m, where m = m 1 + m m n.

Transcription

1 Lecture 7 : Moments nd Centers of Mss If we hve msses m, m 2,..., m n t points x, x 2,..., x n long the x-xis, the moment of the system round the origin is M 0 = m x + m 2 x m n x n. The center of mss of the system is x = M 0 m, where m = m + m m n. Exmple We hve mss of 3 kg t distnce 3 units to the right the origin nd mss of 2 kg t distnce of unit to the left of the origin on the rod below, find the moment bout the origin. Find the center of mss of the system. 2kg 3kg x By Lw of Archimedes if we hve msses m nd m 2 on rod (of negligible mss) on opposite sides of fulcrum, t distnces d nd d 2 from the fulcrum, the rod will blnce if m d = m 2 d 2. (or in generl if we plce msses m, m 2,..., m n t distnces d, d 2,... d n from the fulcrum the rod will blnce if the center of mss is t the fulcrum. ) If we plce fulcrum t the center of mss of the rod bove, we see tht the rod will blnce: Verify tht m d = m 2 d 2 or 3 x 3 = 2 x ( ). Note tht system with ll of the mss plced t the center of mss, hs the sme moment s the originl system. Check this for the system bove. (Tht is, check tht system with mss of 5 kg plced t x = 7 5 hs the sme moment nd center of mss)

2 For two dimensionl system, we use x nd y xes for reference. We now hve moments bout ech xis. If we hve system with msses m, m 2,..., m n t points (x, y ), (x 2, y 2 ),..., (x n, y n ) respectively, then the moment bout the y xis is given by nd the moment bout the x xis is given by M y = m x + m 2 x m n x n M x = m y + m 2 y m n y n. These moments mesure the tendency of the system to rotte bout the x nd y xes respectively. The Center of Mss of the system is given by ( x, ȳ) where x = M y m nd ȳ = M x m for m = m + m m n. Exmple Find the moments nd center of mss of system of objects tht hve msses kg 2 6 position (7, ) (0, 0) ( 3, 0) Note tht system with ll of the mss plced t the center of mss, hs the sme moments s the originl system. Check this for the system bove. If we hve thin plte (which occupies region R of the plne) with uniform density ρ, we re interested in clculting its moments bout the x nd y xis (M x nd M y respectively) nd its center of mss or centroid. These re defined in wy tht grees with our intuition. A line of symmetry of the plte (or region) is line for which 80 o rottion of the plte bout the line mkes no chnge in the plte s ppernce. The Centroid or center of mss lies on ech line of symmetry of the plte. Hence if we hve 2 different lines of symmetry, the centroid must be t their intersection. (Note tht it is not necessry for the center of mss to lie in the plnr region)

3 The moments of the plte or region re defined so tht if ll of the mss of the plte is centered t the centroid, the system hs the sme moments. Also the moments of the union of two non-overlpping regions should be the sum of the moments of the individul regions. Nturlly :) we use Riemnn sums to clculte the moments of region (or plte shped like tht region with constnt density) nd we will strt with region, R, under curve y = f(x) 0 for x b. In order to estimte the moments for such region (with constnt density ρ), we divide the intervl [, b] into n subintervls [x i, x i ], 0 i n, ech of length x = b. We pproximte the moment n of the region beneth the curve bove the subintervl [x i, x i ] by the moment of rectngle bove the subintervl [x i, x i ] of height f( x i +x i ). The centroid of this rectngle is C 2 i ( x i, f( x 2 i)) where x i = x i +x i. The re of this pproximting rectngle is xf( x 2 i ) nd hence its mss is ρ xf( x i ) For the rectngle R i we clculte its moment bout the y-xis s if it were system with ll of its mss t its center of mss C i ( x i, 2 f( x i)). Hence the moment of this rectngle bout the y-xis is M y (R i ) = mss x i = ρ xf( x i ) x i = ρ x i xf( x i ). We dd the moments for the pproximting rectngles to get n pproximtion of the moment of the entire region R; n M y ρ x i f( x i ) x. Tking limits we get the moment of the region R to be M y = lim n i= n ρ x i f( x i ) x = ρ i= xf(x)dx. Similrly we find the moment bout the x-xis for the pproximting rectngle R i to be M x (R i ) = mss ȳ i = ρ xf( x i ) 2 f( x i) = ρ 2 [f( x i)] 2 x. nd the moment bout the x-xis of the region R is given by M x = lim n n i= ρ 2 [f( x i)] 2 x = ρ 2 [f(x)] 2 dx. 3

4 The Center of Mss or centroid of the region R is given by ( x, ȳ) where m x = M y nd mȳ = M x where m is the mss of the entire region R. nd x = M y m = ρ xf(x)dx ρ f(x)dx = ȳ = M ρ x m = 2 [f(x)]2 dx ρ f(x)dx = Note tht the density hs no effect on the center of mss. xf(x)dx f(x)dx b 2 [f(x)]2 dx f(x)dx Exmple Find the centroid of the region bounded by the curve y = x 2, the x-xis, the line x = nd the line x = 2. If the region R is bounded bove by the curve y = f(x) nd below by the curve y = g(x), where f(x) g(x) 0, we hve the moments of plte with tht shpe nd constnt density ρ re given by M y = ρ x[f(x) g(x)]dx nd M x = ρ 2 [f(x)] 2 [g(x)] 2 dx. 4

5 nd the centroid of the region R is given by x = A where A denotes the re of the region R, x[f(x) g(x)]dx ȳ = 2A A = f(x) g(x)dx. [f(x)] 2 [g(x)] 2 dx Exmple Find the centroid of the region bounded bove by y = x + 2 nd below by the curve y = x 2. 5

6 Extrs Exmple Find the moment M 0 nd center of mss of system, consisting of rod with negligible weight, with mss of 2 kg plced 3 units to the right of the origin, mss of 6 kg plced 5 units to the left of the origin, mss of 0 kg plced 8 units to the right of the origin nd mss of 0 kg plced t the origin. M 0 = m x + m 2 x 2 + m 3 x 3 + m 4 x 4 = 6( 5) + 0(0) + 2(3) + 0(8) = = 56. x = M 0 m = 56 m + m 2 + m 3 + m 4 = = 2. Exmple Find the centroid of the region bounded by the curve y = x 2, the x-xis, the line x = nd the line x = 2. Covered in clss Find the moment bout the x-xis of plte with shpe described s in the previous exmple nd density ρ = /2 kg per unit re. Find the moment bout the y-xis of such plte. Using our clcultions from bove we get M x = ρ 2 M y = ρ 2 2 [ ] [f(x)] 2 dx = xf(x)dx = ln

7 Theorem of Pppus Ler R be plne region tht lies entirely on one side of line l in the plne. If R is rotted round the line l, the volume of the resulting solid is the product of A = the re of R nd the distnce trvelled by the centroid of R, ( x, ȳ). Exmple Find the volume of the solid generted by rotting the region between the curve y = (x 3) 2 nd the curve y = 2 4 (x 3) 2 bout the y xis. The solid looks like the doughnut shown below: The region between the bove curves is circle of rdius 2 stisfying the eqution (x 3) 2 + (y 2) 2 = 4. It is circle of rdius 2 centered t the point (3, 2). The centroid of tht region is on the lines of symmetry of the region which meet t the center ( x, ȳ) = (3, 2). The re of the region is 4π nd the distnce trvelled by the centroid is 2π(3) since the centroid is 3 units from the y xis. Hence by Pppus, the volume of the doughnut is 2π(3) 4π = 24π 2. 7