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Preston Terry
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1 ME 131B Fluid Mechanics Solutions to Week Eight Problem Session: Angular Momentum Principle (3/2/98) 1. In control volume analysis, all governing principles share the same common structure: storage = inow outow + production The main dierence between dierent physical principles is in the production term, P. For the following principles, what is this production term equal to? (a) conservation of mass Mass can neither be created nor destroyed. P mass = (b) conservation of linear momentum According to Newton's law of motion, we can change the momentum of a system by apply an external force on it. P lin:mom: = X ~ F There are two main types of forces: { Surface force It is present along the control surfaces of your selected control volume. The best way to identify the surface forces is to trace along the entire control surface and ask yourself the question \What force does my control volume experience along this surface? Some examples of surface force are pressure force (normal direction) and friction (tangential direction). { Body force It is present due to the contents inside the selected control volume under the inuence of the surrounding force eld. Some examples of body force are gravitational force and electrostatic force. (c) conservation of angular momentum If we draw the analogy between linear momentum in translational motion and angular momentum in rotational motion, external force will be analogous to external torque. Hence, P ang:mom: = X ~ 1

2 Since the angular momentum equation is derived from the linear momentum equation, all the external forces, ~ F, (both surface and body) in the linear momentum equation are capable of generating torque, ~r ~ F, on the same control volume as long as the line of action of the force does not pass through the center of rotation. (d) conservation of energy We can change the total energy of a system by adding heat () or doing work (W ) on the system P energy = in + W in 2. The Reynolds Transport Theorem is the core basis in control volume analysis. It serves as a bridge between the control mass and the control volume approach. We can state it as:! dn ( dv ) + V ~ d A ~ system Most physical laws are Lagrangian in nature, i.e. they are derived for a system with a xed amount of substance (control mass approach). However, this approach is not easy to follow for a uid system simply because a uid can be deformed continuously as it moves around in space. The Reynolds Transport Theorem relates the rate of change of an extensive property N of a control mass system! dn dt system with the rate of change of the same property in a ( dv ) + {z } storage ~ V d ~ A {z } outow - inow There are two major components in the above equation: { The rst one is the storage term which accounts for the rate of increase in property N within the control volume. { The second one is the net outow term which accounts for the loss of property N due to the uid motion in and out of the control volume. Only with the Reynolds Transport Theorem, we can then relate the physical laws to what we measure in a xed region in space (control volume approach). 2

3 For the following principles, what are the quantities N and? The quantity N is an extensive property of the system whereas the quantity is its intensive counterpart. (a) conservation of mass N = M (total mass of system), = 1 (b) conservation of linear momentum N = ~ M (total linear momentum of system), = ~ V (c) conservation of angular momentum N = ~ A (total angular momentum of system), = ~r ~ V (d) conservation of energy N = E (total energy of system), = u + j~ V j g z Remarks: With the results of uestion 1 and 2, we can summarize all the conservation laws in the ( dv ) + ~ V d ~ A = P 3. What is the main criterion in choosing a suitable control volume in problem solving? We should put the control surfaces at places where { we know how the ow behaves, for example, ( ~ V ; P ) { we want to know something about, for example, frictional shear, exit pressure. 4. From what physical principle is the angular momentum equation derived? The angular momentum equation is derived by taking the cross product between the position vector, ~r, and the linear momentum equation. Hence, its main physics comes from Newton's law of motion. The main dierence is that the linear momentum equation governs the translational motion while the the angular momentum equation governs the rotational motion of the system. 3

4 5. Choose the best answer in the following question: Pressure is always directed into the control volume of interest. Pressure is a compressive force. Hence, it is always directed into the system of interest. 6. The angular momentum principle can be expressed in the following two forms: Form 1: Form @t ~r ~ F s + ~r ~ V ~r ~ F s + (~r ~g) ( dv ) + ~ T shaft ( dv ) + ~r ~ V (~r ~g) ( dv ) + ~ T shaft ~ V d ~ A ~r h 2~! V ~ + ~! (~! ~r) + ~! _ i ~r ( dv ) ~r V ~ ( dv ) + ~r V ~ V ~ d A ~ (a) What is the main dierence between the above two forms? Form 1 is derived in an inertial frame. Form 2 is derived in a rotating (non-inertial) frame. We need to make sure that the velocity vector, ~ V, is consistent with the corresponding choice of reference frame when we invoke the angular momentum principle. As long as we use the two forms in a consistent manner, they should give identical results. (b) Give a verbal description to each term in the equations. ~r F ~ s is the torque generated by surface force, F ~ s. R (~r ~g)( dv ) is the torque generated by gravitational force. R ~r (2~! V ~ ) ( dv ) is the torque generated by Coriolis force. R ~r [~! (~! ~r)] ( dv ) is the torque generated by centripetal force. R ~r ( ~! _ ~r) ( dv ) is the \ctitious torque due to angular acceleration of the rotating reference (~r V ~ ) ( dv ) is the rate of increase in angular momentum within the control volume. R (~r V ~ ) ( V ~ d A) ~ is the net outow of angular momentum caused by uid motion in and out of the control volume. 4

