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Topic 2.1 Extended Q – Center of mass 4

y. y = f ( x ). L. 0. y. x. y = f ( x ). y cm. x. 0. L. x cm. Topic 2.1 Extended Q – Center of mass 4.  In the last section you learned how to find the cm of bodies built of symmetric plates.  In this section you’ll learn a general way to find the cm of a plate of any shape.

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Topic 2.1 Extended Q – Center of mass 4

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  1. y y = f (x) L 0 y x y = f (x) ycm x 0 L xcm Topic 2.1 ExtendedQ – Center of mass 4 In the last section you learned how to find the cm of bodies built of symmetric plates. In this section you’ll learn a general way to find the cm of a plate of any shape. Consider the following generalized plate, which has an edge cut in the shape of the function y = f(x): Suppose we lay our plate out flat, and slide rulers to find xcm and ycm as we did before: Note that xcm is found using a ruler that is perpendicular to the x-axis. And ycm uses a ruler that is perpendic-ular to the y-axis.

  2. y M A M L M V λ = ρ = σ = y = f (x) L 0 xcm x ycm mass density Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). Now we need to talk about three types of mass density. Linear mass densityλismass per unit length: Area mass densityσismass per unit area: Volume mass densityρismass per unit volume: this is the one you used in chemistry

  3. y y = f (x) L 0 xcm x ∫ L area of plate f(x)dx ycm A = 0 Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). The first step in solving asymmetric plates is to find the area of the plate. This is given by

  4. ∫ L f(x)dx 0 y M A σ = M y = f (x) L 0 xcm x mass density of plate ycm = Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). Why did we use σinstead of λorρ? The second step is to find the mass density of the plate. This is given by

  5. y x2 y = f (x) x5 x4 x3 L 0 x1 Δ m14 Δ m3 Δ m5 Δ m7 Δ m9 Δ m11 Δ m2 Δ m4 Δ m13 Δ m6 Δ m8 Δ m15 Δm1 Δ m10 Δ m12 x Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). etc. f(xi) Δx To find xcm we divide our plate into many vertically oriented rectangles of material having mass Δmi. Since our rectangles are vertical, all of the mass Δmi in each rectangle is, on average, the same distance xi from the origin. Each rectangle has dimensions Δx by f(xi) so that = σf(xi)Δx. Δmi = σΔAi

  6. y x2 y = f (x) x4 x5 x3 L 0 x1 m14 m3 m5 m7 m9 m11 m2 m4 m13 m6 m8 m15 m1 m10 m12 x 15 1 M ∑ 15 Δmixi xcm= 1 M ∑ σxif(xi)Δx = i=1 i=1 ∫ ∫ xcmfor asymmetric plate 15 1 M L L ∑ xiΔmi = 1 M 1 M xdm σ xf(x)dx xcm = = i=1 0 0 Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). etc. f(xi) Δx Now we invoke our xcm formula for discrete masses, and tailor it slightly with the substitution Δmi = σf(xi)Δx: As Δx→ 0,Δmi = σf(xi)Δx becomes dm = σf(x)dx so that

  7. y Δm7 Δm6 Δm5 Δm5 y = f (x) Δm4 y4 y3 Δm3 L 0 y2 Δm2 y1 Δm1 x Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, 0 ≤ y ≤ f(L). etc… Δy L - xi To find ycm we divide our plate into many horizontally oriented rectangles of material having mass Δmi. Since our rectangles are horizontal, all of the mass Δmi in each rectangle is, on average, the same distance yi from the origin. Each rectangle has dimensions Δy by L - xi so that = σ(L-xi)Δy. Δmi = σΔAi

  8. y Δm7 Δm6 Δm5 Δm5 y = f (x) Δm4 y4 y3 Δm3 L 0 Δm2 y1 Δm1 x 7 1 M ∑ 7 Δmiyi ycm= 1 M ∑ σyi(L-xi)Δy = i=1 i=1 ∫ ∫ ycmfor asymmetric plate 7 1 M f(L) f(L) ∑ yiΔmi = 1 M 1 M ydm σ y(L-x)dy ycm = = i=1 0 0 Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, 0 ≤ y ≤ f(L). Δy L - xi y2 Now we invoke our ycm formula for discrete masses, and tailor it slightly with the substitution Δmi = σ(L-xi)Δy: As Δy→ 0,Δmi = σ(L-xi)Δy becomes dm = σ(L-x)dy so that

  9. y y = f (x) L 0 xcm x the mass increment ycm Topic 2.1 ExtendedQ – Center of mass 4 Note that 0 ≤x≤L, 0 ≤ y ≤ f(L). The formulas, and their derivations, are rather complicated and hard to memorize. So let’s attack this problem from a different direction: Now we define the mass increment dm. We will use dm in place of the point mass: Linear mass increment: dm = λdℓ Area mass increment: dm = σdA Volume mass increment: dm = ρdV

  10. y y = f (x) L 0 xcm x n n ∫ L miyi mixi 1 M ∑ ∑ xdm xcm= i=1 i=1 0 ycm 1 M 1 M f(L) ∫ xcm= ycm= 1 M ydm ycm= 0 cm, discrete cm, continuous Topic 2.1 ExtendedQ – Center of mass 4  Note that 0 ≤x≤L, and 0 ≤ y ≤ f(L). It is crucial that horizontal rectangles are used for ycm. Why? It is crucial that vertical rectangles are used for xcm. Why? We then rewrite the discrete formulas for xcm and ycm: dm dm

