1.1 Introduction, nd some definitions. 1 CHAPTER 1 CENTRES OF MASS This chpter dels with the clcultion of the positions of the centres of mss of vrious odies. We strt with rief eplntion of the mening of centre of mss, centre of grvity nd centroid, nd very few rief sentences on their physicl significnce. Mny students will hve seen the use of clculus in clculting the positions of centres of mss, nd we do this for Plne res i for which the eqution is given in -y coordintes; ii for which the eqution is given in polr coordintes. Plne curves i for which the eqution is given in -y coordintes; ii for which the eqution is given in polr coordintes. Three dimensionl figures such s solid nd hollow hemispheres nd cones. There re some figures for which interesting geometric derivtions cn e done without clculus; for emple, tringulr lmins, nd solid tetrhedr, pyrmids nd cones. And the theorems of Pppus llow you to find the centres of mss of semicirculr lmins nd rcs in your hed with no clculus. First, some definitions. Consider severl point msses in the -y plne: m 1 t ( 1, y 1 ) m t (, y ) etc. The centre of mss is point (, y) whose coordintes re defined y m M i i y mi yi 1.1.1 M where M is the totl mss Σ m i. The sum m i i is the first moment of mss with respect to the y is. The sum m y i i is the first moment of mss with respect to the is.
If the msses re distriuted in three dimensionl spce, with m 1 t ( 1, y 1, z 1 ), etc,. the centre of mss is point (,, ) y z such tht m M i i y m y M i i z mi zi 1.1. M In this cse, mi i, mi yi, mi zi re the first moments of mss with respect to the y-z, z- nd -y plnes respectively. In either cse we cn use vector nottion nd suppose tht r 1, r, r re the position vectors of m 1, m, m with respect to the origin, nd the centre of mss is point whose position vector r is defined y m i r i r. 1.1. M In this cse the sum is vector sum nd m i r i, vector quntity, is the first moment of mss with respect to the origin. Its sclr components in the two dimensionl cse re the moments with respect to the es; in the three dimensionl cse they re the moments with respect to the plnes. Mny erly ooks, nd some contemporry ones, use the term "centre of grvity". Strictly the centre of grvity is point whose position is defined y the rtio of the first moment of weight to the totl weight. This will e identicl to the centre of mss provided tht the strength of the grvittionl field g (or grvittionl ccelertion) is the sme throughout the spce in which the msses re situted. This is usully the cse, though it need not necessrily e so in some contets. For plne geometricl figure, the centroid or centre of re, is point whose position is defined s the rtio of the first moment of re to the totl re. This will e the sme s the position of the centre of mss of plne lmin of the sme size nd shpe provided tht the lmin is of uniform surfce density. Clculting the position of the centre of mss of vrious figures could e considered s merely mke-work mthemticl eercise. However, the centres of grvity, mss nd re hve importnt pplictions in the study of mechnics. For emple, most students t one time or nother hve done prolems in sttic equilirium, such s ldder lening ginst wll. They will hve dutifully drwn vectors indicting the forces on the ldder t the ground nd t the wll, nd vector indicting the weight of the ldder. They will hve drwn this s single rrow t the centre of grvity of the ldder s if the entire weight of the ldder could e "considered to ct" t the centre of grvity. In wht sense cn we tke this lierty nd "consider ll the weight s if it were concentrted t the centre of grvity"? In fct
the ldder consists of mny point msses (toms) ll long its length. One of the equilirium conditions is tht there is no net torque on the ldder. The definition of the centre of grvity is such tht the sum of the moments of the weights of ll the toms out the se of the ldder is equl to the totl weight times the horizontl distnce to the centre of grvity, nd it is in tht sense tht ll the weight "cn e considered to ct" there. Incidentlly, in this emple, "centre of grvity" is the correct term to use. The distinction would e importnt if the ldder were in nonuniform grvittionl field. In dynmics, the totl liner momentum of system of prticles is equl to the totl mss times the velocity of the centre of mss. This my e "ovious", ut it requires forml proof, leit one tht follows very quickly from the definition of the centre of mss. 1 1 Likewise the kinetic energy of rigid ody in two dimensions equls MV + Iω, where M is the totl mss, V the speed of the centre of mss, I the rottionl inerti nd ω the ngulr speed, oth round the centre of mss. Agin it requires forml proof, ut in ny cse it furnishes us with nother emple to show tht the clcultion of the positions of centres of mss is more thn merely mke-work mthemticl eercise nd tht it hs some physicl significnce. If verticl surfce is immersed under wter (e.g. dm wll) it cn e shown tht the totl hydrosttic force on the verticl surfce is equl to the re times the pressure t the centroid. This requires proof (redily deduced from the definition of the centroid nd elementry hydrosttic principles), ut it is nother emple of physicl ppliction of knowing the position of the centroid. 1. Plne tringulr lmin Definition: A medin of tringle is line from verte to the mid point of the opposite side. Theorem I. The three medins of tringle re concurrent (meet t single, unique point) t point tht is two-thirds of the distnce from verte to the mid point of the opposite side. Theorem II. The centre of mss of uniform tringulr lmin (or the centroid of tringle) is t the meet of the medins. The proof of I cn e done with nice vector rgument (figure I.1): Let A, B e the vectors OA, OB. Then A + B is the digonl of the prllelogrm of which OA 1 nd OB re two sides, nd the position vector of the point C 1 is (A + B). To get C, we see tht C A + (AM ) A + (M A) A + ( 1 B A) 1 (A + B)
4 FIGURE I.1 Thus the points C 1 nd C re identicl, nd the sme would e true for the third medin, so Theorem I is proved. Now consider n elementl slice s in figure I.. The centre of mss of the slice is t its midpoint. The sme is true of ny similr slices prllel to it. Therefore the centre of mss is on the locus of the mid-points - i.e. on medin. Similrly it is on ech of the other medins, nd Theorem II is proved. FIGURE 1. Tht needed only some vector geometry. We now move on to some clculus.
5 1. Plne res. Plne res in which the eqution is given in -y coordintes FIGURE I. We hve curve y y() (figure I.) nd we wish to find the position of the centroid of the re under the curve etween nd. We consider n elementl slice of width δ t distnce from the y is. Its re is yδ, nd so the totl re is A yd 1..1 The first moment of re of the slice with respect to the y is is yδ, nd so the first moment of the entire re is yd. Therefore yd yd 1.. yd A
1 For y we notice tht the distnce of the centroid of the slice from the is is therefore the first moment of the re out the is is 1 y.yδ. 6 y, nd Therefore y y d 1.. A Emple. Consider semicirculr lmin, + y, > 0, see figure I.4: FIGURE I.4 We re deling with the prts oth ove nd elow the is, so the re of the semicircle is A yd nd the first moment of re is 0 yd. You should find 0 4 /(π) 0.444. Now consider the lmin + y, y > 0 (figure I.5): FIGURE I.5
7 The re of the elementl slice this time is yδ (not yδ), nd the integrtion limits re from to +. To find y, use eqution 1.., nd you should get y 0.444. Plne res in which the eqution is given in polr coordintes. FIGURE I.6 We consider n elementl tringulr sector (figure I.6) etween θ nd θ + δθ. The "height" of 1 the tringle is r nd the "se" is rδθ. The re of the tringle is r δθ. Therefore the whole re 1 r d θ. 1..4 β The horizontl distnce of the centroid of the elementl sector from the origin (more correctly, from the "pole" of the polr coordinte system) is r cosθ. The first moment of re of the sector with respect to the y is is 1 1 r cos θ r δθ cos θδθ r so the first moment of re of the entire figure etween θ nd θ β is 1 β r d cosθ θ.
β r cosθdθ Therefore. 1..5 β r dθ 8 β r sin θdθ Similrly y. 1..6 β r dθ Emple: Consider the semicircle r, θ π/ to +π/. +π / cosθdθ π / + π / 4 cos. +π / θdθ 1..7 π / dθ π π π / The reder should now try to find the position of the centroid of circulr sector (slice of pizz!) of ngle. The integrtion limits will e to +. When you rrive t formul (which you should keep in noteook for future reference), check tht it goes to 4/(π ) if π/, nd to / if 0. 1.4 Plne curves Plne curves in which the eqution is given in -y coordintes FIGURE I.7
Figure I.7 shows how n elementl length δs is relted to the corresponding increments in nd y: 9 ( δ + δy ) 1/ 1+ ( dy / d) [ ] 1/ 1/ δ [( d / dy) + 1] δy. δ s 1.4.1 Consider wire of mss per unit length (liner density) λ ent into the shpe y y( ) etween nd. The mss of n element ds is λ δs, so the totl mss is ds λ [ 1 + ( dy / d) ] 1/ d. λ 1.4. The first moments of mss out the y- nd -es re respectively [ ( dy / d) ] [ ]. 1/ λ 1 + d nd y 1 ( dy / d) ) 1/ d λ + 1.4. If the wire is uniform nd λ is therefore not function of or y, λ cn come outside the integrl signs in equtions 1.4. nd 1.4., nd we hence otin [ 1 + ( dy / d) ] 1 + ( dy / d) 1/ [ 1 + ( dy / d) ] 1 + ( dy / d) d y d, y, 1.4.4 1/ 1/ [ ] d [ ] d 1/ the denomintor in ech of these epressions merely eing the totl length of the wire. Emple: Consider uniform wire ent into the shpe of the semicircle + y, > 0. First, it might e noted tht one would epect > 0.444 (the vlue for plne semicirculr lmin). The length (i.e. the denomintor in eqution 1.4.4) is just π. Since there re, etween nd + δ, two elementl lengths to ccount for, one ove nd one elow the is, the numertor of the first of eqution 1.4.4 must e 0 + In this cse y ( ) [ 1 ( dy / d) ] 1/ d. The first moment of length of the entire semicircle is dy d 1/, 1/ ( ).
