the focus of this class is particle impact. we will not discuss impact for rigid bodies. impact for...
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The focus of this class is particle impact. We will not discuss impact for rigid bodies.
Impact for Particles
Terminology:
1. Central Impact: The incident and departure velocities of the two particles are collinear. (Afterimpact, A and Bmove along sameline.)
A B
P lane of Contac t
L ine of Impac t
vA 1 vB 1
Centra l Impac t
2. Oblique Impact:(See next page)
BvA 1
vB 1
A
Oblique Impac t
Terminology (cont’d): Oblique Impact
2. Oblique Impact: Initial velocities of A and B are NOT collinear. Particles A and B strike a glancing blow and their departure velocities(at least one, A or B) are at angles to their initial velocities.
B
L ine of Impac t
vA 1x
vB 1x
vA 2x
vB 2x
vA 1y
vB 1y
vA 2y
vB 2y
x
y
A
B
vA1
vA 2
vB 1
vB 2
P lane of Contac t
A
Oblique Impac t
EQUA L !
EQUAL !
Important!!
You must drawand label theplane of contactand line of impact.
Assume nofriction impulsealong the planeof contact, thus:
vAy2 = vAy1
vBy2 = vBy1
Terminology (cont’d): Oblique Impact
Labeling the plane of contact vs. the line of impact: Some texts label these tangential (t) and normal (n), but these confuse me.I always have to think about which is which. So, I usually just
B
L ine of Impac t
vA 1x
vB 1x
vA 2x
vB 2x
vA 1y
vB 1y
vA 2y
vB 2y
x
y
A
B
vA1
vA 2
vB 1
vB 2
P lane of Contac t
A
Oblique Impac t
EQUA L !
EQUAL !
draw the pictureand label themx and y. You may do either,whichever makessense to you.The point is this:Draw them andlabel them withsomething!Then, resolve thevectors into com-ponents alongthese axes.
B
L ine of Impac t
vA 1x
vB 1x
vA 2x
vB 2x
vA 1y
vB 1y
vA 2y
vB 2y
x
y
A
B
vA1
vA 2
vB 1
vB 2
P lane of Contac t
A
Oblique Impac t
EQUA L !
EQUAL !
T his looks c omplic ated,but onc e you see that
vA 1yvB 1y
vA 2yvB 2y
=
=then the problem bec omesa s tra ight l ine one in thex -direc tion.
B
L ine of Impac t
vA 1x
vB 1x
xy
P lane of Contac t
A
Onc e you rec ognize that a long the P lane of C ontac t,vA 1yvB 1y
vA 2yvB 2y
=
=all that remains is a
problem in the x -. s imple s tra ight l ine
direc tion only
Resolving vectors along the plane of contact greatly aids solving the problem!
Details of Impact; Coefficient of Restitution
What happens during impact? An Up-Close View:
A BB efore impac t, re lative veloc ity vA 1
v
v B 1
vA 2 vB 2
v = v - v B 1/A 1 B 1 A 1
K inetic energy s tored as deformation
momentarily s hare s ame veloc ity, v.
deformation c onver ted bac k to K E .
Rel Veloc ity: v = v - v B 2/A2 B 2 A 2
M ax deformation: A and B
Deformation phas e:
Res titution phase: E nergy s tored as
A fter impac t: I f e < 1, s ome K E los t.
A C lose V iew of Impac tof P artic les A and B
Des c ription of P hase of Impac t
e =vB 2/A 2
v B 1 / A 1
v - vB 2 A 2
v - v B 1 A 1=
( )( )
v - vB 2 A 2
v - v A 1 B 1=
( )( )
Coefficient ofRestitution, e:
Details of Impact; Coefficient of Restitution
Coefficient of Restitution, e: (Abbreviated 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 text for a derivation…)
Relative Departure Veloc ity
Relative Inc ident Veloc ity
e = =vB 2/A 2
v B 1 / A 1
v - vB 2 A 2
v - v B 1 A 1=
( )( )
v - vB 2 A 2
v - v A 1 B 1=
( )( )
Cases: e = 1 “Perfectly Elastic” Rel Departure Velocity = Rel Incident Velocity
e = 0 “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
More on Coefficient of Restitution (COR)
Kinetic Energy recovered after impact is approx e2 .A COR (e) of 0.8 sounds high. But the KE after impact is (.8)2 = 64% of the original KE, meaning 36% of KE was lost!
Applications:
Golf Drivers: The USGA limits the COR of a driver’s face to be no greater than COR = 0.83. 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. But metal bats have a higher COR (batted balls have a 5-10% higher velocity off of a metal bat) than wooden bats. Batting and slugging averages are inflated because of this. Pitchers are also subject to injury.
Cool high speed videos of balls on bats: http://www.kettering.edu/~drussell/bats-new/ball-bat-0.htmlAlso excellent information at this site about bats and balls. Highly recommended.
Equations for Impact Problems(Let x be the Line of Impact, y be the Plane of Contact)
Along the Plane of Contact:(Assumes no friction impulse along this plane….)
vA 1yvB 1y
vA 2yvB 2y
=
=
B
L ine of Impac t
vA 1x
vB 1x
xy
P lane of Contac t
A
Along the Line of Impact (Conservation of Momentum:
m vA A 1x =m vB B 1x+ m vA A 2x m vB B 2x+
Also along the Line of Impact:
e =v x - B 2 v xA 2( )
( )v x - v xA 1 B 1
Use a consistent sign convention for v’s in these equations.
Along the Line of Impact (x-direction…)
B
L ine of Impac t
vA 1x
vB 1x
xy
P lane of Contac t
A
Write TWO equations to solve for TWO unknowns (vAx2, vBx2 ):
m vA A 1x =m vB B 1x+ m vA A 2x m vB B 2x+
e =v x - B 2 v xA 2( )
( )v x - v xA 1 B 1 We know: vAx1, vBx1
Equation 1: Conservation of Momentum
Equation 2: “e Equation”, Coeff of Restitution Equation
Solve for: vAx2, vBx2
using these two equations.