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: __________________

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IMPACT (Section 15.4)

Today’s objectives: Studentswill be able to

1 Understand and analyzethe mechanics of impact.

2 Analyze the motion ofbodies undergoing acollision, in both centraland oblique cases ofimpact.

In-class activities:• Reading Quiz• Applications• Central Impact• Coefficient of Restitution• Oblique Impact• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 When the motion of one or both of the particles is at anangle 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.

ANS: (b)

2 The ratio of the restitution impulse to the deformationimpulse is called(a) impulse ratio.(b) restitution coefficient.(c) energy ratio.(d) mechanical efficiency.

ANS: (b)

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APPLICATIONS

• The quality of a tennis ball is measured by the height of itsbounce. This can be quantified by the coefficient ofrestitution of the ball.

• If the height from which the ball is dropped and the heightof its resulting bounce are known, how can we determinethe coefficient of restitution of the ball?

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APPLICATIONS (continued)

• In the game of billiards, it is important to be able to predictthe trajectory and speed of a ball after it is struck byanother ball.

• If we know the velocity of ball A before the impact, how canwe determine the magnitude and direction of the velocity ofball B after the impact?

• What parameters do we need to know for this?5 / 34

IMPACT

• Impact occurs when two bodies collide during a very shorttime period, causing large impulsive forces to be exertedbetween the bodies. Common examples of impact are ahammer striking a nail or a bat striking a ball. The line ofimpact is a line through the mass centers of the collidingparticles. In general, there are two types of impact:

• Central impact occurs when thedirections of motion of the twocolliding particles are along the line ofimpact.

• Oblique impact occurs when thedirection of motion of one or both ofthe particles is at an angle to the lineof impact.

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CENTRAL IMPACT

• Central impact happens when the velocities of the twoobjects are along the line of impact (recall that the line ofimpact is a line through the particles’ mass centers).

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vA

line of impact

vB

• Once the particles contact, they may deformif they are non-rigid. In any case, energy istransferred between the two particles.

• There are two primary equations used when solvingimpact problems. The textbook provides extensive detailon their derivation.

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CENTRAL IMPACT (continued)

• In most problems, the initial velocities of the particles,(vA)1 and (vB)1, are known, and it is necessary todetermine the final velocities, (vA)2 and (vB)2. So the firstequation used is the conservation of linear momentum,applied along the line of impact.

(mAvA)1 + (mBvB)1 = (mAvA)2 + (mBvB)2

• This provides one equation, but there are usually twounknowns, (vA)2 and (vB)2. So another equation isneeded. The principle of impulse and momentum is used todevelop this equation, which involves the coefficient ofrestitution, or e.

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CENTRAL IMPACT (continued)

• The coefficient of restitution, e, is the ratio of the particles’relative separation velocity after impact, (vB)2 − (vA)2, tothe particles’ relative approach velocity before impact,(vA)1 − (vB)1. The coefficient of restitution is also anindicator of the energy lost during the impact.

• The equation defining the coefficient of restitution, e, is

e =(vB)2 − (vA)2(vA)1 − (vB)1

=(vB/A)After

(vA/B)Before=BAA

ABB

• If a value for e is specified, this relation provides thesecond equation necessary to solve for (vA)2 and (vB)2.

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COEFFICIENT OF RESTITUTION

• In general, e has a value between zero and one. The twolimiting conditions can be considered:

• Elastic impact (e = 1): In a perfectly elastic collision, noenergy is lost and the relative separation velocity equalsthe relative approach velocity of the particles. In practicalsituations, this condition cannot be achieved.

• Plastic impact (e = 0): In a plastic impact, the relativeseparation velocity is zero. The particles stick together andmove with a common velocity after the impact.

• Some typical values of e are:Steel on steel: 0.5− 0.8

Wood on wood: 0.4− 0.6

Lead on lead: 0.12− 0.18

Glass on glass: 0.93− 0.95

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IMPACT: ENERGY LOSSES

• Once the particles’ velocities before and after the collisionhave been determined, the energy loss during the collisioncan be calculated on the basis of the difference in theparticles’ kinetic energy. The energy loss is

ΣU1−2 = ΣT2 − ΣT1

where Ti = 12mi(vi) (i = 1, 2)

• During a collision, some of the particles’ initial kineticenergy will be lost in the form of heat, sound, or due tolocalized deformation.

• In a plastic collision (e = 0), the energy lost is a maximum.

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OBLIQUE IMPACT

• In an oblique impact, one or both of theparticles’ motion is at an angle to theline of impact. Typically, there will befour unknowns: the magnitudes anddirections of the final velocities.

