1 / 36 CEE 271: Applied Mechanics II, Dynamics Lecture 33: Ch.19, Sec.1 2 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Thursday, December 6, 2012
2 / 36 LINEAR AND ANGULAR MOMENTUM, PRINCIPLE OF IMPULSE AND MOMENTUM Today s objectives: Students will be able to 1 Develop formulations for the linear and angular momentum of a body. 2 Apply the principle of linear and angular impulse and momentum. In-class activities: Reading Quiz Applications Linear and Angular Momentum Principle of Impulse and Momentum Concept Quiz Group Problem Solving Attention Quiz
3 / 36 READING QUIZ 1 The angular momentum of a rotating two-dimensional rigid body about its center of mass G is. (a) mv G (b) I G v G (c) mω (d) I G ω ANS: (d) 2 If a rigid body rotates about a fixed axis passing through its center of mass, the body s linear momentum is. (a) a constant (b) zero (c) mv G (d) I G ω ANS: (b)
4 / 36 APPLICATIONS The swing bridge opens and closes by turning using a motor located under the center of the deck at A that applies a torque M to the bridge. If the bridge was supported at its end B, would the same torque open the bridge in the same time, or would it open slower or faster? What are the benefits of making the bridge with the variable depth (thickness) substructure as shown?
5 / 36 APPLICATIONS(continued) As the pendulum of the Charpy tester swings downward, its angular momentum and linear momentum both increase. By calculating its momenta in the vertical position, we can calculate the impulse the pendulum exerts when it hits the test specimen. As the pendulum rotates about point O, what is its angular momentum about point O? ANS: H = I O ω
6 / 36 APPLICATIONS(continued) The space shuttle has several engines that exert thrust on the shuttle when they are fired. By firing different engines, the pilot can control the motion and direction of the shuttle. If only engine A is fired, about which axis does the shuttle tend to rotate? ANS: +x-axis
7 / 36 LINEAR AND ANGULAR MOMENTUM (Section 19.1) The linear momentum of a rigid body is defined as L = mv G This equation states that the linear momentum vector L has a magnitude equal to (mv G ) and a direction defined by v G. The angular momentum of a rigid body is defined as H G = I G ω Remember that the direction of H G is perpendicular to the plane of rotation.
8 / 36 LINEAR AND ANGULAR MOMENTUM (continued) Translation only When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because ω = 0 Therefore, L = mv G and H G = 0
9 / 36 LINEAR AND ANGULAR MOMENTUM (continued) Rotation about a fixed axis. When a rigid body is rotating about a fixed axis passing through point O, the body s linear momentum and angular momentum about G are: L = mv G and H G = I G ω It is sometimes convenient to compute the angular momentum of the body about the center of rotation O. H O = (r G mv G ) + I G ω = I O ω
10 / 36 LINEAR AND ANGULAR MOMENTUM (continued) General plane motion. When a rigid body is subjected to general plane motion, both the linear momentum and the angular momentum computed about G are required. L = mv G H G = I G ω The angular momentum about point A is then (in magnitude) H A = I G ω + mv G (d)
11 / 36 PRINCIPLE OF IMPULSE AND MOMENTUM (Section 19.2) Linear impulse-linear momentum equation: t2 L 1 + Σ F dt = L 2 t 1 (mv G ) 1 + Σ t2 t 1 F dt = (mv G ) 2 Angular impulse-angular momentum equation: t2 (H G ) 1 + Σ M G dt = (H G ) 2 t 1 I G ω 1 + Σ t2 t 1 M G dt = I G ω 2
12 / 36 PRINCIPLE OF IMPULSE AND MOMENTUM (continued) The previous relations can be represented graphically by drawing the impulse-momentum diagram. To summarize, if motion is occurring in the x y plane, the linear impulse-linear momentum relation can be applied to the x and y directions and the angular momentum-angular impulse relation is applied about a z-axis passing through any point (i.e., G). Therefore, the principle yields three scalar equations describing the planar motion of the body.
13 / 36 PROCEDURE FOR ANALYSIS 1 Establish the x, y, z inertial frame of reference. 2 Draw the impulse-momentum diagrams for the body. 3 Compute I G, as necessary. 4 Apply the equations of impulse and momentum (one vector and one scalar or the three scalar equations). 5 If more than three unknowns are involved, kinematic equations relating the velocity of the mass center G and the angular velocity ω should be used to furnish additional equations.
