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MAE 242 Dynamics – Section I Dr. Kostas Sierros

MAE 242 Dynamics – Section I Dr. Kostas Sierros. Problem 1. Problem 2. Problem 3. Planar kinetics of a rigid body: Impulse and Momentum Chapter 19. Chapter objectives. Develop formulations for the linear and angular momentum of a body

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MAE 242 Dynamics – Section I Dr. Kostas Sierros

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  1. MAE 242 Dynamics – Section I Dr. Kostas Sierros

  2. Problem 1

  3. Problem 2

  4. Problem 3

  5. Planar kinetics of a rigid body: Impulse and Momentum Chapter 19 Chapter objectives • Develop formulations for the linear and angular momentum of a body • Apply the principles of linear and angular impulse and momentum to solve rigid body planar kinetic problems that involve force, velocity and time • To discuss the application of the conservation of momentum

  6. Lecture 21 • Planar kinetics of a rigid body: Impulse and Momentum • Linear and angular momentum • Principle of impulse and momentum • Conservation of momentum • -19.1-19.3

  7. Material covered • Planar kinetics of a rigid body :Impulse and Momentum • Whole of Chapter 19 (except 19.4) • …Next lecture…FinalReview 1

  8. Today’s Objectives Students should be able to: • Develop formulations for the linear and angular momentum of a body. • Apply the principle of linear and angular impulse and momentum. • Understand the conditions for conservation of linear and angular momentum. • Use the condition of conservation of linear/ angular momentum.

  9. Applications 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.

  10. 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.

  11. Linear and angular momentum (19.1) The linear momentum of a rigid body is defined as L = m vG This equation states that the linear momentum vector L has a magnitude equal to (mvG) and a direction defined by vG. Theangular momentumof a rigid body is defined as HG = IG w Remember that the direction of HGis perpendicular to the plane of rotation.

  12. Linear and angular momentum (19.1) continued Translation. When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because w = 0. Therefore: L = m vG HG = 0

  13. Linear and angular momentum (19.1) 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 = m vG HG = IGw It is sometimes convenient to compute the angular momentum of the body about the center of rotation O. HO = ( rG x mvG) + IGw = IO w

  14. Linear and angular momentum (19.1) 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 = m vG HG = IGw The angular momentum about point A is HA = IGw + (d)mvG

  15. Linear impulse-linear momentum equation: L1 + F dt = L2or (mvG)1 + F dt = (mvG)2 t2 t2 ò ò å å t1 t1 Angular impulse-angular momentum equation: (HG)1 + MG dt = (HG)2orIGw1 + MG dt = IGw2 t2 t2 ò ò å å t1 t1 Principle of Impulse and Momentum (19.2) As in the case of particle motion, the principle of impulse and momentum for a rigid body is developed by combining the equation of motion with kinematics. The resulting equations allow adirect solution to problems involving force, velocity, and time.

  16. Principle of Impulse and Momentum (19.2) continued + = The previous relations can be represented graphically by drawing theimpulse-momentum diagram. Tosummarize, 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 (eqs 19-4) describing the planar motion of the body.

  17. Procedure of analysis • 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 19-4). •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 in order to have additional equations.

  18. Applications (section 19.3) 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.

  19. Applications (section 19.3) continued Conservation of angular momentum allows ice skaters to spin faster or slower, cats to land on their feet, and divers to flip, twist, spiral and turn. It also helpsteachersmake their heads spin! Conservation of angular momentum makeswater circle the drain fasteras it gets closer to the drain.

  20. Recall that the linear impulse and momentum relationship is L1 + F dt = L2 or (m vG)1 + F dt = (m vG)2 0 0 t2 t2 ò ò å å t1 t1 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 L1 = L2. Conservation of momentum (section 19.3) This equation is referred to as theconservation of linear momentum. The conservation oflinearmomentum equation can be used if the linear impulses are small or non-impulsive.

  21. Conservation of angular momentum (section 19.3) 0 0 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 the conservation of angular momentum or (HG)1 = (HG)2 . The angular impulse-angular momentum relationship is: (HG)1 + MG dt = (HG)2 or IGw1 + MG dt = IGw2 t2 t2 ò ò å å t1 t1 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.

  22. Procedure of analysis (section 19.3) •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.

  23. Now I will put some music on… …I will leave the room for a few minutes and… …You will fill the evaluation forms

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