Understanding Linear Impulse and Momentum in Mechanics
Explore the principles of linear impulse and momentum, conservation of linear momentum, mechanics of impact, and more in this study module. Learn to analyze forces, solve problems involving fluid streams and propulsion, and apply these concepts to real-world scenarios. Engage in practical problem-solving exercises to deepen your understanding of this fundamental concept in physics.
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Objectives To develop the principle of linear impulse and momentum for a particle. To study the conservation of linear momentum for particles. To analyze the mechanics of impact. To introduce the concept of angular impulse and momentum. To solve problems involving steady fluid streams and propulsion with variable mass. 07/09/67 2 ME212 .
Principle of linear impulse and momentum d v = = F m a m dt t v = 2 2 F dt m d v t v 1 1 t = 2 F dt m v m v 2 1 t 1 Linear Impulse Linear Momentum 07/09/67 3 ME212 .
Principle of linear impulse and momentum t + = 2 v F v m dt m 1 2 t 1 t 2 + = ( ) ( ) m v F dt m v 1 2 x x x t 1 t 2 + = ( ) ( ) m v F dt m v 1 2 y y y t 1 t 2 + = ( ) ( ) m v F dt m v 1 2 z z z 1 t 07/09/67 4 ME212 .
Procedure for Analysis Select the inertial coordinate system Draw FBD to account for all forces Establish direction and sense of acceleration Solve for unknowns 07/09/67 5 ME212 .
Problem 15-10 A man kicks the 200-g ball such that it leaves the ground at an angle of 30 with the horizontal and strikes the ground at the same elevation a distance of 15 m away. Determine the impulse of his foot F on the ball. 07/09/67 6 ME212 .
Problem 15-11 The particle P is acted upon by its weight of 30 N and forces F1and F2, where t is in seconds. If the particle originally has a velocity of v1= (3i + 1j + 6k) m/s, determine its speed after 2 s. 07/09/67 7 ME212 .
Problem 15-27 Block A weighs 100 N and block B weighs 30 N. If B is moving downward with a velocity (vB)1= 1 m/s at t=0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth. 07/09/67 8 ME212 .
Principle of Linear Impulse and Momentum for a System of Particles v d = F i m i i dt ( ) v ( )2 i v t + = 2 F m dt m i i i i 1 t 1 07/09/67 9 ME212 .
Conservation of Linear Momentum for a System of Particles When the sum of the external impulses acting on a system of particles is zero, the equation is ( ) v ( ) v = m m i i i i 1 2 This equation is referred to as the conservation of linear momentum. 07/09/67 10 ME212 .
Example 1 The 15-Mg boxcar A is coasting at 1.5 m/s on the horizontal track when it encounters a 12-Mg tank B coasting at 0.75 m/s toward it. If the cars meet and couple together, determine (a) the speed of both cars just after the coupling, and (b) the average force between them if the coupling takes place in 0.8 s. 07/09/67 11 ME212 .
Problem 15-52 The boy B jumps off the canoe at A with a velocity of 5 m/s relative to the canoe as shown. If he lands in the second canoe C, determine the final speed of both canoes after the motion. Each canoe has a mass of 40 kg. The boy s mass is 30 kg, and the girl has a mass of 25 kg. Both canoes are originally at rest. 07/09/67 12 ME212 .
Impact Central Impact Velocities are colinear (a) Deforms during collision (b) Restore after deformation (b) Coefficient of restitution = Irestitution/Ideform 07/09/67 13 ME212 .
Coefficient of Restitution Consider m1 F e t = t dt r 1 1 [ ( )] m v v v v t = = 1 0 0 0 [ ( )] m v v v v 1 0 1 1 0 F dt d t 0 Consider m2 F e t = t dt 2 1 v v r 2 2 [ ] m v v v v = t = = e 2 0 0 0 [ ] m v v v v v v 2 0 2 0 2 F dt 1 2 d t 0 07/09/67 14 ME212 .
Impact Oblique Impact 2 1 ( ) ( ) v v = n n e ( ) ( ) v v 1 2 n n 07/09/67 15 ME212 .
Problem 15-64 If the girl throws the ball with a horizontal velocity of 3 m/s, determine the distance d so that the ball bounces once on the smooth surface and then lands in the cup at C. Take e=0.8. 07/09/67 16 ME212 .
Angular Momentum If the particle is moving along a space curve, the vector cross product can be used to determine the angular momentum about O. m O = v = + + r i j k r r r x y z H r v = + + i j k v v v x y z H = sin m v r 0 07/09/67 17 ME212 .
Angular Momentum Recall d L = = ( ) F m v dt d H = = ( ) ( ) r F r m v O dt H = M O O The moment about the fixed point O of all forces acting on m equals to the rate of change of angular momentum. 07/09/67 18 ME212 .
Principle of Angular Momentum d H = = O M H O O dt = M dt d H O O t 2 = M dt H H O O O 2 1 t 1 t 2 t + 2 = H M dt H + = H M dt H O O O 1 2 O O O 1 2 t 1 t 1 07/09/67 19 ME212 .
Conservation of Angular Momentum Finally, if all angular impulse is zero. = H H O O 1 2 For a system of particles = H H O O 1 2 07/09/67 20 ME212 .
Example 2 The 2 kg disk rests on a smooth horizontal surface and is attached to an elastic cord that has a stiffness kc= 20 N/m and is initially unstretched. If the disk is given a velocity (vD)1 = 1.5 m/s, perpendicular to the cord, determine the rate at which the cord is being stretched and the speed of the disk at the instant the cord is stretched 0.2 m. 07/09/67 21 ME212 .
Meriam 3-226 The small spheres, which have the masses and initial velocities shown in the figure, strike and become attached to the spiked ends of the rod, which is freely pivoted at O and is initially at rest. Determine the angular velocity of the assembly after impact. Neglect the mass of the rod. 07/09/67 22 ME212 .
Meriam 3-230 The 6-kg sphere and 4-kg block are secured to the arm of negligible mass which rotates in the vertical plane about a horizontal axis at O. The 2-kg plug is released from rest at A and falls into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity 0 = 2 rad/s. Determine the angular velocity of the arm immediately after the plug has wedged itself in the block. 07/09/67 23 ME212 .