14 / 36 EXAMPLE Given: The 300 kg wheel has a radius of gyration about its mass center O of k O = 0.4 meter. The wheel is subjected to a couple moment of M = 300 N m. Find: The angular velocity after 6 seconds if it starts from rest and no slipping occurs. Plan: Time as a parameter should make you think Impulse and Momentum! Since the body rolls without slipping, point A is the center of rotation. Therefore, applying the angular impulse and momentum relationships along with kinematics should solve the problem.
EXAMPLE (Solution) Impulse-momentum diagram: Mt W t I G ω 1 mv G1 + = F t Nt I G ω 2 mv G2 Impulse and Momentum using kinematics: (v G ) 2 = rω 2 : (H A ) 1 + Σ t2 M A dt = (H A ) 2 0 + Mt = m(v G ) 2 r + I G ω 2 = mr 2 ω 2 + m(k O ) 2 ω 2 11.5 rad/s = m [ r 2 + (k O ) 2] ω 2 ω 2 (t = 6 s) = 300(6) 300(0.6 2 + 0.4 2 ) = 15 / 36
CONCEPT QUIZ 1 If a slab is rotating about its center of mass G, its angular momentum about any arbitrary point P is its angular momentum computed about G (i.e., I G ω). (a) larger than (b) less than (c) the same as (d) None of the above ANS: (c) 2 The linear momentum of the slab in question 1 is. (a) constant (b) zero (c) increasing linearly with time (d) decreasing linearly with time ANS: (b) 16 / 36
GROUP PROBLEM SOLVING Given: A gear set with: m A = 10 kg m B = 50 kg k A = 0.08 meter k B = 0.15 meter M = 10 N m Find: The angular velocity of gear B after 5 seconds if the gears start turning from rest. Plan: Time is a parameter, thus Impulse and Momentum is recommended. First, relate the angular velocities of the two gears using kinematics. Then apply angular impulse and momentum to both gears. 17 / 36
18 / 36 GROUP PROBLEM SOLVING(Solution) Impulse-momentum diagrams: Note that the initial momentum is zero for both gears. Gear A: Mt Gear B: A y t W A t F t F t A x t B x t = W B t B y t r A = I A ω A r B I B ω B y x
19 / 36 GROUP PROBLEM SOLVING (continued) Kinematics: r A ω A = r B ω B Angular impulse and momentum relation: For gear A: Mt (F t)r A = I A ω A For gear B: (F t)r B = I B ω B so that (F t) = (I B ω B )/r B Combining the two equations yields: Mt = I A ω A + (r A /r B )I B ω B Substituting from kinematics for ω A = (r B /r A )ω B, yields Mt = ω B [(r B /r A )I A + (r A /r B )I B ] (1)
GROUP PROBLEM SOLVING (continued) I A = m A (k A ) 2 = 10(0.08) 2 = 0.064kg m 2 I B = m B (k B ) 2 = 50(0.15) 2 = 1.125kg m 2 Using Eq. (1), Mt = ω B [(r B /r A )I A + (r A /r B )I B ] 10(5) = ω B [(0.2/0.1)0.064 + (0.1/0.2)1.125] 50 = 0.6905ω B (2) Therefore, ω B = 72.4 rad/s and ω A = (r B /r A )ω B = (0.2/0.1)72.4 = 144 rad/s 20 / 36
21 / 36 ATTENTION QUIZ 1. If a slender bar rotates about end A, its angular momentum with respect to A is? l 01 A 00 11 00 11 00 11 G 00 11 00 11ω 00 11 00 11 01 (a) 1 12 ml2 ω (b) 1 6 ml2 ω (c) 1 3 ml2 ω (d) ml 2 ω ANS: (c)
22 / 36 ATTENTION QUIZ (Continued) 2. As in the principle of work and energy, if a force does no work, it does not need to be shown on the impulse and momentum diagram/equation. (a) False (b) True (c) Depends on the case (d) No clue! ANS: (a)
23 / 36 CONSERVATION OF MOMENTUM Today s objectives: Students will be able to 1 Understand the conditions for conservation of linear and angular momentum. 2 Use the condition of conservation of linear/ angular momentum. In-class activities: Reading Quiz Applications Conservation of Linear and Angular Momentum Concept Quiz Group Problem Solving Attention Quiz
24 / 36 READING QUIZ 1 If there are no external impulses acting on a body. (a) only linear momentum is conserved (b) only angular momentum is conserved (c) both linear momentum and angular momentum are conserved (d) neither linear momentum nor angular momentum are conserved ANS: (c) 2 If a rigid body rotates about a fixed axis passing through its center of mass, the body s linear momentum is. (a) a constant (b) zero (c) mv G (d) I G ω ANS: (b)
25 / 36 APPLICATIONS A skater spends a lot of time either spinning on the ice or rotating through the air. To spin fast, or for a long time, the skater must develop a large amount of angular momentum. If the skater s angular momentum is constant, can the skater vary her rotational speed? How? The skater spins faster when the arms are drawn in and slower when the arms are extended. Why?
