APPLICATIONS CEE 271: Applied Mechanics II, Dynamics Lecture 17: Ch.15, Sec.4 7 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Date: The quality of a tennis ball is measured by the height of its bounce. This can be quantified by the coefficient of restitution of the ball. If the height from which the ball is dropped and the height of its resulting bounce are known, how can we determine the coefficient of restitution of the ball? 1 / 34 4 / 34 IMPACT (Section 15.4) APPLICATIONS (continued) Today s objectives: Students will be able to 1 Understand and analyze the mechanics of impact. 2 Analyze the motion of bodies undergoing a collision, in both central and oblique cases of impact. In-class activities: Reading Quiz Applications Central Impact Coefficient of Restitution Oblique Impact Concept Quiz Group Problem Solving Attention Quiz In the game of billiards, it is important to be able to predict the trajectory and speed of a ball after it is struck by another ball. If we know the velocity of ball A before the impact, how can we determine the magnitude and direction of the velocity of ball B after the impact? What parameters do we need to know for this? 2 / 34 5 / 34 READING QUIZ 1 When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be (a) central impact. (b) oblique impact. (c) major impact. (d) None of the above. 2 The ratio of the restitution impulse to the deformation impulse is called (a) impulse ratio. (b) restitution coefficient. (c) energy ratio. (d) mechanical efficiency. IMPACT Impact occurs when two bodies collide during a very short time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. The line of impact is a line through the mass centers of the colliding particles. In general, there are two types of impact: Central impact occurs when the directions of motion of the two colliding particles are along the line of impact. Oblique impact occurs when the direction of motion of one or both of the particles is at an angle to the line of impact. 3 / 34 6 / 34
PROCEDURE FOR ANALYSIS In most impact problems, the initial velocities of the particles and the coefficient of restitution, e, are known, with the final velocities to be determined. Define the x y axes. Typically, the x-axis is defined along the line of impact and the y-axis is in the plane of contact perpendicular to the x-axis. For both central and oblique impact problems, the following equations apply along the line of impact (x-dir.): Σm(v x ) 1 = Σm(v x ) 2 and e = [(v Bx ) 2 (v Ax ) 2 ]/[(v Ax ) 1 (v Bx ) 1 ] For oblique impact problems, the following equations are also required, applied perpendicular to the line of impact (y-dir.): m A (v Ay ) 1 = m A (v Ay ) 2 and m B (v By ) 1 = m B (v By ) 2 CONCEPT QUIZ 1 Two balls impact with a coefficient of restitution of 0.79. Can one of the balls leave the impact with a kinetic energy greater than before the impact? (a) Yes (b) No (c) Impossible to tell (d) Don t pick this one! ANS: (a) 2 Under what condition is the energy lost during a collision maximum? (a) e = 1.0 (b) e = 0.0 (c) e = 1.0 (d) Collision is non-elastic. 13 / 34 16 / 34 EXAMPLE GROUP PROBLEM SOLVING Top view Given: The ball strikes the smooth wall with a velocity (v b ) 1 = 20 m/s. The coefficient of restitution between the ball and the wall is e = 0.75. Find: The velocity of the ball just after the impact. Given: A 2 kg crate B is released from rest, falls a distance h = 0.5 m, and strikes plate P (3 kg mass). The coefficient of restitution between B and P is e = 0.6, and the spring stiffness is k = 30N/m. Find: The velocity of crate B just after the collision. Plan: The collision is an oblique impact, with the line of impact perpendicular to the plane (through the relative centers of mass). Thus, the coefficient of restitution applies perpendicular to the wall and the momentum of the ball is conserved along the wall. Plan: 1 Determine the speed of the crate just before the collision using projectile motion or an energy method. 2 Analyze the collision as a central impact problem. 14 / 34 17 / 34 EXAMPLE (Solution) Solve the impact problem by using x y axes defined along and perpendicular to the line of impact, respectively: The ball momentum is conserved in the y-dir: m(v b ) 1 sin30 = m(v b ) 2 sinθ (1) 10m/s = (v b ) 2 sinθ (2) The coefficient of restitution applies in the x-dir: e = 0.75 = [0 (v bx) 2 ] [(v bx ) 1 0] = [0 ( v b) 2 cosθ] [20cos30 0] (v b ) 2 cosθ = 12.99m/s (3) Using Eqs. (1) and (2) and solving for the velocity and θ yields: (v b ) 2 = (12.99 2 +10 2 ) 0.5 = 16.4m/s θ = tan 1 (10/12.99) = 37.6 15 / 34 GROUP PROBLEM SOLVING (continued) Determine the speed of block B just before impact by using conservation of energy (why?). Define the gravitational datum at the initial position of the block (h 1 = 0) and note the block is released from rest (v 1 = 0): T 1 +V 1 = T 2 +V 2 (4) 1 2 m(v 1) 2 +mgh 1 = 1 2 m(v 2) 2 +mgh 2 0+0 = 1 2 (2)(v 2) 2 +(2)(9.81)( 0.5) v 2 = 3.132m/s( ) (5) This is the speed of the block just before the collision. Plate (P) is at rest, velocity of zero, before the collision. 18 / 34
Again, EXAMPLE (Solution) 23.4 10 6 = 0.5(v B ) 2 (3.99 1014 ) r B (6) Now use Conservation of Angular Momentum. (r A m s v A )sinφ A = r B m s v B sin90 r B v B = (15 10 6 )(10000)sin70 or r B = (140.95 109 ) v B (7) Solving the two equations of Eq. (6) and (7) for r B and v B yields r B = 13.8 10 6 m (8) v B = 10.2km/s (9) ATTENTION QUIZ 1 A ball is traveling on a smooth surface in a 3 ft radius circle with a speed of 6ft/s. If the attached cord is pulled down with a constant speed of 2ft/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. (c) z direction. (b) y direction. (d) z y direction. ANS: (a) 31 / 34 34 / 34 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). 32 / 34 GROUP PROBLEM SOLVING (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 = 0.3727v 2 Angular impulse: t2 t 1 Mdt = t2 t 1 (0.5t+0.8)dt = [ (0.5/2)t 2 +0.8t ] 4 0 (10) = 7.2lb ft s (11) Apply the principle of angular impulse and momentum. 0+7.2 = 0.3727v 2 v 2 = 719.4ft/s 33 / 34