Impact for Particles. The focus of this class is particle impact. We will not discuss impact for rigid bodies.

Transcription

1 Impact for Particles The focus of this class is particle impact. We will not discuss impact for rigid bodies. Terminology: 1. Central Impact: The incident and departure velocities of the two particles are collinear. (fter impact, and move along same line.) Central Impact Plane of Contact v 1 v 1 2. Oblique Impact: (See net page) Oblique Impact v 1 v 1

2 Terminology (cont d): Oblique Impact 2. Oblique Impact: Initial velocities of and are NOT collinear. Particles and strike a glancing blow and their departure velocities (at least one, or ) are at angles to their initial velocities. Important!! You must draw and label the plane of contact and line of impact. ssume no friction impulse along the plane of contact, thus: v y2 = v y1 Oblique Impact v y1 v 1 v 1 v 2 y v y2 v 2 v 2 v 1 v 1 EQUL! v y1 v y2 = v y1 EQUL! v 2 v y2 Plane of Contact

3 Terminology (cont d): Oblique Impact Labeling the plane of contact vs. the line of impact: Some tets label these tangential (t) and normal (n), but these confuse me. I always have to think about which is which. So, I usually just draw the picture and label them and y. You may do either, whichever makes sense to you. The point is this: Draw them and label them with something! Then, resolve the vectors into components along these aes. Oblique Impact v y1 EQUL! v 1 v 1 v 2 v 2 y v y2 v y2 v 2 v 2 v 1 v 1 Plane of Contact EQUL! v y1

4 Oblique Impact v y1 EQUL! v 1 v 1 v 2 v 2 y v y2 v y2 v 2 v 2 v 1 v 1 Plane of Contact EQUL! v y1 Resolving vectors along the plane of contact greatly aids solving the problem! This looks complicated, but once you see that v y2 v y2 = = v y1 v y1 then the problem becomes a straight line one in the -direction. Once you recognize that along the Plane of Contact, v y2 v y2 = = v y1 v y1 all that remains is a simple straight line problem in the -. direction only v 1 y v 1 Plane of Contact

5 Details ; Coefficient of Restitution What happens during impact? n Up-Close View: Description of Phase efore impact, relative velocity v 1/1 = v 1 - v 1 Deformation phase: Kinetic energy stored as deformation Ma deformation: and momentarily share same velocity, v. Close View of Particles and v 1 v v 1 Restitution phase: Energy stored as deformation converted back to KE. fter impact: If e < 1, some KE lost. Rel Velocity: v 2/2 = v 2 - v 2 v 2 v 2 Coefficient of Restitution, e: e = v 2/2 v 1/1 v 2 - v2 = ( ) v - v ( ) 1 1 v 2 - v2 = ( ) v - v ( ) 1 1

6 Details ; Coefficient of Restitution Coefficient of Restitution, e: (bbreviated as COR ) COR, e, is a measure of the energy stored in deformation during impact which is recovered back to kinetic energy. More precisely. (see your tet for a derivation ) Relative Departure Velocity Relative Incident Velocity e = = v 2/2 v 1/1 v 2 - v2 = ( ) v - v ( ) 1 1 v 2 - v2 = ( ) v - v ( ) 1 1 Cases: e = 1 e = 0 Perfectly Elastic Rel Departure Velocity = Rel Incident Velocity Perfectly Plastic (Particles stick together ) Rel Departure Velocity = 0 0 < e < 1 Range for e is between zero and one. Typical values:.5 to.8 for balls

7 More on Coefficient of Restitution (COR) Kinetic Energy recovered after impact is appro e 2. COR (e) of 0.8 sounds high. ut the KE after impact is (.8) 2 = 64% of the original KE, meaning 36% of KE was lost! pplications: Golf Drivers: The USG limits the COR of a driver s face to be no greater than COR = They have on-site testing facilities to test compliance (if requested). High school and college metal baseball bats: Using metal bats saves money because wood bats break. ut metal bats have a higher COR (batted balls have a 5-10% higher velocity off of a metal bat) than wooden bats. atting and slugging averages are inflated because of this. Pitchers are also subject to injury. Cool high speed videos of balls on bats: lso ecellent information at this site about bats and balls. Highly recommended.

8 Champaign News-Gazette, pril 19, 2003

9 Various alls: Rebound Height (function of energy recovered from bounce collision)

10 Equations for Impact Problems (Let be the, y be the Plane of Contact) long the Plane of Contact: (ssumes no friction impulse along this plane.) v y2 v y2 = = v y1 v y1 long the (Conservation of Momentum: m v 1 + m v m v = m v 2 lso along the : v 1 y v 1 Plane of Contact ( e = v 2 - v 2) ( ) v - v 1 1 Use a consistent sign convention for v s in these equations.

11 long the (-direction ) Write TWO equations to solve for TWO unknowns (v 2, v 2 ): Equation 1: Conservation of Momentum m v 1 + m v m v = m v 2 Equation 2: e Equation, Coeff of Restitution Equation ( e = v 2 - v 2) ( ) v - v 1 1 We know: v 1, v 1 y Solve for: v 2, v 2 using these two equations. v 1 v 1 Plane of Contact

12