26 / 36 APPLICATIONS(continued) Conservation of angular momentum allows cats to land on their feet and divers to flip, twist, spiral and turn. It also helps teachers make their heads spin! Conservation of angular momentum makes water circle the drain faster as it gets closer to the drain.
27 / 36 CONSERVATION OF LINEAR MOMENTUM (19.3) Recall that the linear impulse and momentum relationship is L 1 + t 2 t 1 F dt = L 2 where L 1 = (mv G ) 1 and L 2 = (mv G ) 2. If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L 1 = L 2. This equation is referred to as the conservation of linear momentum. The conservation of linear momentum equation can be used if the linear impulses are small or non-impulsive.
28 / 36 CONSERVATION OF ANGULAR MOMENTUM Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved H G1 + t 2 M G dt = H G2 where H G1 = I G ω 1 and H G2 = I G ω 2. t 1 The resulting equation is referred to as the conservation of angular momentum or H G1 = H G2. If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the final linear or angular velocity of a body just after an event occurs.
29 / 36 PROCEDURE FOR ANALYSIS 1 Establish the x, y, z inertial frame of reference and draw FBDs. 2 Write the conservation of linear momentum equation. 3 Write the conservation of angular momentum equation about a fixed point or at the mass center G. 4 Solve the conservation of linear or angular momentum equations in the appropriate directions. 5 If the motion is complicated, use of the kinematic equations relating the velocity of the mass center G and the angular velocity ω may be necessary.
30 / 36 EXAMPLE Given: A 10 kg wheel (I G = 0.156 kg m 2 ) rolls without slipping and does not rebound. Find: The minimum velocity, v G, the wheel must have to just roll over the obstruction at A. Plan: Since no slipping or rebounding occurs, the wheel pivots about point A. The force at A is much greater than the weight, and since the time of impact is very short, the weight can be considered non-impulsive. The reaction force at A is a problem as we don Äôt know either its direction or magnitude. This force can be eliminated by applying the conservation of angular momentum equation about A.
EXAMPLE (Solution) Impulse-momentum diagram: Conservation of angular momentum H A1 = H A2 r mv G1 + I G ω 1 = rmv G2 + I G ω 2 (0.2 0.03)(10)v G1 + 0.156ω 1 = 0.2(10)v G2 + 0.156ω 2 Kinematics: Since there is no slip, ω = v G /r = 5v G. Substituting and solving the momentum equation yields v G2 = 0.892v G1 31 / 36
32 / 36 EXAMPLE To complete the solution, conservation of energy can be used. Since it cannot be used for the impact (why?), it is applied just after the impact. In order to roll over the bump, the wheel must go to position 3 from 2. When v G2 is a minimum, v G3 is zero. Why? Energy conservation equation: T 2 + V 2 = T 3 + V 3 1 2 (10)v G,22 + 1 2 (0.156)ω 22 + 0 = 0 + 98.1(0.03) Substituting ω 2 = 5v G2 and v G2 = 0.892v G1 and solving yields v G1 = 0.729 m/s
33 / 36 GROUP PROBLEM SOLVING Given: Two children (m A = m B = 30 kg) sit at the edge of the merry-go-round, which has a mass of 180 kg and a radius of gyration of k z = 0.6 m. Find: The angular velocity of the merry-go-round if A jumps off horizontally in the +t direction with a speed of 2 m/s, measured relative to the merry-go-round. Plan: Draw an impulse-momentum diagram. The conservation of angular momentum can be used to find the angular velocity.
34 / 36 GROUP PROBLEM SOLVING(Solution) B M A Apply the conservation of angular momentum equation: H1 = H 2 180(0.6) 2 (2) + 2 [(30)2(0.75)2] = 180(0.6)2ω + (30)ω(0.75)2 v A/M A +(30)(0.75ω + 2)(0.75) B M Now we solve 197.1 = 98.55ω + 45 ω = 1.54 rad/s
35 / 36 ATTENTION QUIZ 1. Using conservation of linear and angular momentum requires that (a) all linear impulses sum to zero (b) all angular impulses sum to zero (c) both linear and angular impulses sum to zero (d) None of the above ANS: (c) 2. The angular momentum of a body about a point A that is the fixed axis of rotation but not the mass center (G) is (a) I A ω (b) I G ω (c) r G (mv G ) + I G ω (d) Both (a) and (c) ANS: (d)
Note 36 / 36