5 7. A total water discharge of 2 cm 3 is issued from a sprinkler as shown in the following gure: A j =.1 cm 2 1 cm 1 cm 1 cm W j Wj W j o 4 W j W j W j 1 cm 1 cm 1 cm A o 4 Assume that the jet speed is the same from all the holes. We rst choose a control volume to include the entire sprinkler arm as indicated above. Let us solve this angular momentum problem using an inertial reference frame and see how the analysis works. The corresponding angular momentum equation is X ~r V ~ ( dv ) + ~r V ~ V ~ d A ~ We then examine every individual term in the above equation: { The sources of external torque in this problem come from shaft torque, ~ shaft frictional torque, ~ f { The storage term is zero because we are dealing with a steady problem. { The net angular momentum outow term can be evaluated by rst considering the jet out of one hole only: ω W j r i For this single jet case, r i ω ~r V ~ V ~ d A ~ = (r Vt ) ( W j A j ) ~e k = (r V t ) ~e k 6 where V t is the tangential velocity component measured in an inertial frame. 5

6 { Consider the relative motion equation ~ V = ~ U + ~ W, we can resolve it in the tangential direction as V t = U t + W t = r i! + W j sin ) r V t = r 2 i! + r i W j sin = r 2 i! + r i sin { Hence, the angular momentum outow from this hole is equal to r i sin r 2 i!! 6 { The total angular momentum outow can be obtained by summing the contribution from all six holes together ~r V ~ V ~ d A ~ = 2 3X i = 1 r i ~e k! sin r 2 i 6 A! # ~e k j 6 where r 1 = 1 cm, r 2 = 2 cm, r 3 = 3 cm. The entire angular momentum equation can then be simplied to shaft + f = 3 # (r 1 + r 2 + r 3 ) sin! r r r 2 3 This general equation of motion forms the common basis for the following special cases of interests. (a) Static Case: Determine the torque that must be applied to the sprinkler arms to hold them from rotating. In this static case, we have {! = { f = The restraining torque is shaft = 2 18 A j! (r 1 + r 2 + r 3 ) sin (b) Frictionless Case: Determine the angular speed if the arms are free to rotate and there is no friction. In this frictionless case, we have 6

7 { shaft = { f = The angular speed of the sprinkler arm is! = r 1 + r 2 + r 3 r r r 2 3 sin (c) Frictional Case: Determine the angular speed if there is a constant frictional torque of 1 N-m resisting rotation of the arms. In this frictional case, we have { shaft = The angular speed of the sprinkler arm is! = 1 r r r 2 3 (r 1 + r 2 + r 3 ) sin which reduces to the results in Part (b) for f =. # 3 f (d) Relative Motion: Determine the absolute velocity of the uid leaving Hole A in Part (b) and (c). In both cases, we can apply the relative motion equation V ~ = U ~ + W ~ to analyze the velocity components. { Radial component: { Tangential component: V r = U r + W r = + W j cos = cos V t = U t + W t = r 3! + W j sin = sin r 3! The absolute velocity of the uid is given by V = = = q vu u t vu u t V 2 t + Vr 2! 2 sin r 3! +! 2 cos! 2 + (r 3!) 2 3 A j r 3! sin 7

8 (e) Sketch the corresponding velocity vector diagrams for Part (d). W = V r = W r V V t = U W t U = r 3 ω 8. Refer to the schematic below, a \wye joint splits a pipe ow into two equal amounts, =2, which exit at a distance R from the x-axis. The system rotates about the x axis at a rate. /2 Τ, Ω R >> D pipe R x /2 (a) Inertial Frame Analysis: Apply the angular momentum principle in an inertial frame to i. determine the torque required to turn the pipe (constant speed). ii. determine the additional torque which is required to generate an angular acceleration _ on the existing system (constant acceleration). The angular momentum equation in an inertial frame is X ~r ~ V ( dv ) + ~r ~ V ~ V d ~ A Let us rst consider a control volume which includes the upper tube only: 8

9 1 /2 d r Τ, Ω r R y x z According to the coordinates system chosen, we can express the position and velocity vectors as ~r = r cos + sin ~j cos + sin ~j + r sin ~ k ~V = 2 A We can then perform the cross product evaluation as follows: ~r V ~ = r cos + sin ~j cos + sin ~j + r sin ~ k 2 A r = 2 A sin cos ~ i ~j + r2 sin cos + r 2 A sin cos ~ j + r2 sin 2 = r 2 sin sin cos ~j ~ j ~ k ~ i ~ k The total angular momentum stored in the upper tube can be obtained by integrating the above expression along the entire upper tube length ~r V 1 ~ R = sin h i ( dv ) = r 2 sin sin cos ~j ( A dr) = A R3 sin cos ~j 2 The storage term can then be obtained by take the time derivative of the above expression. (Notice that only the angular speed is a function of ~r 1 ~ A R 3 _ ( dv ) = sin cos ~j 2 The angular momentum outow term can be obtained by evaluating the ~r V ~ expression at r = R = sin 1 ~r ~ V ~ V d ~ A = R 2 sin sin cos ~j # 2 9

