Exam (elaborations) TEST BANK FOR Dynamics with Dynamics of Rigid Bodies By S L Loney (Solution Manual)-Converted

If r be the perpendicular distance from any given line of any element m of the mass of a body, then the quantity åmr2 is called the moment of inertia of the body about the given line. In other words, the moment of inertia is thus obtained; take each element of the body, multiply it by the square of its perpendicular distance from the given line; and add together all the quantities thus obtained. If this sum be equal to Mk2, where M is the total mass of the body, then k is called the Radius of Gyration about the given line. It has sometimes been called the Swing-Radius. If three mutually perpendicular axes Ox;Oy;Oz be taken, and if the coordinates of any element m of the system referred to these axes be x;y and z, then the quantities åmyz;åmzx; and åmxy are called the products of inertia with respect to the axes y and z; z and x; and x and y respectively. Since the distance of the element from the axis of x is p y2+z2; the moment of inertia about the axis of x =åm(y2+z2): 1 2 Chapter 11: Moments and Products of Inertia: Principal Axes 145. Simple cases of Moments of Inertia. I. Thin uniform rod of mass M and length 2a: Let AB be the rod, and PQ any element of it such that AP = x and PQ =d x: The mass of PQ is d x 2a :M: Hence the moment of inertia about an axis through A perpendicular to the rod =åd x 2a :M:x2 = M 2a Z 2a 0 x2dx = M 2a : 1 3 [2a]3 = M: Ma2 3 : Similarly, if O be the centre of the rod, OP = y and PQ = d y; the moment of inertia of the rod about an axis through O perpendicular to the rod =åd y 2a :M:y2 = M 2a Z +a ¡a y2dy = M 2a : 1 3 [y3]+a ¡a = M: a2 3 : II. Rectangular lamina. Let ABCD be the lamina, such that AB = 2a and AD=2b; whose centre is O: By drawing a large number of lines parallel to AD we obtain a large number of strips, each of which is ultimately a straight line. The moment of inertia of each of these strips about an axis through O parallel to AB is (by I) equal to its mass multiplied by b2 3 : Hence the sum of the moments of the strips, i.e. the moment of inertia of the rectangle, about the same line is M b2 3 : So its moment of inertia, about an axis through O parallel to the side 2b; is M a2 3 : If x and y be the coordinates of any point P of the lamina referred to axes through O parallel to AB and AD respectively, these results give LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 3 åmy2 = moment of inertia about Ox = M b2 3 and åmx2 = M a2 3 : The moment of inertia of the lamina about an axis through O perpendicular to the lamina =åm:OP2 =åm(x2+y2) = M a2+b2 3 : III. Rectangular Parallelepiped. Let the lengths of its sides be 2a;2b; and 2c: Consider an axis through the centre parallel to the side 2a; and conceive the solid as made up of a very large number of thin parallel rectangular slices all perpendicular to this axis; each of these slices has sides 2b and 2c and hence its moment of inertia about the axis is its mass multiplied by b2+c2 3 : Hence the moment of inertia of the whole body is the whole mass multiplied by b2+c2 3 ; i:e:; M b2+c2 3 : IV. Circumference of a circle. Let Ox be any axis through the centre O, P any point of the circumference such that xOP = q ;PQ an element adq ; then the moment of inertia about Ox =å · adq 2pa ¸ a2 sin2q = Ma2 2p Z 2p 0 sin2q dq =4£ Ma2 2p Z 2=p 0 sin2q dq = 2Ma2 p : 1 2 : p 2 = M a2 2 : V. Circular disc of radius a. The area contained between concentric circles of radii r and r +d r is 2prd r and its mass is thus 2prdr pa2 :M; its moment of inertia about a diameter by the previous article = 2rdr a2 M: r2 2 : 4 Chapter 11: Moments and Products of Inertia: Principal Axes Hence the required moment of inertia = M a2 Z a 0 r3dr = M a2 : a4 4 = M: a2 4 : So for the moment about a perpendicular diameter. The moment of inertia about an axis through the centre perpendicular to the disc = (as in II) the sum of these = M: a2 2 : Elliptic disc of axes 2a;2b: Taking slices made by lines parallel to the axis of y, the moment of inertia about the axis of x clearly =2 Z 0 p=2 · 2bsinf d(acosf ) pab M ¸ : b2 sin2f 3 = 4 3 Mb2 p Z p=2 0 sin4f df =M: b2 4 : So the moment of inertia about the axis of y = M: a2 4 : VI. Hollow sphere. Let it be formed by the revolution of the circle of IV about the diameter. Then the moment of inertia about the diameter =å · adq :2pasinq 4pa2 M ¸ a2 sin2q = Ma2 2 Z p 0 sin3q :dq =2: Ma2 2 : 2 3:1 = 2Ma2 3 : VII. Solid sphere. The volume of the thin shell included between spheres of radii r and r+d r is 4pr2d r; and hence its mass is LONEY’S DYNAMICS