Chapter 15 Kinematics of a Particle: Impulse and Momentum Lecture Notes for Section 15-5~7
ANGULAR MOMENTUM, MOMENT OF A FORCE AND PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM Today s Objectives: Students will be able to: 1. Determine the angular momentum of a particle and apply the principle of angular impulse & momentum. 2. Use conservation of angular momentum to solve problems. In-Class Activities: Check Homework Reading Quiz Applications Angular Momentum Angular Impulse and Momentum Principle Conservation of Angular Momentum Concept Quiz Group Problem Solving Attention Quiz
READING QUIZ 1. Select the correct expression for the angular momentum of a particle about a point. A) r v B) r (m v) C) v r D) (m v) r 2. The sum of the moments of all external forces acting on a particle is equal to A) angular momentum of the particle. B) linear momentum of the particle. C) time rate of change of angular momentum. D) time rate of change of linear momentum.
APPLICATIONS Planets and most satellites move in elliptical orbits. This motion is caused by gravitational attraction forces. Since these forces act in pairs, the sum of the moments of the forces acting on the system will be zero. This means that angular momentum is conserved. If the angular momentum is constant, does it mean the linear momentum is also constant? Why or why not?
APPLICATIONS (continued) The passengers on the amusement-park ride experience conservation of angular momentum about the axis of rotation (the z-axis). As shown on the free body diagram, the line of action of the normal force, N, passes through the z-axis and the weight s line of action is parallel to it. Therefore, the sum of moments of these two forces about the z-axis is zero. If the passenger moves away from the z- axis, will his speed increase or decrease? Why?
ANGULAR MOMENTUM (Section 15.5) The angular momentum of a particle about point O is defined as the moment of the particle s linear momentum about O. i j k H o = r mv = r x r y r z mv x mv y mv z The magnitude of H o is (H o ) z = mv d
RELATIONSHIP BETWEEN MOMENT OF A FORCE AND ANGULAR MOMENTUM (Section 15.6) The resultant force acting on the particle is equal to the time rate of change of the particle s linear momentum. Showing the time derivative using the familiar dot notation results in the equation F = L = mv We can prove that the resultant moment acting on the particle about point O is equal to the time rate of change of the particle s angular momentum about point O or M o = r F = H o
PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM (Section 15.7) Considering the relationship between moment and time rate of change of angular momentum t t M o = H o = dh o /dt By integrating between the time interval t 1 to t 2 2 1 ( H H M dt ( o ) 2 o ) or o ( H o ) 1 1 This equation is referred to as the principle of angular impulse and momentum. The second term on the left side, M o dt, is called the angular impulse. In cases of 2D motion, it can be applied as a scalar equation using components about the z-axis. t 2 + dt t 1 M o ( H o ) 2
CONSERVATION OF ANGULAR MOMENTUM When the sum of angular impulses acting on a particle or a system of particles is zero during the time t 1 to t 2, the angular momentum is conserved. Thus, (H O ) 1 = (H O ) 2 An example of this condition occurs when a particle is subjected only to a central force. In the figure, the force F is always directed toward point O. Thus, the angular impulse of F about O is always zero, and angular momentum of the particle about O is conserved.
EXAMPLE Given:A satellite has an elliptical orbit about earth. m satellite = 700 kg m earth = 5.976 10 24 kg v A = 10 km/s r A = 15 10 6 m A = 70 Find: The speed, v B, of the satellite at its closest distance, r B, from the center of the earth. Plan: Apply the principles of conservation of energy and conservation of angular momentum to the system.
Solution: EXAMPLE (continued) Conservation of energy: T A + V A = T B + V B becomes 1 m s v A2 G m s m e = 1 m s v B2 G m s m e 2 r A 2 r B where G = 66.73 10-12 m 3 /(kg s 2 ). Dividing through by m s and substituting values yields: 0.5 (10,000 ) 2 66.73 10-12 (5.976 15 x 10 6 10 24 ) or 0.5 23.4 10 6 = 0.5 (v B ) 2 (3.99 10 14 )/r B v 2 B 66.73 10-12 (5.976 r B 10 24 )
Solution: EXAMPLE (continued) Now use Conservation of Angular Momentum. (r A m s v A ) sin A = r B m s v B (15 10 6 )(10,000) sin 70 = r B v B or r B = (140.95 10 9 )/v B Solving the two equations for r B and v B yields r B = 13.8 10 6 m v B = 10.2 km/s
CONCEPT QUIZ 1. If a particle moves in the x - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) x - y direction. 2. If there are no external impulses acting on a particle 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.
