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Calculus methods for determining the centre of mass of a uniform plane lamina In M2 we consideredhow to determinethe position of the centre of mass of a uniform plane lamina using its symmetries. r-\ f \ Fr ["]=F*,1'' I ,\y, L/ Ll '\, ) ) We must now considerhow to deal with lamina which have no symmetries. When a lamina or plane region is boundedby curves for which the equationsare known, calculus methodscan be used to find the position of its centreof mass. When a lamina occupiesa region betweena curve and the x-axis as shown above, it can be approximatedby elementarystrips of width 6x and length y. One such strip, situatedat the point (x, y) on the curve is shown. The dot indicatesthe centre of mass of this elementarystrip. Let p (Greek letter 'rho') be the massper unit areaof the lamina. Massof eachstrip,6M: Jfturr* Qew>n1 t Art-eno? Centreof massof strip@(r.:) \ ( Yde? Ll Hence centre of mass of lamina (t,y) can be determined: (Zsn) x = t(ril',) (ZKz6*)i' 2(yyt,*) (zf:Aq' Z(tWb_)) (ztt*)A ' /!z.Vt-) (zlti7'+Zy"[x tn lirit| (z;n1, =/Pnt{)) \_ )nlo l>* ,5uplu^l;,*, ,"t"?r'I,'b' fJr4 4F- zI * 1r * "b $4 ;1l# "- f,.+ o ()rJlq=+Jv'J/nb \ l, -b In summarv: (b /x\l= l | oI** , Ml \y)ll"l*,* \d b wherethe total mass,U = lWdx c Examples 1. Find the position of the centre of massof a uniform semi-circular lamina of radius r. --i \_, 2. Find the coordinates of the centre of mass of the region between the curve y:4-* andthepositivexandyaxis. ,: Exercise 5A Paee 171 Odd's r -y ;M t. t,3 3 t(z- | tr* 7- L-(tt +32 35 t3,