  11. a b a b y = x y = x. Topic 2.1 ExtendedQ – Center of mass 4 y a Note 0 ≤x≤b, 0 ≤ y ≤ a. y = f (x) x b 0 The important thing is that we can use these formulas correctly. Let’s look at a concrete example: Suppose we wish to find the center of mass of a right triangle having a mass M, a base b and an altitude a: The equation of the line y = f(x) is

  12. a b ∫ b = x 1 M xdm xcm= 0 a b = σ xdx a b y = x Topic 2.1 ExtendedQ – Center of mass 4 y a Note 0 ≤x≤b, 0 ≤ y ≤ a. y dA = ydx x dx b 0 To find xcm use vertical rectangles : dm, where dm = σdA. Then… dm = σdA = σydx

  13. M A 1 3 σ= - 03 σa bM 1 3 b3 = ∫ σa bM b x2dx 2M ba = 2Mab2 3baM a b ∫ = σ= b 1 M xσ xdx σab2 3M 0 1 2 = xcm= A= ba 2b 3 xcm triangle a b ∫ 0 σab3 3bM b 1 M xcm = y = x xdm xcm= = 0 Topic 2.1 ExtendedQ – Center of mass 4 y a Note 0 ≤x≤b, 0 ≤ y ≤ a. x b 0 Then and But . Thus so that

  14. x ∫ y is the variable a 1 M ydm ycm= 0 a b b a y = x, x = y b a a b dm = σ(b - y)dy. y = x Topic 2.1 ExtendedQ – Center of mass 4 y a Note 0 ≤x≤b, 0 ≤ y ≤ a. Always match the variables before integrating. b - x dy dA = (b - x) dy x b 0 To find ycm use horizontal rectangles : dm, where dm = σdA. Then… since dm = σdA so that = σ(b - x)dy

  15. ∫ a b a 1 M yσ(b - y)dy = ∫ 1 a a σb M a (y - y2)dy σb M 1 3a = 0 - y3 = 0 0 M A σ= 1 2 y2 2M ba 2Mba2 6baM = σ= σba2 6M 1 2 ycm= A= ba 1a 3 ycm triangle a b ∫ σba2 6M a 1 M ycm = y = x ydm ycm= = 0 Topic 2.1 ExtendedQ – Center of mass 4 y a Note 0 ≤x≤b, 0 ≤ y ≤ a. x b 0 Then and But . Thus so that

  16. a 3 a 3 a 3 xcm b 3 b 3 b 3 ycm 1a 3 2b 3 xcm triangle ycm triangle xcm = ycm = Now you may use triangular plates just like squares, rectangles, circles, etc. to solve symmetric plate sums and differences. Topic 2.1 ExtendedQ – Center of mass 4 y a x b 0 To summarize… The cm of a right triangle is located one third of the way from the right angle along each side.

  17. H H 2 Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (a) Where is the cm before the soda begins to drain? Since the cm of the can, and the cm of the soda are located in the center, h = H/2. (b) Where is the cm after the soda has finished draining? Since the cm of the can is still located in the center, h = H/2.

  18. H x But the mass of the soda in the can is proportional to x so that msoda(x) = x. m H Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (c) If x is the height of the remaining soda at any time, at what value of x is the cm of the can-soda combo the lowest? Note that the cm of the soda is at x/2, and the cm of the can is at H/2. Note that mass of the can is always mcan = M. You can check that the latter works by substituting 0 and H for x.

  19. H x m H m H ·x· ·x Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (c) If x is the height of the remaining soda at any time, at what value of x is the cm of the can-soda combo the lowest? Now we can obtain the cm h(x)of the two-body combo: x 2 H 2 + M· h(x) = + M (multiply top and bottom by 2H) + mx2 MH2 h(x) = + 2MH 2mx

  20. H x Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (c) If x is the height of the remaining soda at any time, at what value of x is the cm of the can-soda combo the lowest? Now we can obtain the cm h(x)of the two-body combo: (get into product form) h(x) = (MH2 + mx2)·(2MH + 2mx)-1 (take the derivative) (2mx) ·(2MH + 2mx)-1 h'(x) = (MH2 + mx2) (-1)(2MH + 2mx)-2 + ·2m

  21. H x Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (c) If x is the height of the remaining soda at any time, at what value of x is the cm of the can-soda combo the lowest? Now we can obtain the cm h(x)of the two-body combo: (set the derivative equal to zero to minimize) 0 = 2mx(2MH + 2mx)-1 2m + (MH2 + mx2)·(-1)(2MH + 2mx)-2· -1 2m(MH2 + mx2)(2MH + 2mx)-2 = 2mx(2MH + 2mx)-1 (MH2 + mx2) = x(2MH + 2mx)

  22. H a b c x -2MH ± √4M2H2 + 4mMH 2 x = 2m -2MH ± 2H√M(M+ m) H m √M(M+m) - M = = 2m Topic 2.1 ExtendedQ – Center of mass 4 EXERCISE 12*: A can of mass M and height H is filled with soda of mass m. We punch small holes in the top and bottom so that the soda drains slowly. As it drains, the height h of the cm of the can-soda combo will change. (c) If x is the height of the remaining soda at any time, at what value of x is the cm of the can-soda combo the lowest? Now we can obtain the cm h(x)of the two-body combo: (MH2 + mx2) = x(2MH + 2mx) MH2 + mx2 = 2MHx + 2mx2 mx2+ 2MHx - MH2 = 0

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