• The four equations required to solve for the unknowns are:

1. Conservation of momentum and thecoefficient of restitution equation areapplied along the line of impact (x-axis):mA(vAx)1 +mB(vBx)1

= mA(vAx)2 +mB(vBx)2e = [(vBx)2 − (vAx)2]/[(vAx)1 − (vBx)1]

2. Momentum of each particle is conserved in the directionperpendicular to the line of impact (y-axis):mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2

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PROCEDURE FOR ANALYSIS

• In most impact problems, the initial velocities of theparticles 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 definedalong the line of impact and the y-axis is in the plane ofcontact perpendicular to the x-axis.

• For both central and oblique impact problems, the followingequations apply along the line of impact (x-dir.):

Σm(vx)1 = Σm(vx)2 ande = [(vBx)2 − (vAx)2]/[(vAx)1 − (vBx)1]

• For oblique impact problems, the following equations arealso required, applied perpendicular to the line of impact(y-dir.):

mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2

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EXAMPLE

Top view• Given: The ball strikes the smooth wall

with a velocity (vb)1 = 20 m/s. Thecoefficient of restitution between theball and the wall is e = 0.75.

• Find: The velocity of the ball just afterthe impact.

• Plan: The collision is an oblique impact, with the line ofimpact perpendicular to the plane (through the relativecenters of mass). Thus, the coefficient of restitution appliesperpendicular to the wall and the momentum of the ball isconserved along the wall.

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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(vb)1 sin 30◦ = m(vb)2 sin θ (1)10 m/s = (vb)2 sin θ (2)

• The coefficient of restitution applies in the x-dir:

e = 0.75 =[0− (vbx)2]

[(vbx)1 − 0]=

[0− (−vb)2 cos θ]

[20 cos 30◦ − 0]

(vb)2 cos θ = 12.99 m/s (3)• Using Eqs. (1) and (2) and solving for the velocity and θ

yields:(vb)2 = (12.992 + 102)0.5 = 16.4 m/s

θ = tan−1(10/12.99) = 37.6◦

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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 energygreater 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 collisionmaximum?(a) e = 1.0(b) e = 0.0(c) e = −1.0(d) Collision is non-elastic.

ANS: (b)

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GROUP PROBLEM SOLVING

• Given: A 2 kg crate B is released fromrest, falls a distance h = 0.5 m, andstrikes plate P (3 kg mass). Thecoefficient of restitution between B andP is e = 0.6, and the spring stiffness isk = 30 N/m.

• Find: The velocity of crate B just afterthe collision.

• Plan:

1 Determine the speed of the crate just before the collisionusing projectile motion or an energy method.

2 Analyze the collision as a central impact problem.17 / 34

GROUP PROBLEM SOLVING (continued)

• Determine the speed of block B just before impactby using conservation of energy (why?).

• Define the gravitational datum at the initial positionof the block (h1 = 0) and note the block is releasedfrom rest (v1 = 0):

T1 + V1 = T2 + V2 (4)12m(v1)

2 +mgh1 = 12m(v2)

2 +mgh2

0 + 0 = 12(2)(v2)

2 + (2)(9.81)(−0.5)

v2 = 3.132 m/s(↓) (5)

• This is the speed of the block just before thecollision. Plate (P ) is at rest, velocity of zero, beforethe collision.

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GROUP PROBLEM SOLVING(continued)

• Analyze the collision as a central impact problem.

vB2

vP2 vP1 = 0

vB1 = 3.132m/s

B (2kg)

P (3kg)

• Apply conservation of momentum to thesystem in the vertical direction (+ ↑):

mB(vB)1 +mP (vP )1 = mB(vB)2 +mP (vP )2

(2)(−3.132) + 0 = (2)(vB)2 + (3)(vP )2

• Using the coefficient of restitution (+ ↑):

e = [(vP )2 − (vB)2]/[(vB)1 − (vP )1]

0.6 = [(vP )2 − (vB)2]/[−3.132− 0]

−1.879 = (vP )2 − (vB)2

• Solving the two equations simultaneously yields(vB)2 = −0.125 m/s ↓ and (vP )2 = −2.00 m/s ↓

• Both the block and plate will travel down after the collision.19 / 34

ATTENTION QUIZ

1. Block B (1 kg) is moving on the smooth surface at 10 m/swhen it squarely strikes block A (3 kg), which is at rest. Ifthe velocity of block A after the collision is 4 m/s to theright, (vB)2 is

(a) 2 m/s→

(b) 7 m/s←

(c) 7 m/s→

(d) 2 m/s←

ANS: (d)

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3kgB1kg

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ANGULAR MOMENTUM, MOMENT OF A FORCEAND PRINCIPLE OF ANGULAR IMPULSE AND

MOMENTUM (section 15.5)

Today’s objectives: Studentswill be able to

1 Determine theangular momentum of aparticle and apply theprinciple of angular impulseand momentum.

2 Use conservation of angularmomentum to solveproblems.

In-class activities:• Reading Quiz• Applications• Angular Momentum• Angular Impulse and

Momentum Principle• Conservation of Angular

Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 Select the correct expression for the angular momentum ofa particle about a point.(a) r × v(b) r × (mv)(c) v × r(d) (mv)× r

ANS: (b)

2 The sum of the moments of all external forces acting on aparticle is equal to(a) angular momentum of the particle.(b) linear momentum of the particle.(c) time rate of change of angular momentum.(d) time rate of change of linear momentum.

