [Solve the integral on your own.]
PHYSCS LOCUS We can often simplify the calculation of moments of inertia for various bodies by using general theorems relating the moment of inertia about one axis of the body to that about another axis. Steiner s theorem or parallel axis theorem, relates the moment of inertia about an axis through the centre of mass of a body to that about a second parallel axis. Let cm be the moment of inertia about an axis through the centre of mass of a body and be that about a parallel axis a distance h away. The parallel axis theorem states that mh cm...(7.0) where m is the mass of the body, Consider the situation shown in figure 7.9 (a). A body of mass M is shown with its centre of mass coinciding with the origin of the reference frame. The moment of inertia of an element of mass dm at the point (x, y, z), as shown in figure 7.9(b), about the axis coinciding with the z-axis and hence through the centre of mass of the body is cm square of perpendicular distance from the z-axis dm dm ( x y ) Now, consider another axis parallel to the z-axis and meeting the x-y plane at the point (a, b) as shown in figure 7.9 (b). The moment of inertia of the same element about this axis is dm ( x a ) ( y b) Using change of refrence from method dm( x y ) dm(a b ) a dm x b dm y Therefore, the moment of inertia of the whole body about this axis is Top view of x-y plane. dm( x cm y ) dm(a b ) a dm x b dm d dm( x a dm x b dm y cm cm m d 0 0 dm y dm y cm ; dm m ycm y) and a b m; 0. dm x cm d m xcm 0;
PHYSCS LOCUS 4 Some applications of parallel axis theorem are shown in figure 7.0. A thin uniform rod: mh cm ml l m 1 ml ml 1 4 ml A uniform ring: mh cm a uniform disc: mh cm Applications of parallel axis theorem
PHYSCS LOCUS 5 The plane figure theorem or perpendicular axis theorem relates the moments of inertia about two perpendicular axes in a plane figure to the moment of inertia about a third axis perpendicular to the figure. f x, y, z, are perpendicular axes for a figure which lies in the x-y plane, the moment of inertia about the z-axis equals the sum of the moments of inertia about the x and y axes. Therefore, we have,...(7.1) Before applying this theorem you must make sure that the body is in x-y plane only and the third axis (z-axis in this case) must pass perpendicularly through the intersection points of x and y axes. The moment of inertia of the chosen element in figure 7.1, about the x-axis is x dm x The moment of inertia of the element about the y-axis is y dm y The moment of inertia of the element about the z-axis is z dm r dm ( x y ) dm x dm y x Therefore, the moment of inertia of the whole planar body about the z-axis is y z z x y x y
PHYSCS LOCUS 6 figure 7.: Some applications of perpendicular axis theorem are shown in 0 0 / Finding moment of inertia of a ring about its diameter 0 0 4 Finding moment of inertia of a disc about its diameter. 0 1 ma mb 1 1 a m b 1 Finding moment of inertia of a rectangular plate about an axis passing through its axis and perpendicular to its plane. For a square plate, put a = b.