GROUP PROBLEM SOLVING Given: The four 5 lb spheres are rigidly attached to the crossbar frame, which has a negligible weight. A moment acts on the shaft as shown, M = 0.5t + 0.8 lb ft). Find: The velocity of the spheres after 4 seconds, starting from rest. Plan: Apply the principle of angular impulse and momentum about the axis of rotation (z-axis).
GROUP PROBLEM SOLVING (continued) Solution: Angular momentum: H Z = r mv reduces to a scalar equation. (H Z ) 1 = 0 and (H Z ) 2 = 4 {(5/32.2) (0.6) v 2 } = 0.3727 v 2 Angular impulse: t 2 t 2 M dt = (0.5t + 0.8) dt = [(0.5/2) t 2 + 0.8 t] = 7.2 lb ft s t t 0 1 1 Apply the principle of angular impulse and momentum. 0 + 7.2 = 0.3727 v 2 v 2 = 19.4 ft/s 4
ATTENTION QUIZ 1. A ball is traveling on a smooth surface in a 3 ft radius circle with a speed of 6 ft/s. If the attached cord is pulled down with a constant speed of 2 ft/s, which of the following principles can be applied to solve for the velocity of the ball when r = 2 ft? A) Conservation of energy B) Conservation of angular momentum C) Conservation of linear momentum D) Conservation of mass 2. If a particle moves in the z - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) z - y direction.
Example 15.12 The box has a mass m and is traveling down the smooth circular ramp such that when it is at the angle θ it is a speed v. Determine its angular momentum about point O at this instant and the rate of increase in its speed, i.e., a t.
Example 15.12 Solution Since v is tangent to the path, the angular momentum is HO rmv From the FBD, the weight W = mg contributes a moment about O ; ( sin ) d M ( ) O H O mg r rmv dt Since r and m are constant, dv dv mgr sin rm g sin dt dt
Example 15.14 The 0.4 kg ball B is attached to a cord which passes through a hole at A in a smooth table. When the ball is r 1 = 0.5 m from the hole, it is rotating around in a circle such that its speed is v 1 = 1.2 m/s. By applying a force F the cord is pulled downward through the hole with a constant speed v c = 2 m/s.
Example 15.14 Determine (a) the speed of the ball at the instant it is r 2 = 0.2 m from the hole, and (b) the amount of work done by F in shortening the radial distance from r 1 to r 2.
Example 15.14 Solution Part (a) Free-Body Diagram As the ball the cord force F on the ball passes through the z axis. Weight and N B are parallel and the conservation of angular momentum applies about the z axis. Conservation of Angular Momentum H1 H2 rm 1 Bv1 rm 2 Bv 2 (0.5)(0.4)(1.2) (0.2)(0.4) v v 3m/s 2 2
Example 15.14 Solution Conservation of Angular Momentum The speed of the ball is thus 2 2 v2 (3.0) (2) 3.606m/s Part (b) The only force that does work on the ball is F. T U T 1 12 2 1 2 1 2 (0.4)(1.2) UF (0.4)(3.606) UF 2.313J 2 2
Chapter 15 Kinematics of a Particle: Impulse and Momentum Lecture Notes for Section 15-8~9
15.8 Steady Flow of a Fluid Stream Consider the diversion of a steady stream of fluid by a fixed pipe The impulse and momentum diagrams for the fluid are as shown below
15.8 Steady Flow of a Fluid Stream The force F represents the resultant of all the external forces acting on the fluid stream Since flow is steady, F will be constant during the time interval dt Applying the principle of linear impulse and momentum to the fluid stream, A B dm v mv Fdt dm v mv
15.8 Steady Flow of a Fluid Stream Principle of Impulse and Momentum Solving for the resultant force yields dm F ( vb va) dt It is convenient to express vector equation in the form of two scalar component equations dm Fx ( vbx vax) dt dm Fy ( vby vay) dt
15.8 Steady Flow of a Fluid Stream Principle of Impulse and Momentum Since flow is steady in the x-y plane, hence we have dm + M O ( dobvb doava) dt Once the velocity of the fluid flowing onto the device is determined, the mass flow is calculated using dm AvA A A BvA B B AQA BQB dt
15.8 Steady Flow of a Fluid Stream Procedure of Analysis Free-Body Diagram Draw a FBD of the device which is directing the fluid Equations of Steady Flow Apply the equations of steady flow, dm M O ( dobvb doava) dt
Example 15.6 Determine the components of reaction which the fixed pipe joint at A exerts on the elbow. If water flowing through the pipe is subjected to a static gauge pressure of 100 kpa at A. The discharge at B is QB = 0.2 m 3 /s. Water has a density ρw = 1000kg/m 3, and the water-filled elbow has a mass of 20 kg and center of mass at G.
