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Mechanics

Mechanics. 8.7 Impulse-Momentum Method. Chapter 8. Planar Kinetics of Rigid Bodies. Section 18.6,18.6, 18.7. Mechanics. 8.7 Impulse-Momentum Method. Linear and angular momentum Principle of impulse and momentum Conservation of momentum Rigid body impact.

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Mechanics

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  1. Mechanics 8.7 Impulse-Momentum Method Chapter 8. Planar Kinetics of Rigid Bodies Section 18.6,18.6, 18.7

  2. Mechanics 8.7 Impulse-Momentum Method • Linear and angular momentum • Principle of impulse and momentum • Conservation of momentum • Rigid body impact Chapter 8. Planar Kinetics of Rigid Bodies

  3. m(vG)y P = mvG m(vG)x LG = IGw G Momentum diagram 1. Linear and angular momentum The linear momentum of a rigid body is defined as P = mvG This equation states that the linear momentum vector P has a magnitude equal to (mvG) and a direction defined by vG. The angular momentumof a rigid body is defined as LG = IG w Remember that the direction of LGis perpendicular to the plane of rotation.

  4. y´ P=mvG G x´ rG d A 1. Linear and angular momentum Translation. When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because w = 0. Therefore: P= m vG LG = 0

  5. y´ P=mvG LG = IG G x´ rG A(fixed) 1. Linear and angular momentum Rotation about a fixed axis. When a rigid body is rotating about a fixed axis passing through point A, the body’s linear momentum and angular momentum about G are: P = m vG LG = IGw It is sometimes convenient to compute the angular momentum of the body about the center of rotation A. LA = ( rG × mvG) + IGw = IA w

  6. y´ LG = IG G P=mvG x´ rG d A 1. Linear and angular momentum 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. p = mvG LG = IGw The angular momentum about point A is LA = IGw + (d)mvG

  7. Mechanics 8.7 Impulse-Momentum Method • Linear and angular momentum • Principle of impulse and momentum • Conservation of momentum • Rigid body impact Chapter 8. Planar Kinetics of Rigid Bodies

  8. Linear impulse-linear momentum equation: or • Angular impulse-angular momentum equation: For plane motion only or A: fixed point of mass center 2. Principle of impulse and momentum Principle of impulse and momentum: • Relates the impulse of the external forces to the change in the body’s momentum. • Allow adirect solution to problems involving force, velocity, and time.

  9. IG2 + = G G m(vG)2 Final Momentum diagram Impulse diagram IG1 For plane motion of a rigid body: G m(vG)1 Initial Momentum diagram The previous relations can be represented graphically by drawing theimpulse-momentum diagram. 2. Principle of impulse and momentum

  10. PROCEDURE FOR ANALYSIS 2. Principle of impulse and momentum • Establishthe x, y, z inertial frame of reference. • Draw the impulse-momentumdiagramsfor the body. • Compute IG, as necessary. • Apply theequations of impulse and momentum(one vector and one scalar or the three scalar equations). • If more than three unknowns are involved,kinematic equations relating the velocity of the mass center G and the angular velocity w should be used to furnish additional equations.

  11. P = 10N r=0.2m Example 8.7-1 2. Principle of impulse and momentum Given:A disk weighing 25kg has a rope wrapped around it. The rope is pulled with a force P equaling 10N. Find:The angular velocity of the cylinder after 4 seconds if it starts from rest and rolls without slipping. Plan:Time as a parametershould 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.

  12. y Wt x Pt G + = G (mvG)1 (mvG)2 r A F t IGw1 IG w2 N t Impulse & Momentum: (P t) 2 r = (mvG)2 r + IGw2 = m r2 w2 + ½ m r2 w2 = 1.5 m r2 w2 4 Pt 4(10)(4) = = = w2 10.7 rad/s 3mr 3(25)(0.2) Solution: 2. Principle of impulse and momentum Impulse-momentum diagram: Kinematics:(vG)2 = rw2

  13. Example 8.7-2 2. Principle of impulse and momentum Given:A gear set with: mA = 7kg mB = 5kg kA = 0.15m kB = 0.1m M = 2(1 – e-0.5t) N·m 0.25m 0.15m Find:The angular velocity of gear A 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.

  14. y x WAt Gear A: rA Axt A = Ayt IA wA F t Gear B: WBt Ft IB wB = Bx t rB t2 ò M dt By t t1 Solution: 2. Principle of impulse and momentum Impulse-momentum diagrams:Note that the initial momentum is zero for both gears.

