Chapter 18 Planar Kinetics of a Rigid Body

Chapter 18: Planar Kinetics of a Rigid Body: Impulse and Momentum

18. 1 Linear and Angular Momentum Linear Momentum • The linear momentum of a rigid body is determined by summing vectorially the linear momenta of all particles of the body, L = ∑mivi • Since ∑mivi = mv. G, we can also write, • This equation states that the body’s linear momentum is a vector quantity having a magnitude mv. G, and a direction defined by v. G © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • Consider the body as shown, which is subjected to general planar motion • At the instant shown, the arbitrary point P has a velocity v. P, and the body has an angular velocity ω © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • If the velocity of the ith particle of the body is to be determined, then • The angular momentum of particle i about point P is equal to the “moment” of the particle’s linear momentum about P, thus © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • Expressing vi in term of v. P and using Cartesian vectors, we have • Letting mi → dm and integrating over the entire mass m of the body, we obtain © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • HP represents the angular momentum of the body about an axis (the z axis) perpendicular to the plane of motion and passing through point P. • The integrals for the first and second terms on the right are used to locate the body’s center of mass G with respect to P © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • The last integral represents the body’s moment of inertia computed about the z axis, IP = ∫r 2 dm • Thus, • This equation reduces to a simpler form it point P coincides with the mass center G for the body, in which case © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • Hence, • This equation states that the angular momentum of the body computed about G is equal to the product of moment of inertia of the body about an axis passing through G and the body’s angular velocity. © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • The equation can be also written as © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • With reference to the figure, the equation indicates that when the angular momentum of the body is computed about point P, it is equivalent to the moment of the linear momentum mv. G or its components m(v. G)x and m(v. G)y about P plus the angular momentum IGω • Since ω is a free vector, HG can act at any point on the body provided it preserves its same magnitude and direction © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • Furthermore, since angular momentum is equal to the “moment” of the linear momentum, the line of action of L must pass through the body’s mass center G in order to preserve the correct magnitude of HP when “moments” are computed about P. • As the result of this analysis , we will now consider three types of motion. © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum Translation • When a rigid body of mass m is subjected to either rectilinear or curvilinear translation, its mass center has a velocity of v. G = v and ω = 0 for the body. • Hence the linear momentum and the angular momentum computed about G become © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • If the angular momentum is computed about any other point A on or off the body, the “moment” of the linear momentum L must be computed about the point • Since d is the moment arm as shown in the figure, then HA = (d)(mv. G) © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum Rotation About a Fixed Axis • When a rigid body is rotating about a fixed axis passing through point O, the linear momentum and the angular momentum computed about G are © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • It is sometimes convenient to compute the angular momentum of the body about point O. • In this case it is necessary to account for the “moment” of both L and HG about O • Noting that L (or v. G) is always perpendicular to r. G, we have © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • This equation may be simplified by first substituting v. G = r. Gω, in which case HO = (IG + mr. G 2)ω, and, by parallel-axis theorem, noting that the terms inside the parentheses represent the moment of inertia IO of the body about an axis perpendicular to the plane of motion and passing through point O. • Hence, © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum General Plane Motion • When a rigid body is subjected to general plane motion, the linear momentum and the angular momentum about G become © 2007 Pearson Education South Asia Pte Ltd

18. 1 Linear and Angular Momentum • If the angular momentum is computed about a point A located either on or off the body, it is necessary to find the moments of both L and HG about this point. • In this case, • Here d is the moment arm, as shown © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 At a given instant, the 10 kg disk and a 5 kg bar have the motions as shown. Determine their angular momenta about point G and about point B for the disk and about G and about the IC for the bar at this instant. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 Disk • Since the disk is rotating about a fixed axis (through point B), then v. G = (8)(0. 25) = 2 m/s • Hence, © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 • Also from the table on the inside back cover, IB = (3/2) mr 2, so that Bar • The bar undergoes general plane motion • The IC is established as shown in figure © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 • So that ω = (2)(3. 464) = 0. 5774 rad/s and v. G = (0. 5774)(2) = 1. 155 m/s • Thus, • Moments of IGω and mv. G about IC yield © 2007 Pearson Education South Asia Pte Ltd