Example 15.16 Solution Since the density of water is constant, Q B = Q A = Q, dm wq 200 kg/s dt Q 0.2 Q vb 25.46 m/s ; v 6.37 m/s 2 A A (0.05) A B Free-Body Diagram Since 1 kpa = 1000 N/m 2, F p A A A A 3 2 [100(10 )][ (0.1) ] 3141.6N A
Example 15.16 Solution Equations of Steady Flow dm Fx ( vbx vax); dt Fx 3141.6 200(0 6.37) Fx 4.41kN dm Fy ( vby vay); dt Fy 20(9.81) 200( 25.46 0) F 4.90kN y
Example 15.16 Solution Equations of Steady Flow If moments are summed about point O, then F x, F y, and static pressure F A are eliminated, as well as moment of momentum of water entering at A, dm MO ( dobvb doava) dt M O 20(9.81)(0.125) 200[(0.3)(25.46) 0] M 1.50kN m O
15.9 Propulsion with Variable Mass A Control Volume That Loses Mass Consider a device such as a rocket which at an instant of time has a mass m and is moving forward with a velocity v The closed system included both the mass m of the device and the expelled mass m e
15.9 Propulsion with Variable Mass A Control Volume That Loses Mass Applying the principle of impulse and momentum to the system, mv m v F dt ( m dm )( v dv) ( m dm ) v e e CV e e e e F dt vdm m dv dm dv v dm CV e e e e The relative velocity of the device as seen by the observer moving with the particles of the ejected mass is v D/e = (v + v e ), dv FCV m v dt D/ e dm dt e
15.9 Propulsion with Variable Mass A Control Volume That Gain Mass A device such as a scoop or a shovel may gain mass as it moves forward The device shown has a mass m and is moving forward with a velocity v
15.9 Propulsion with Variable Mass A Control Volume That Gain Mass Applying the principle of impulse and momentum to the system, mv m v F dt ( m dm )( v dv) ( m dm ) v i i CV i i i i We may write this equation as dv dmi FCV m ( vvi ) dt dt
Example 15.18 The initial combined mass of a rocket and its fuel is m 0. A total mass mf of fuel is consumed as a constant rate of dm e /dt = c and expelled at a constant speed of u relative to the rocket. Determine the max velocity of the rocket i.e., at the instant the fuel runs out.
Example 15.18 Solution The rocket is losing mass as it moves upward. The only external force acting on the system consisting of the rocket and a portion of the expelled mass is the weight W dv dme dv Fs m vd/ e ; W m uc dt dt dt Since W = mg, ( m0 ct) g ( m0 ct) dv uc dt
Example 15.18 Solution Separating the variables and integrating, realizing that v = 0 at t = 0, we have uc m v t t 0 dv ln( 0 ) ln 0 g dt v u m ct gt u gt 0 0 m0 ct m0 ct The time t needed to consume all the fuel is given by By sub, we get dme mf t ct t mf / c dt v max uln m gm 0 m0 m f c f
End section 15-1 _01: 15-2, 15-8, 15-13, 15-20 End section 15-2 _02: 15-31, 15-48, 15-64 End section 15-3 _03: 15-82, 15-88 CH015 HW