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

Exam (elaborations) TEST BANK FOR Dynamics of Rigid Bodies By S L Loney (Solution Manual) LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL BY S. L. LONEY Professor of Mathematics Royal Holloway College University of London, Englifield Green, Surrey, UK Fellow, Sidney Sussex College, Cambridge, UK Kindle Edition M-A-T-H VALLEY v CONTENTS DYNAMICS OF RIGID BODIES 11 MOMENTS AND PRODUCTS OF INERTIA: PRINCIPAL AXES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 12 D’ ALEMBERT’S PRINCIPLE THE GENERAL EQUATIONS OF MOTION. . . . . . . . . . 31 13 MOTION ABOUT A FIXED AXIS . . . . . . . . . . . . . . . . . . 43 14 MOTION IN TWO DIMENSIONS. FINITE FORCES. 83 15 MOTION IN TWO DIMENSIONS. IMPULSIVE FORCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 16 INSTANTANEOUS CENTRE. ANGULAR VELOCITIES. MOTION IN THREE DIMENSIONS . . . 161 17 ON THE PRINCIPLES OF THE CONSERVATION OF MOMENTUM AND CONSERVATION OF ENERGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 18 LAGRANGE’S EQUATIONS IN GENERALISED COORDINATES . . . . . . . . . . . . . . . . . . . 241 ix x CONTENTS 19 SMALL OSCILLATIONS: INITIAL MOTIONS. TENDENCY TO BREAK . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 20 MOTION OF A TOP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 22 MISCELLANEOUS EXAMPLES II . . . . . . . . . . . . . . . . . 307 ON THE SOLUTION OF SOME OF THE MORE COMMON FORMS OF DIFFERENTIAL EQUATIONS . . . . . . . . . . . . . . . . . . . . . 327 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 SOLUTION MANUAL 11 MOMENTS AND PRODUCTS OF INERTIA: PRINCIPAL AXES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 12 D’ ALEMBERT’S PRINCIPLE THE GENERAL EQUATIONS OF MOTION. . . . . . . . . . 31 13 MOTION ABOUT A FIXED AXIS . . . . . . . . . . . . . . . . . . 43 14 MOTION IN TWO DIMENSIONS. FINITE FORCES. 83 15 MOTION IN TWO DIMENSIONS. IMPULSIVE FORCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 CONTENTS xi 16 INSTANTANEOUS CENTRE. ANGULAR VELOCITIES. MOTION IN THREE DIMENSIONS . . . 161 17 ON THE PRINCIPLES OF THE CONSERVATION OF MOMENTUM AND CONSERVATION OF ENERGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 18 LAGRANGE’S EQUATIONS IN GENERALISED COORDINATES . . . . . . . . . . . . . . . . . . . 241 19 SMALL OSCILLATIONS: INITIAL MOTIONS. TENDENCY TO BREAK . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 20 MOTION OF A TOP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 22 MISCELLANEOUS EXAMPLES II . . . . . . . . . . . . . . . . . 307 Chapter 11 MOMENTS AND PRODUCTS OF INERTIA: PRINCIPAL AXES 144. 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 OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 5 4pr2d r 4p a3 3 M; i:e: 3r2d r a3 M: Hence, by VI, the required moment of inertia about a diameter = Z a 0 3r2d r a3 M: 2r2 3 = 2M a3 : a5 5 = M: 2a2 5 : VIII. Solid ellipsoid about any principal axis. Let the equation to the ellipsoid be x2 a2 + y2 b2 + z2 c2 = 1: Consider any slice included between planes at distances x and x+d x from the centre and parallel to the plane through the axes of y and z: The area of the section through PMP0 is p:MP:MP0: A B C P O M A' P' Now PM2 OC2 + OM2 OA2 = 1; so that MP = c r 1¡ x2 a2 : So MP0 = b r 1¡ x2 a2 : Hence the volume of the thin slice =pbc µ 1¡ x2 a2 ¶ :d x: Also its moment of inertia