10 1 + 0 1/ d d 0 1/ ( ). From this point the student is left to his or her own devices to derive / π 0.666. Plne curves in which the eqution is given in polr coordintes. FIGURE I.8 Figure I.8 shows how n elementl length δs is relted to the corresponding increments in r nd θ : δs [( ) ( ) ] 1/ dr δr + rδθ ( ) [ ] 1/ dθ + r δθ [ 1+ ( r ) ] 1/ δr. 1.4.5 dθ dr The mss of the curve (etween θ nd θ β) is dr [( ) + r ] 1/ θ β λ d θ d.
11 The first moments out the y- nd -es re (reclling tht r cosθ nd y rsinθ ) β λr cos θ + θ [( ) ] dr r 1/ nd ( dr λr sin θ ) d [ + r ] 1/ θ. β θ d d If λ is not function of r or θ, we otin dr 1/ 1 β dr [( ) + r ] dθ, y r θ ( ) [ + r ] 1/ θ β d dθ L dθ 1 r cos θ sin 1.4.6 L where L is the length of the wire. Emple: Agin consider the uniform wire of figure I.8 ent into the shpe of semicircle. The eqution in polr coordintes is simply r, nd the integrtion limits re θ π / to θ +π /. The length is π. 1 + / 1/ θ + π / π π Thus cos [ 0 ] d. θ π The reder should now find the position of the centre of mss of wire ent into the rc of circle of ngle. The epression otined should go to /π s goes to π/, nd to s goes to zero.
1 1.5 Summry of the formuls for plne lmins nd curves SUMMARY Uniform Plne Lmin y y( ) r r( θ) 1 A yd β r β cos θdθ r dθ y 1 A y d y β r β sin θdθ r dθ Uniform Plne Curve y y( ) r r( θ) y 1 1 L + dy ( ) 1/ d 1 dr r cos θ ( ) 1 y 1 L + d dy ( ) 1/ d d β L 1 dr y r sin θ ( ) β L [ + r ] 1/ dθ dθ [ + r ] 1/ dθ dθ
1.6 The Theorems of Pppus. (Pppus Alendrinus, Greek mthemticin, pproimtely rd or 4th century AD.) 1. I. If plne re is rotted out n is in its plne, ut which does not cross the re, the volume swept out equls the re times the distnce moved y the centroid. II. If plne curve is rotted out n is in its plne, ut which does not cross the curve, the re swept out equls the length times the distnce moved y the centroid. These theorems enle us to work out the volume of solid of revolution if we know the position of the centroid of plne re, or vice vers; or to work out the re of surfce of revolution if we know the position of the centroid of plne curve or vice vers. It is not necessry tht the plne or the curve e rotted through full 60 o. We prove the theorems first. We then follow with some emples. z FIGURE I.9 A δa φ y
14 Consider n re A in the z plne (figure I.9), nd n element δa within the re t distnce from the z is. Rotte the re through n ngle φ out the z is. The length of the rc trced y the element δa in moving through n ngle φ is φ, so the volume swept out y δa is φδa. The volume swept out y the entire re is φ da. But the definition of the centroid of A is such tht its distnce from the z is is given y A da. Therefore the volume swept out y the re is φ A. But φ is the distnce moved y the centroid, so the first theorem of Pppus is proved. z FIGURE I.10 δs φ y Consider curve of length L in the z plne (figure I.10), nd n element δs of the curve t distnce from the z is. Rotte the curve through n ngle φ out the z is. The length of the rc trced y the element ds in moving through n ngle φ is φ, so the re swept out y δs is φδs. The re swept out y the entire curve is φ ds. But the definition of the centroid is such tht its distnce from the z is is given y L ds. Therefore the re swept out y the curve is φ L. But φ is the distnce moved y the centroid, so the second theorem of Pppus is proved.