18. 2 Conservation of Momentum Conservation of Linear Momentum • If the sum of all the linear impulses acting on the system of connected rigid is zero, the linear momentum of the system is constant, or conserved • This equation is referred to as the conservation of linear momentum. © 2007 Pearson Education South Asia Pte Ltd

18. 3 Conservation of Momentum Conservation of Angular Momentum • The angular momentum of a system of connected rigid bodies is conserved about the system’s center of mass G, or a fixed point O, when the sum of all the angular impulses created by the external forces acting on the system is zero or small (nonimpulsive) when computed about these points. © 2007 Pearson Education South Asia Pte Ltd

18. 3 Conservation of Momentum • This equation is referred to as the conservation of angular momentum. • In the case of a single rigid body, this equation applied to point G becomes (IGω)1 = (IGω)2 • Provided the initial linear or angular velocity of the body is known, the conservation of linear or angular momentum us used to determine the respective final linear or angular velocity of the body just after the time period considered. © 2007 Pearson Education South Asia Pte Ltd

18. 3 Conservation of Momentum • Furthermore, by applying these equations to a system of bodies, the internal impulses acting within the system, which may be unknown, are eliminated from the analysis, since they occur in equal but opposite collinear pairs. • If it is necessary to determine an internal impulsive force acting on only one body of a system of connected bodies, the body must be isolated (free-body diagram) and the principle of linear and angular impulse and momentum applied to the body © 2007 Pearson Education South Asia Pte Ltd

18. 3 Conservation of Momentum • After the impulse ∫F dt is calculated; then, provided the time Δt for which the impulse acts is known, the average impulsive force Favg can be determined from Favg = (∫F dt )/ Δt © 2007 Pearson Education South Asia Pte Ltd

PROCEDURE FOR ANALYSIS Free-body Diagram • Establish the x, y inertial frame of reference and draw the FBD for the body or system of bodies during the time of impact. From this diagram, classify each of the applied forces as being either “impulsive” or “nonimpulsive”. • By inspection of the FBD, the conservation of linear momentum applies in a given direction when no external impulsive forces act on the body or system in that direction. © 2007 Pearson Education South Asia Pte Ltd

PROCEDURE FOR ANALYSIS • The conservation of angular momentum applies about a fixed point O or at the mass center G of a body or system of bodies when all the external impulsive forces acting on the body or system create zero moment ( or zero angular impulse) about O or G. • As an alternative procedure, draw the impulse and momentum diagrams for the body or system of bodies. © 2007 Pearson Education South Asia Pte Ltd

PROCEDURE FOR ANALYSIS • These diagrams are helpful in order to visualize the “moment” terms used in the conservation of angular momentum equation, when it has been decided that angular momenta are to be computed about point other than the body’s mass center G. Conservation of Momentum • Apply the conservation of linear or angular momentum in the appropriate directions © 2007 Pearson Education South Asia Pte Ltd

PROCEDURE FOR ANALYSIS Kinematics • If the motion appears to be complicated, kinematics (velocity) diagrams may be helpful in obtaining the necessary kinematics relations. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 6 The 10 kg wheel has a moment of inertia IG = 0. 156 kg. m 2. Assuming that the wheel does not slip or rebound, determine the minimum velocity v. G it must have to just roll over the obstruction at A. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 6 Impulse and Momentum Diagrams • Since no slipping or rebounding occurs, the wheel pivots about point A during contact • The figure shown respectively , the momentum of the wheel just before impact, the impulses given to the wheel during impact, and the momentum of the wheel just after impact. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 6 • Only two impulses (forces) act on the wheel • By comparison, the force at A is much greater than that of the weight, and since the time of impact is very short, the weight can be considered as nonimpulsive. • The impulsive force F at A has both an unknown magnitude and an unknown direction θ • To eliminate this force from the analysis, angular momentum about A is conserved © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 6 Conservation of Angular momentum Kinematics • Since no slipping occurs, in general ω = v. G/r = v. G/0. 2 = 5 v. G • Substituting this into the above equation, (1) © 2007 Pearson Education South Asia Pte Ltd

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