about the perpendicular to its plane 6 Chapter 11: Moments and Products of Inertia: Principal Axes = its mass £ MP2+MP02 4 =pbcd x: µ 1¡ x2 a2 ¶2 : b2+c2 4 :r: Hence the required moment of inertia =pbc b2+c2 4 Z +a ¡a µ 1¡ x2 a2 ¶2 :d x:r =pabc: b2+c2 4 : 16 15 r = µ 4 3 pabcr ¶ £ b2+c2 5 =M£ b2+c2 5 : 146. Dr Routh has pointed out a simple rule for remembering the moments of inertia of many of the simpler bodies, viz. The moment of inertia about an axis of symmetry is Mass £ the sum of the squares of the perpendicular semi-axes 3;4; or 5 ; the denominator to be 3, 4, or 5 according as the body is rectangular, elliptical (including circular) or ellipsoidal (including spherical). 147. If the moments and products of inertia about any line, or lines, through the centre of inertia G of a body are known, to obtain the corresponding quantities for any parallel line or lines. Let GX;GY;GZ be any three axes through the centre of gravity, and OX0;OY0, OZ0 parallel axes through any point O: Let the coordinates of any element m of the body be x;y; z referred to the first three LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 7 axes, and x0;y0 and z0 referred to the second set. Then if f ;g and h be the coordinates of G referred to OX0;OY0, and OZ0, we have x0 = x+ f ; y0 = y+g; and z0 = z+h: Hence the moment of inertia of the body with regard to OX0 =åm(y02+z02) =åm[y2+z2+2yg+2zh+g2+h2] :::(1) Now åm:2yg = 2g:åmy: Also, by Statics, åmy åm = the y-coordinate of the centre of inertia referred to G as origin = 0: Hence åm:2yg = 0 and similarly åm:2zh = 0: Hence, from (1), the moment of inertia with regard to OX0 =åm(y2+z2)+M(g2+h2) = the moment of inertia with regard to GX + the moment of inertia of a mass M placed at G about the axis OX0: Again, the product of inertia about the axes OX0 and OY0 =åmx0y0 =åm(x+ f )(y+g) =åm[xy+g:x+ f y+ f g] =åmxy+Mf g = the product of inertia about GX and GY+ the product of inertia of a mass M placed at G about the axes OX0 and 0Y0: COR. It follows from this article that of all axes drawn in a given direction the one through the centre of inertia is the one such that the moment of inertia about it is a minimum. 8 Chapter 11: Moments and Products of Inertia: Principal Axes EX. The moment of inertia of the arc of a complete circle about a tangent = M a2 2 +Ma2 = 3M 2 a2: The moment of inertia of a solid sphere about a tangent = M: 2a2 5 +Ma2 = 7 5 Ma2: 148. If the moments and products of inertia of a body about three perpendicular and concurrent axes are known, to find the moment of inertia about any other axis through their meeting point. Let OX;OY;OZ be the three given axes, and let A;B and C be the moments of inertia with respect to them, and D;E and F the products of inertia with respect to the axes of y and z, of z and x; and x and y respectively. Let the moment of inertia be required about OQ; whose directioncosines with respect to OX;OY and OZ are l;m and n: O Y P Q M K X L Z Take any element m0 of the body at P whose coordinates are x;y; and z; so that OK = x;KL = y and LP = z: Draw PM perpendicular to the axis OQ: Then PM2 = OP2¡OM2: LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 9 Now OP2 = x2+y2+z2; and OM = the projection on OQ of the straight line OP = the projection on OQ of the broken line OKLP =l:OK+m:KL+n:LP = lx+my+nz: Hence the required moment of inertia about