15 Applictions of the Theorems of Pppus. Rotte plne semicirculr figure of re 1 swept out is 4 Pppus, 4 /( π ). π through 60 o out its dimeter. The volume π, nd the distnce moved y the centroid is π. Therefore y the theorem of Rotte plne semicirculr rc of length rgument to show tht / π. π through 60 o out its dimeter. Use similr FIGURE I.11 Consider right-ngled tringle, height h, se (figure I.11). Its centroid is t distnce / from the height h. The re of the tringle is h/. Rotte the tringle through 60 o out h. The distnce moved y the centroid is π/. The volume of the cone swept out is h/ times π/, equls π h/. Now consider line of length l inclined t n ngle to the y is (figure I.1). Its centroid is t distnce 1 l sin from the y is. Rotte the line through 60 o out the y is. The distnce moved y the centroid is π 1 l sin πl sin. The surfce re of the cone swept out is l πl sin πl sin.
16 FIGURE I.1 l The centre of circle of rdius is t distnce from the y is. It is rotted through 60 o out the y is to form torus ( figure I.1). Use the theorems of Pppus to show tht the volume nd surfce re of the torus re, respectively, π nd 4π. V π A 4π. FIGURE I.1
17 1.7 Uniform solid tetrhedron, pyrmid nd cone. Definition. A medin of tetrhedron is line from verte to the centroid of the opposite fce. Theorem I. The four medins of tetrhedron re concurrent t point /4 of the wy from verte to the centroid of the opposite fce. Theorm II. The centre of mss of uniform solid tetrhedron is t the meet of the medins. Theorem I cn e derived y similr vector geometric rgument used for the plne tringle. It is slightly more chllenging thn for the plne tringle, nd it is left s n eercise for the reder. I drw two digrms (figure I.14). One shows the point C 1 tht is /4 of the wy from the verte A to the centroid of the opposite fce. The other shows the point C tht is /4 of the wy from the verte B to the centroid of its opposite fce.. You should e le to show tht C 1 (A + B + D)/4. FIGURE I.14
18 In fct this suffices to prove Theorem I, ecuse, from the symmetry etween A, B nd D, one is ound to rrive t the sme epression for the three-qurter wy mrk on ny of the four medins. But for ressurnce you should try to show, from the second figure, tht C (A + B + D)/4. The rgument for Theorem II is esy, nd is similr to the corresponding rgument for plne tringles. Pyrmid. A right pyrmid whose se is regulr polygon (for emple, squre) cn e considered to e mde up of severl tetrhedr stuck together. Therefore the centre of mss is /4 of the wy from the verte to the mid point of the se. Cone. A right circulr cone is just specil cse of regulr pyrmid in which the se is polygon with n infinite numer of infinitesiml sides. Therefore the centre of mss of uniform right circulr cone is /4 of the wy from the verte to the centre of the se. We cn lso find the position of the centre of mss of solid right circulr cone y clculus. We cn find its volume y clculus, too, ut we'll suppose tht we lredy know, from the theorem of Pppus, tht the volume is 1 se height. FIGURE I.15
19 Consider the cone in figure I.15, generted y rotting the line y /h (etween 0 nd h) through 60 o out the is. The rdius of the elementl slice of thickness δ t is /h. Its volume is π δ / h. Since the volume of the entire cone is π h/, the mss of the slice is π δ M h π h M h δ, where M is the totl mss of the cone. The first moment of mss of the elementl slice with respect to the y is is M δ/h. The position of the centre of mss is therefore h h d h 0 4. 1.8 Hollow cone. The surfce of hollow cone cn e considered to e mde up of n infinite numer of infinitesimlly slender isosceles tringles, nd therefore the centre of mss of hollow cone (without se) is / of the wy from the verte to the midpoint of the se. 1.9 Hemispheres. Uniform solid hemisphere Figure I.4 will serve. The rgument is ectly the sme s for the cone. The volume of the elementl slice is π y δ π( ) δ, nd the volume of the hemisphere is π /, so the mss of the slice is M π M ( ) ( ) δ δ π / ), ( where M is the mss of the hemisphere. The first moment of mss of the elementl slice is times this, so the position of the centre of mss is 0 ( ) d. 8
0 Hollow hemisphericl shell. We my note to egin with tht we would epect the centre of mss to e further from the se thn for uniform solid hemisphere. Agin, figure I.4 will serve. The re of the elementl nnulus is πδ (NOT πyδ!) nd the re of the hemisphere is π. Therefore the mss of the elementl nnulus is M πδ (π ) Mδ /. The first moment of mss of the nnulus is times this, so the position of the centre of mss is 0 d. 1.10 Summry. SUMMARY Tringulr lmin: / of wy from verte to midpoint of opposite side Solid Tetrhedron, Pyrmid, Cone: opposite fce. /4 of wy from verte to centroid of Hollow cone: / of wy from verte to midpoint of se. Semicirculr lmin: 4/(π ) Lmin in form of sector of circle, ngle : ( sin )/() Semicirculr wire: /π Wire in form of n rc of circle, ngle : ( sin ) / Solid hemisphere: /8 Hollow hemisphere: /