  15. Angular impulse & momentumrelation for gear A about point A yields: (Ft)rA = IAwAorFt = IAwA/rA For gear B:Mdt – (Ft)rB = IBwB t2 ò t1 t2 Combining the two equations: ò t1 M dt = (rB/rA)IAwA + IBwB 2. Principle of impulse and momentum Kinematics:rAwA = rBwB

  16. Substituting from kinematics forwA = (rB/rA)wB,yields Where M dt = 2(1 – e-0.5t) dt = 2t + 4e-0.5t = (10.328) – (4) = 6.328 N·m·s t2 ò M dt = wB[(rB/rA)2IA + IB] t2 5 t1 ò ò t1 0 Therefore, and 2. Principle of impulse and momentum IA = mA(kA)2 = (7)(0.25)2 = 0.4375 kg·m2 IB = mB(kB)2 = (5)(0.15)2 = 0.1125 kg·m2

  17. Mechanics 8.7 Impulse-Momentum Method • Linear and angular momentum • Principle of impulse and momentum • Conservation of momentum • Rigid body impact Chapter 8. Planar Kinetics of Rigid Bodies

  18. or 0 0 3. Conservation of momentum 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, orconserved. So P1 = P2. This equation is referred to as theconservation of linear momentum.The conservation oflinear momentum equation may also be used if the linear impulses are small or non-impulsive.

  19. 0 0 3. Conservation of momentum For plane motion only or 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 . The resulting equation is referred to as theconservation of angular momentumor (LA)1 = (LA)2 . 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 bodyjust afteran event occurs.

  20. PROCEDURE FOR ANALYSIS 3. Conservation of momentum • Establish the x, y, z inertial frame of reference and draw FBDs. • Write the conservation of linear momentum equation. • Write the conservation of angular momentum equation about a fixed point or at the mass center G. • Solve the conservation of linear or angular momentum equations in the appropriate directions. • If the motion is complicated, kinematic equations relating the velocity of the mass center G and the angular velocity w may be necessary.

  21. 3. Conservation of momentum Example 8.7-3 Given:A 10 kg wheel (IG = 0.156 kg·m2) rolls without slipping and does not bounce at A. Find:The minimum velocity,vG, the wheel must have to just roll over the obstruction at A.

  22. 3. Conservation of momentum Plan: • Sinceno slipping or bouncing occurs,the wheelpivotsabout point A. • The force at A is much greater than the weight, and since the time of impact is very short, theweight can be considered non-impulsive. • The reaction force at A is a problem as we don’t know either its direction or magnitude. This force can be eliminated by applying the conservation of angular momentum equationabout point A.

  23. y Impulse-momentum diagram: x + = Solution: 3. Conservation of momentum Conservation of angular momentum: (LA)1 = (LA)2 r ' (mvG)1 + IG w1 = r (mvG)2 + IG w2 (0.2 - 0.03)(10)(vG)1 + 0.156 w1 = 0.2(10)(vG)2 + 0.156 w2 Kinematics:Since there is no slip,w= vG / r = 5 vG . Substituting and solving the momentum equation yields (vG)2 = 0.892 (vG)1

  24. To complete the solution,conservation of energycan 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 (vG)2 is a minimum, (vG)3 is zero.Why? 3. Conservation of momentum Ek2 + Ep2 = Ek3 + Ep3 {½(10)(vG)22 + ½(0.156)w22 } + 0 = 0 + 98.1 (0.03) Substituting w2 = 5 (vG)2 and (vG)2 = 0.892 (vG)1 and solving yields (vG)1 = 0.729 m/s

  25. A 1m B B v=304.8m/s 0.4m 0.4m Example 8.7-4 3. Conservation of momentum Given:A slender rod (mr = 2kg) has a wood block (mw = 4kg) attached. A bullet (mb = 0.09kg) is fired into the block at 304.8/s. Assume the pendulum is initially at rest and the bullet embeds itself into the block. Find:The angular velocity of the pendulum just after impact. Plan:The force due to impact is internal to the system (the pendulum and bullet), so the impulses sum to zero. Thus, angular momentum is conserved and the conservation of angular momentum can be used to find the angular velocity.

  26. First draw a FBD. Ay Ax A mrg 1m B mbg B F 0.4m F 0.4m mwg 3. Conservation of momentum Solution: To use conservation of angular momentum, the mass moment of inertia of the pendulum and the embedded bullet about A must be found.

  27. Ay Ax A mrg 1m B mbg B F 0.4m F 0.4m mwg 3. Conservation of momentum Apply theconservation of angular momentumequation: (LA)1 = (LA)2 (mbvb)1 (rb) = (IA )2 w2 (0.09)(304.8)(1.2) = 7.33 w2 Solving yields: w2 = 4.5 rad/s

  28. Mechanics 8.7 Impulse-Momentum Method • Linear and angular momentum • Principle of impulse and momentum • Conservation of momentum • Rigid body impact Chapter 8. Planar Kinetics of Rigid Bodies

  29. v2 n t Line of impact m2 m1 v1 v20 v10 4. Rigid body impact Impact(碰撞): A collision between two bodies whichoccurs in a veryshort time period, and during which the two bodies exert relatively large impulsive forces on each other. Central impact(中心碰撞) eccentric impact(偏心碰撞) v2 n t Line of impact m2 v1 m1 v20 v10 Impact between two rigid bodies

  30. n Line of impact n Line of impact B A B A vB v´B v´A vA 4. Rigid body impact Consider the impact between two rigid bodies: • More complex than the impact between two particles: • Depends on the geometries of the impacting bodies and their surface characteristics, as well as their relative velocities • The impact is eccentric: affect both the translation and rotation of the impact bodies