OM =åm0:PM2 = åm0[x2+y2+z2¡(lx+my+nz)2] =åm0: £ x2(m2+n2)+y2(n2+l2)+z2(l2+m2) ¡2mnyz¡2nlzx¡2lmxy] ; since l2+m2+n2 = 1 =l2åm0(y2+z2)+m2åm0(z2+x2)+n2:åm0(x2+y2) ¡2mnåm0yz¡2nlåm0zx¡2lmåm0xy =Al2+Bm2+Cn2¡2Dmn¡2Enl¡2Flm: 149. As a particular case of the preceding article consider the case of a plane lamina. Let A, B be its moments of inertia about two lines OX and OY at right angles, and F its product of inertia about the same two lines, so that A =åmy2;B =åmx2;F =åmxy: If (x0;y0) are the coordinates of a point P referred to new axes OX0 and OY0; where XOX0 =q ; then x = x0 cosq ¡y0 sinq ; and y = x0 sinq +y0 cosq : ) x0 = x cosq +ysinq and y0 = ycosq ¡x sinq : 10 Chapter 11: Moments and Products of Inertia: Principal Axes O X P Y Y' X' q Hence the moment of inertia about OX0 =åmy02 = åm(ycosq ¡x sinq )2 =cos2q :åmy2+sin2q :åmx2 ¡2sinq cosq :åmxy =Acos2q +Bsin2q ¡2F sinq cosq : The product of inertia about OX0 and OY0 =åmx0y0 = åm(x cosq +y sinq )(y cosq ¡x sinq ) =åm[y2 sinq cosq ¡x2 sinq cosq +xy(cos2q ¡sin2q )] =(A¡B) sinq cosq +F cos2q : In the case of a plane lamina, if A and B be the moments of inertia about any two perpendicular lines lying in it, the moment of inertia about a line through their intersection perpendicular to the plane åm(x2+y2) =åmy2+åmx2 = A+B: LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 11 150. EX. 1. Find the moment of inertia of an elliptic area about a line CP inclined at q to the major axis, and about a tangent parallel to CP: The moments of inertia, A and B; about the major and minor axes are, as in Art. 145, M b2 4 and M a2 4 : Hence the moment of inertia about CP = M b2 4 cos2q +M a2 4 sin2q ; since F = 0; by symmetry. The perpendicular CY upon a tangent parallel to CP = ab CP : Hence, by Art. 147, the moment of inertia about this tangent =M b2 4 cos2q +M a2 4 sin2q +M a2b2 CP2 =M b2 4 cos2q +M a2 4 sin2q +Ma2b2 · cos2q a2 + sin2q b2 ¸ = 5M 4 (a2 sin2q +b2 cos2q ): EX. 2. The moment of inertia of a uniform cube about any axis through its centre is the same. For A = B =C; and D = E = F = 0; Therefore, from Art. 148, I = A(l2+m2+n2) = A: EXAMPLES Find the moments of inertia of the following: 1. A rectangle about a diagonal and any line through the centre. 2. A circular area about a line in its own plane whose perpendicular distance from its centre is c: 12 Chapter 11: Moments and Products of Inertia: Principal Axes 3. The arc of a circle about (1) the diameter bisecting the arc, (2) an axis through the centre perpendicular to its plane, (3) an axis through its middle point perpendicular to its plane. 4. An isosceles triangle about a perpendicular from the vertex upon the opposite side. 5. Any triangular area ABC about a perpendicular to its plane through A. · Result. M 12 (3b2+3c2¡a2) ¸ 6. The area bounded by r2 = a2 cos2q abou·t its axis. Result. Ma2 16 µ p ¡ 8 3 ¶¸ 7. A right circular cylinder about (1) its axis, (2) a straight line through its centre of gravity perpendicular to its axis. 8. A rectangular parallelepiped about an edge. 9. A hollow sphere about a diameter, its external and internal radii being a and b: · Result. 2M 5 a5¡b5 a3¡b3 ¸ 10. A truncated cone about its axis, the radii of ·its ends being a and b Result. 3M 10 a5¡b5 a3¡b3 ¸ 11. Show that the moment of inertia of a right solid cone, whose height is h and the radius of whose base is a; is 3Ma2 20 : 6h2