CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM

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1 Chapter 4 CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM Figure 4.1: 4.1 Conservation of Angular Momentum Angular momentum is defined as the moment of the linear momentum about some spatial reference point. Thus, angular momentum is the cross product of position and linear momentum: (angular momentum) = r (linear momentum) Like linear momentum, angular momentum is conserved (or, to be more eact, balanced). Thus, like conservation of linear momentum, the same spatial and time derivatives eist; ecept the operate 97

2 98 CHAPTER 4. CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM on the angular momentum. Recall the equation describing the conservation of linear momentum: { (ρv) (ρv v) = + (ρv v) + (ρv } v) + (i) + (j) + (k) + ρg (4.1) The control volume (differential volume element) is located b the radius vector r = i + j + k. Bod forces act on it, tractions are applied to its boundar, and linear momentum enters it and leaves it, as shown in Figures 4.2 and 4.3. r 0 Figure 4.2: Control Volume at Position r The control volume is assumed to be at a fied reference point in time, i.e., it does not move, so that r is independent of time. Angular momentum of the sstem can now be determined with respect to the origin. To construct the conservation of angular momentum statement, we follow a procedure analogous to the one we used in Chapter 3 to develop the conservation of linear momentum. In the absence of applied bod moments, the differential balance of conservation of angular momentum reduces to (r ρv) = [ (r ρv v)+ (r ρv v)+ (r ρv v) + (r t (i))+ (r t (j))+ (r t (k))+ρg B epanding the partial derivatives of the product terms, one obtains ( ) r ρv + r (ρv) [ r = ρv v + r ρv v + r ρv v + r (ρv v) + r t (i) + r + r + r (ρv v) t (j) + r ( (i) + (j) + (k) ] + r (ρv v) t (k) ) + r ρg Note that the double-underlined term on the left-hand side of Equation (4.3) equal the doubleunderlined terms on the right-hand side: we know this b crossing r with the left- and right-hand ] (4.2) (4.3)

3 4.1. CONSERVATION OF ANGULAR MOMENTUM 99 ( ρ v v) + t() j + ( ρ v v) t( k) t( i) ( v) ρ v + ( ρ v v) ρ g t() i + t( k) + ( v) ρ v + ( ρ v v) t( j) Figure 4.3: 3-D Control Volume with Linear Momentum, Tractions and Bod Forces sides of the angular momentum conservation equation. We can thus eliminate these terms. B assuming that the position vector r is fied relative to time so that r = 0, Equation (4.3) then reduces to: 0 = r ρv v + r ρv v + r ρv v + r t (i) + r t (j) + r t (k) (4.4) Recall that the position vector for differential volume located at (,, )isgiven b: Using (4.5), equation (4.4) becomes 0 = r = i + j + k = r = i, r = j, r = k (4.5) 0 = i ρv v + j ρv v + k ρv v + i t (i) + j t (j) + k t (k) (4.6) Epanding the cross-products in the brackets gives the following: ρ [v v k + v v ( j)+v v ( k)+v v i + v v j ++v v ( i)] = ρ [(v v v v )k)+( v v + v v )j +(v v v v )i)] = 0 (4.7) Therefore, equation (4.6) reduces to 0 = i t (i) + j t (j) + k t (k) (4.8)

4 100 CHAPTER 4. CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM Cauch s formula, Equation (3.32), which relates tractions to stresses, can now be substituted into equation (4.8) to obtain 0 = i (iσ + jσ + kσ )+j (iσ + jσ + kσ )+k (iσ + jσ + kσ ). (4.9) Distributing the cross-products over addition ields 0 = kσ jσ kσ + iσ + jσ iσ =(σ σ )k (σ σ )j +(σ σ )i. (4.10) From equation (4.10), observe that conservation of angular momentum is satisfied if and onl if the coefficients of the unit vectors are ero. Thus, one obtains three relations that must be satisfied: Conservation of Angular Momentum Requirement = σ = σ, σ = σ, σ = σ. (4.11) From this result, one can tell that the stress matri is smmetric. Hence, it has onl si independent components: σ, σ, σ, σ (or σ ), σ (or σ ), and σ (or σ ). Eample 4.1 In order to visualie the conservation of angular momentum without all the cross products, the following eample uses onl what ou have learned to this point and the conservation of angular momentum equation used in ENGR 211 to derive the same result. Below is a differential volume element. We will assume static case (all components of velocit are ero), and sum the moments about the center of the element. Assume the bod forces act through the center of the differential volume element. σ σ σ σ ρg ρg σ σ σ σt Figure 4.4: M =0 Since the normal components and the bod forces act through the center and we are summing moments about that point the will not contribute and we are left with

5 4.1. CONSERVATION OF ANGULAR MOMENTUM 101 σ ( ) + σ 2 In the above equation, we multipl the stress force area b its moment arm (distance from the center). Dividing b gives ( ) ( σ ) ( σ ) = σ = σ b the area it acts over to get force and then which is the same result found before. Note: ifabod moment per unit volume m [ ] N m m is applied to the bod, then the conservation of angular momentum takes the final form: 0 = i(σ σ )+im +j(σ σ )+jm +k(σ σ )+km = σ σ = m σ σ = m σ σ = m (4.12) In the presence of bod moments, therefore the stress matri is not smmetric! For the remainder of this course, the assumption will alwas be made that there are no bod moments applied, and therefore the stress matri will alwas be smmetric. The stress components for 2-D and 3-D Cartesian coordinates and for clindrical coordinates are summaried in Figure 4.5.

6 102 CHAPTER 4. CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM Stress Components in Cartesian Coordinates σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ = σ σ = σ σ = σ σ σ σ Stress Components in Clindrical Coordinates σ σ θ σ θ σ θθ θ σ rθ σ θr σ rr σ r σ r σ r σ rθ σ θr σ θθ σ θ σ r σ rr r σ θ σ θ r σ rθ = σ θr σ r = σ r σ θ = σ θ Figure 4.5: Stress Components in Cartesian and Clindrical Coordinates

7 4.1. CONSERVATION OF ANGULAR MOMENTUM 103 Deep Thought Conservation of angular momentum helps those with skill and grace perform beautiful moves on ice. Unfortunatel for the rest of us, it is a nightmare.

8 104 CHAPTER 4. CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM 4.2 Questions 4.1 (a) What is the result from the conservation of angular momentum? (b) In reaching this result, what other conservation laws are emploed? Eplain how these laws plaed a role. (c) Does the conservation of angular momentum need to be eplicitl considered in solving problems with fluid flow if the stress matri is assumed to be smmetric? Wh? 4.3 Problems 4.2 A fluid with a densit of 25 lbm ft 3 at point (,, )=(4ft, 7 ft, 2ft),has a velocit of: Find the fluid s: v =2i +6j 5k, ft s a) mass flu b) linear momentum flu in relation to the three surfaces each defined b a unit normal aligned with their respective coordinate aes, c) the,, and components of each vector in (b), d) the angular momentum flu in relation to the three surfaces each defined b a unit normal aligned with their respective coordinate aes, and e) the,, and components of each of the three vectors in (d). 4.3 Same instructions as in 4.3, at a point (,, )=( 3 m, 1m, 4 m), fluid having densit of 33 kg m with a velocit of 3 v = i 6j +7k m s 4.4 Given the components of the stress matri in Cartesian coordinates as a function of (,, ): [σ] = MPa a) Show whether the equations of conservation of angular momentum are satisfied. b) Find the components of the traction vector on the plane with normal n = cos θi + sin θj for the material point located at (,, )=(1, 1, 1).

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