  31. n Line of impact n Line of impact B A B A vB v´B v´A vA 4. Rigid body impact Consider the impact between two rigid bodies: • Simplification: • The motion is impulsive: t0, neglect all displacements of the bodies during the impact • The bodies are frictionlessthe forces they exert on each other are directed along the line of impact • Plane motion only

  32. Deformation stage: 4. Rigid body impact Once the two bodies contact, they will be compressed by the impact force and will begin to deform. At the end of the period of deformation, the velocities uAand uB of A and B will have equal components along the line of impact n n -P B A P uB uA Restitution stage: n During this stage, the bodies return totally or partially to its undeformed shape. At the end, the velocities of A and B reach to v´A and v´B respectively. -R B A R v´B v´A

  33. Define: 4. Rigid body impact Coefficient of restitution(恢复系数): the ratio of the magnitudes of the impulse corresponding, respectively, to the restitution stage and to the deformation stage

  34. m(vG)t m(vG)n m(uG)t n m(uG)n n G Pdt n G IGw A G A Igw* A 4. Rigid body impact Proof: (1) The motion of each of the two colliding bodies is unconstrained Consider the body to which point A belongs: Deformation stage: the impulsive force acting on A is the force P exerted by B = + r

  35. m(uG)t m(uG)n m(v´G)t n m(v´G)n n G Rdt n G Igw* A G A Igw A 4. Rigid body impact Restitution stage: R is the force exerted by B on A = + r 

  36. 4. Rigid body impact A similar analysis of the body to which point B belongs

  37. Qydt G o O IOw* Qxdt IOw n n n Pdt r = + A A A Qydt G o O IOw Qxdt Iow* n n n Rdt r = + A A A 4. Rigid body impact (2) The motion of each of the two colliding bodies is unconstrained

  38. 4. Rigid body impact A similar analysis of the body to which point B belongs

  39. 4. Rigid body impact Summary: • We need two equations to determine the two velocities, • but cannot use angular momentum equation of each body • since we do not know restitution force. • So: • Conservation of angular momentum applied to system • (restitution force is internal so does not appear) • 2) Coefficient of restitution

  40. 4. Rigid body impact Note: Consider the two colliding bodies as a system, then The angular momentum of the system is conserved. O Linear momentum is conserved Linear momentum is NOT conserved

  41. O O L L B B vB=0 vs v´s v´B 4. Rigid body impact The analysis discussed above can be applied to the analysis of impact between a particle and a rigid body Angular momentum conservation+coefficient of restitution

  42. 4. Rigid body impact

  43. QUIZ 2. Principle of impulse and momentum 1. The angular momentum of a rotating two-dimensional rigid body about its center of mass G is A) m vG. B) IG vG. C) m w. D) IG w. 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) m vG. D) IG w.

  44. QUIZ 2. Principle of impulse and momentum 1. The angular momentum of a rotating two-dimensional rigid body about its center of mass G is A) m vG. B) IG vG. C) m w.D) IG w. 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) m vG. D) IG w.

  45. 1. If a slender bar rotates about end A, its angular momentum with respect to A is A) (1/12) m l2 w B) (1/6) m l2 w C) (1/3) m l2 w D) m l2 w A w G l QUIZ 2. Principle of impulse and momentum 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

  46. 1. If a slender bar rotates about end A, its angular momentum with respect to A is A) (1/12) m l2 w B) (1/6) m l2 w C) (1/3) m l2w D) m l2 w A w G l QUIZ 2. Principle of impulse and momentum 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) FalseB) True

  47. QUIZ 2. Principle of impulse and momentum 3. If a slab is rotating about its center of mass G, its linear momentum is A) constant. B) zero. C) increasing linearly with D) decreasing linearly time. with time.

  48. 2. Principle of impulse and momentum If a slab is rotating about its center of mass G, its linear momentum is A) constant.B) zero. C) increasing linearly with D) decreasing linearly time. with time. QUIZ

  49. 1. A slender rod (mass = M) is at rest. If a bullet (mass = m) is fired with a velocity of vb, the angular momentum of the bullet about A just before impact is A) 0.5 m vb2 . B) m vb . C) 0.5 m vb . D) zero. A w2 0.5 m G 1.0 QUIZ 3. Conservation of momentum 2. For the rod in question 1, the angular momentum about A of the rod and bullet just after impact will be A) m vb + M(0.5)w2 B) m(0.5)2w2 + M(0.5)2w2 C) m(0.5)2w2 + M(0.5)2w2 D) zero. + (1/12) M w2

  50. 1. A slender rod (mass = M) is at rest. If a bullet (mass = m) is fired with a velocity of vb, the angular momentum of the bullet about A just before impact is A) 0.5 m vb2 . B) m vb . C) 0.5 m vb .D) zero. A w2 0.5 m G 1.0 QUIZ 3. Conservation of momentum 2. For the rod in question 1, the angular momentum about A of the rod and bullet just after impact will be A) m vb + M(0.5)w2 B) m(0.5)2w2 + M(0.5)2w2 C) m(0.5)2w2 + M(0.5)2w2 D) zero. + (1/12) M w2

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