10 These procedures conclude the analysis of the upper tube. We can then proceed to analyze the lower tube by following the same procedures. Τ, Ω y r R z x d r 2 /2 The main dierences will be in the expressions of position and velocity vectors ~r = r cos sin ~j ~V = cos sin ~j r sin 2 A ~ k After algebraic manipulations, we ~r 2 ~ A R 3 _ ( dv ) = sin + cos ~j 2 and ~r V 2 ~ V ~ d A ~ R 2 = sin To obtain the global conservation equation for the entire system, we need to sum up the results from the upper and the ~r ~ V ( dv ) + ~r ~ V sin + cos ~j # 2 ~ V d ~ A = R A R3 _ From the angular momentum equation in an inertial frame, we deduce that the applied torque is ~ = R A R3 _ The rst term corresponds to the torque required to turn the pipe at constant speed,, while the second term corresponds to the additional torque required to produce an angular acceleration, _, on the existing system. Hence, ~ = R 2 ~ _ = 2 A _ R3!! 1

11 (b) Rotating Frame Analysis: Repeat the analysis in Part (a) in a rotating frame. The angular momentum equation in a rotating frame is ~ ~r h 2~! ~ V + ~! (~! ~r) + _ ~! ~r i ( dv ) ~r V ~ ( dv ) + ~r V ~ V ~ d A Let us rst consider a control volume which includes the upper tube only: 1 /2 d r Τ, Ω r R y x z According to the coordinates system chosen, we can express the position, velocity and angular velocity vectors as ~r = r cos + sin ~j ~V = cos + sin ~j 2 A ~! = Since ~r is collinear to ~ V, their cross product is equal to zero ~r ~ V = ~ Hence, the storage term and the net outow term both equal zero and do not contribute to the angular momentum balance in the rotating frame. Let us evaluate each cross product term in the \ctitious torque carefully: 2 ~! ~ V = 2 2 A = A sin ~ i ~j = A sin ~ k cos + sin ~j ~! ~r = h i r cos + sin ~j ~ i ~j = r sin = r sin ~ k 11

12 These combine to give and ~! (~! ~r) = r sin ~ k ~ i k ~ = r 2 sin = r 2 sin ~j _~! ~r = _ h i r cos + sin ~j ~ i ~j = r _ sin = r _ sin ~ k 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r = sin ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ i ~r = r sin 2 A A + r _ ~ k r 2 sin ~j + r _ r sin cos A + r _ ~j r 2 2 sin cos ~ k The combined \ctitious torque can be obtained by integrating the above expression along the entire upper tube length i ( dv ) = 1 ~r h 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r R = sin = ( A) R 2 r sin 2 A i r 2 2 sin cos ~ k ( A dr) 2 A + R _! # R 3 2 cos ~ k 2 + r _ r sin cos R 2 cos sin A + r _ ~j 2 A + R _! ~j These procedures conclude the analysis of the upper tube. We can then proceed to analyze the lower tube by following the same procedures. 12

13 Τ, Ω y r R z x d r 2 /2 The main dierences will be in the expressions of position and velocity vectors ~r = r cos sin ~j ~V = cos sin ~j 2 A After algebraic manipulations, we obtain = ( A) ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ ~r 2 R 2 2 A + R _ # ~ k + R3 2 cos 2! + R2 cos sin i ( dv ) 2 A + R _! ~j To obtain the global conservation equation for the entire system, we need to sum up the results from the upper and the lower tubes. = ( A) ~r h 2 ~! ~ V + ~! (~! ~r) + _ ~! ~r i ( dv ) 2 R 2 2 A + R _! # From the angular momentum equation in a rotating frame, we deduce that the applied torque is ~ = = ( A) ~r h 2 ~! V ~ + ~! (~! ~r) + ~! _ i ~r ( dv ) 2 R 2 2 A + R _! # The rst term corresponds to the torque required to turn the pipe at constant speed,, while the second term corresponds to the additional torque required to produce an angular acceleration, _, on the existing system. Hence, ~ = R 2 ~ _ = 2 A _ R3 13

14 This example demonstrates that we can obtain the same results by choosing either the inertial or rotating reference frame to analyze angular momentum problems as long as the velocity vector is consistent with the corresponding chosen reference frame. In this problem, { the production, the storage and the net outow terms are all active in the inertial frame analysis; { the production term and the \ctitious torque are both active but the storage and the net outow terms are both zero in the rotating frame analysis. 14

Chapter 8. Centripetal Force and The Law of Gravity

Chapter 8. Centripetal Force and The Law of Gravity Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration