LLG Paris-Abu Dhabi Advanced Math and Science Pilot Class
MATHEMATICS Gr11 Lesson 7.
Barycentre. 1. Barycentre of two points. We call weighted point, a couple of a point √ and a real number. Example : If A,B,C are points, (A, 3), B, 3 , (C, −π) are weighted points. In all what will follow, capital letters will be points and normal letters numbers, A and B are two given points and a and b two given real numbers. −→ −−→ → − Considering the vectorial equality aAG + bBG = 0 , let’s try to get more informations about G. −→ −−→ → − aAG + bBG = 0 −−→ −→ → −→ − ⇔ aAG + b BA + AG = 0 −→ −→ → − ⇔ (a + b) AG + bBA = 0 If a + b = 0, then it comes two cases b = 0 or A = B for which G can be anywhere, if b 6= 0 and A 6= B, there does not exist any such G point. If a + b 6= 0, −→ −b −→ b −→ AG = BA = AB a+b a+b We just prove the following theorem:
Theorem: If a + b 6= 0, there exists a unique point G satisfying −→ −−→ → − aAG + bBG = 0
Definition: We call then this point, barycentre of the two weighted points set {(A, a) , (B, b)}
Remark: The condition a + b 6= 0 is very important. If a + b = 0 then the set of weighted points does not have barycentre.
Theorem: A set of two weighted points and its barycentre (when it exists) are always collinear. −→ −b −→ Proof: The equality AG = a+b BA gives the proof.
Properties : −→ −−→ → −→ −−→ −→ −−→ → − → − − • If λ ∈ R∗ , aAG + bBG = 0 ⇔ λaAG + λbBG = λ 0 ⇔ λaAG + λbBG = 0 . So if a + b 6= 0 and λ 6= 0 then the barycentre of {(A, a) , (B, b)} is the barycentre of {(A, λa) , (B, λb)} . −→ −−→ → − Remark: This property applied with λ = −1, give the equality aGA + bGB = 0 used as often as the definition equality. • If a + b 6= 0 then, G is the center of mass of {(A, a) , (B, b)} is equivalent to B is the center of mass of {(A, −a) , (G, a + b)} and to A is the center of mass of {(B, −b) , (G, a + b)}
1
Proof : G is the center of mass of {(A, a) , (B, b)} is equivalent to −→ −−→ → − aAG + bBG = 0 −→ −→ −→ → − ⇔ aAG + bBA + bAG = 0 −→ −→ → − ⇔ (a + b) AG − bAB = 0 −→ −−→ −−→ → − ⇔ aAB + aBG + bBG = 0 −→ −−→ → − ⇔ −aBA + (a + b) BG = 0
(1)
(2)
And by (2) and (3) we get the proof !
Drawing of
G.
−→ Considering the equality AG = two weighted points.
b −→ AB, it is easy to draw the center of mass of a given set of a+b
If a and b are both positive or both negative, then 0 ≤
b ≤ 1 so G belongs to the segment a+b
line [AB] b is positive and greater than 1 and G belongs to the line (AB) a+b b outside of [AB] on the same side as B. If b < −a then is negative and G belongs to the line a+b (AB) outside of [AB] on the same side as A. By repeating the same thing in the case b < 0, we get the rule : If a.b > 0, G belongs to the segment line [AB], else to the straight line (AB) outside of [AB], anyway closer to the point with the greater absolute value of weight than to the other. If a < 0 and b > −a then
|a| > |b| A
|b| > |a| M
B a=b
a.b > 0
Barycentre (centroid) Definition: If a = b then the the barycentre of {(A, a) , (B, b)} is named centroid of A and B. −→ −−→ → − AG + BG = 0 −→ −−→ ⇔ AG = GB The barycentre of A and B is the midpoint of [AB]
Fundamental theorem Theorem: For a + b 6= 0, the system {(A, a) , (B, b)} admit a center of mass G if and only if whatever the point M −−→ −−→ −−→ aM A + bM B = (a + b) M G Proof: Introducing the G point in the different vectors, using the Chasles’s rule, it comes: −−→ −−→ −−→ aM A + bM B = (a + b) M G −−→ −→ −−→ −−→ −−→ ⇔ a M G + GA + b M G + GB = (a + b) M G −−→ −→ −−→ −−→ ⇔ (a + b) M G + aGA + bGB = (a + b) M G −→ −−→ → − ⇔ aGA + bGB = 0
2. Barycentre of three points (or more) Consider A, B and C three points and a, b and c three real numbers such that a + b + c 6= 0, the −→ −−→ −→ → − vectorial equality aGA + bGB + cGC = 0 is equivalent to −→ −−→ −→ −→ → −→ − aGA + b GA + GB + c GA + AC = 0 −→ −→ −→ → − ⇔ (a + b + c) GA + bAB + cAC = 0 −→ −→ −→ c b AB + AC ⇔ AG = (a + b + c) (a + b + c) So there is only one G point satisfying this equality and we can draw it easily.
Definition:
−→ −−→ −→ → − If a + b + c 6= 0 then it exists only one point G satisfying aGA + bGB + cGC = 0 The G point is named barycentre of the set {(A, a) , (B, b) , (C, c)}
Remark: By the same way, we can define the center of mass a set {(A1 , a1 ) , (A2 , a2 ) , (A3 , a3 ) · · · (An , an )} n X when ai 6= 0 i=1
Property:
−→ −−→ −→ → −→ −−→ −→ − → − If λ ∈ R∗ , aAG + bBG + cCG = 0 ⇔ λaAG + λbBG + λcCG = λ 0 . So if a + b 6= 0 and λ 6= 0 then the barycentre of {(A, a) , (B, b) , (C, c)} is the barycentre of {(A, λa) , (B, λb) , (C, c)}
Drawing the barycentre of three points. Some examples : • The barycentre of
−→ −→ −→ −→ A, 21 , B, 21 , (C, 2) using the equality 3AG = 12 AA + 21 AB + 2AC
A
Theorem of sub-mass centration. Theorem: If a + b + c 6= 0 and a + b 6= 0, then the barycentre of {(A, a) , (B, b) , (C, c)} is the center of mass of {(G1 , a + b) , (C, c)} where G1 is the barycentre of {(A, a) , (B, b)} In other words, in a calculation of a barycentre, we can replace a subset of weighted points by its center of mass weighted with the total weight of the weighted points replaced. Proof : If a + b 6= 0, the set {(A, a) , (B, b)} have a barycentre G1 satisfying −−→ −−→ −−−→ aM A + bM B = (a + b) M G1
∀M, and then the equality
−−→ −−→ −−→ −−→ aM A + bM B + cM C = (a + b + c) M G
becomes
−−−→ −−→ −−→ (a + b) M G1 + cM C = (a + b + c) M G.
So it is equivalent to write ”G is the barycentre of {(A, a) , (B, b) , (C, c)} and G is the barycentre of {(G1 , a + b) , (C, c)} Remark: In one way, this theorem is very useful to determine the barycentre of a set of more than 2 weighted points without vectorial calculation. In the other way, it helps us to prove that a point belongs to different straight lines.
Example: Using this theorem, considering the barycentre G of a triangle ABC, we have G is the barycentre of {(C 0 , 2) , (C, 1)} where C 0 is the barycentre of {(A, 1) , (B, 1)} so the midpoint of [AB] . Therefore, C 0 , C and G are collinear. This proves that G belongs to the median passing through C. By an analog method, we prove that G belongs to the other two medians of the triangle, and that it is the intersection points of the median of the triangle ABC. Moreover, G is the barycentre of {(C 0 , 2) , (C, 1)} is equivalent to −→ 2 −−→0 d CG = CC 3 2 which proves that the centroid lies on the median at . 3 B e
C0 G A
B0
C
Example: The barycentre G of {(A, 3) , (B, 2) , (C, 1)} is the barycentre of {(G1 , 5) , (C, 1)} where G1 is the barycentre of {(A, 3) , (B, 2)} So we can draw G1 satisfying −→ 2 −→ AG1 = AB 5 and then G such that
−→ 5 −−→ CG = CG1 6
A smarter choice would be to draw G2 the center of mass of {(B, 2) , (C, 1)} using −−→ 1 −−→ BG2 = BC 3 and then G as the midpoint of [AG2 ]. B G2 G1 G A
C
3. In a frame. − − Consider the plane framed by (O, → u ,→ v ) then using the fundamental theorem for O in M , we have (a + b + c 6= 0) −→ −−→ −→ −→ aOA + bOB + cOC = (a + b + c) OG and the coordinates of G are given by : −→ −−→ −→ −→ a b c OA + OB + OC = OG (a + b + c) (a + b + c) (a + b + c) So
G
axA + bxB + cxC ayA + byB + cyC , a+b+c a+b+c
In other words, the coordinates of the center of mass are the weighted average of the coordinates of the involved points.
4. Centre of mass of a homogeneous plate. The problem of this part is to find the place of the center of mass of a plate with an homogeneous plate. First, a briefly reminder of some physics principles: 1. The centroid of a triangle is the intersection point of its medians. 2. If the plate has a centre of symmetry I, then I is its center of mass. 3. If the plate has an axe of symmetry ∆, then its center of mass belongs to ∆ 4. The center of mass of the plate P is the barycentre of {(G1 , m1 ) , (G2 , m2 )} where G1 is the center of mass of P1 with mass m1 and G1 is the center of mass of P2 with mass m2 .
G1 m1
G2 m2 Then two worked examples:
• Centre of mass of a quadrilateral homogeneous plate ABCD Generally, the center of mass of the plate is not the barycenter of the four vertex. Let’s name G1 , G2 , G3 and G4 the centroids of ABD, BCD, ABC and ACD respectively. By the principle 4, the centroid of the plate is the centroid of G1 and G2 on one hand and the centroid of G3 and G4 on the other hand. So G belongs to the straight lines (G1 G2 ) and (G3 G4 ). We can now draw it as the intersection point of this two lines. Then we know that the centroid I of the set {A, B, C, D} is centroid of the midpoints of [AC] and [BD] . In other words, it is the midpoint of the segment lines joining the midpoints of the diagonals. B C G3 M1
G2
I G G1
A
G4
M2
D
• Centre of mass of a wheel with a hole
A homogeneous wheel with a radius 4 has a circular hole with radius 2 as shown on the figure. How to find its center of mass ? First, the principle 2 tell us that the center of mass G is on the straight line of the centers. Then we can consider this wheel has a plain wheel from which we take out the circular hole. If we fill the hole by a disk of center B and radius 2 made of the same material as the wheel, we will get A center of mass of the obtained wheel (center of symmetry) and barycentre of {(G, mT − mH ) , (B, m mT and mH are respectively the mass of the plain disk and of the removed disk (the hole) So G is the barycentre of {(A, mT ) , (B, −mH )} Because the wheel is homogeneous, the mass are proportional to the areas and we just have to find G the barycentre of {(A, 16π) , (B, −4π)} or of {(A, 4) , (B, −1)} and −→ −1 −→ AB AG = 3
a
GA
B