ANS: (a)

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APPLICATIONS

• Planets and most satellites move in elliptical orbits. Thismotion is caused by gravitational attraction forces. Sincethese forces act in pairs, the sum of the moments of theforces acting on the system will be zero. This means thatangular momentum is conserved.

• If the angular momentum is constant, does it mean thelinear momentum is also constant? Why or why not?

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APPLICATIONS(continued)

• The passengers on theamusement-park ride experienceconservation of angular momentum aboutthe axis of rotation (the z-axis). Asshown on the free body diagram, theline of action of the normal force, N ,passes through the z-axis and theweight’s line of action is parallel to it.Therefore, the sum of moments of thesetwo forces about the z-axis is zero.

• If the passenger moves away from thez-axis, will his speed increase ordecrease? Why?

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ANGULAR MOMENTUM (Section 15.5)

• The angular momentum of a particle about point O isdefined as the ‘moment’ of the particle’s linear momentumabout O.

• Ho = r ×mv =

i j krx ry rzmvx mvy mvz

• The magnitude of Ho is (Ho)z = mvd,

where d is the shortest distance betweenthe object and the axis.

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RELATIONSHIP BETWEEN MOMENT OF A FORCEAND ANGULAR MOMENTUM (Section 15.6)

• The resultant force acting on the particle is equal to thetime rate of change of the particle’s linear momentum.Showing the time derivative using the familiar ‘dot’ notationresults in the equation

ΣF = L = mv

• We can prove that the resultant moment acting on theparticle about point O is equal to the time rate of change ofthe particle’s angular momentum about point O or

ΣMo = r × F = Ho

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PRINCIPLE OF ANGULAR IMPULSE ANDMOMENTUM (Section 15.7)

• Considering the relationship between moment and timerate of change of angular momentum

ΣMo = Ho =dHo

dt• By integrating between the time interval t1 to t2

Σ

∫ t2

t1

Modt = (Ho)2 − (Ho)1

or

(Ho)1 + Σ

∫ t2

t1

Modt = (Ho)2

• This equation is referred to as the principle ofangular impulse and momentum. The second term on theleft side, Σ

∫Modt, is called the angular impulse. In cases

of 2D motion, it can be applied as a scalar equation usingcomponents about the z-axis. 27 / 34

CONSERVATION OF ANGULAR MOMENTUM

• When the sum of angular impulses acting on a particle or asystem of particles is zero during the time t1 to t2, theangular momentum is conserved. Thus,

(HO)1 = (HO)2

• An example of this condition occurs when aparticle is subjected only to a central force.In the figure, the force F is always directedtoward point O. Thus, the angular impulse ofF about O is always zero, and angularmomentum of the particle about O isconserved: r ‖ F so that

r × F ∝ r × (−r) = 0

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EXAMPLE

• Given: A satellite has an elliptical orbitabout earth.msatellite = 700 kgmearth = 5.976× 1024 kgvA = 10 km/srA = 15× 106 mφA = 70◦

• Find: The speed, vB, of the satellite at its closest distance,rB, from the center of the earth.

• Plan: Apply the principles of conservation of energy andconservation of angular momentum to the system.

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EXAMPLE (Solution)

• Conservation of energy: TA + VA = TB + VB becomes

1

2msv

2A −G

msme

rA=

1

2msv

2B −G

msme

rB

where G = 66.73× 10−12m3/(kg · s2). Dividing through byms and substituting values yields:

0.5(10000)2 −66.73× 10−12

(5.976× 1024

)15× 106

=

0.5(vB)2 −66.73× 10−12

(5.976× 1024

)rB

or

23.4× 106 = 0.5(vB)2 − (3.99× 1014)

rB

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EXAMPLE (Solution)

Again,

23.4× 106 = 0.5(vB)2 − (3.99× 1014)

rB(6)

• Now use Conservation of Angular Momentum.

(rAmsvA) sinφA = rBmsvB sin 90◦

rBvB = (15× 106)(10000) sin 70◦ or

rB =(140.95× 109)

vB(7)

• Solving the two equations of Eq. (6) and (7) for rB and vByields

rB = 13.8× 106m (8)vB = 10.2 km/s (9)

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GROUP PROBLEM SOLVING

• Given: The four 5 lb spheres are rigidlyattached to the crossbar frame, whichhas a negligible weight. A moment actson 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 andmomentum about the axis of rotation (z-axis).

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GROUP PROBLEM SOLVING (Solution)

• Angular momentum: HZ = r ×mv reduces to a scalarequation. (HZ)1 = 0 and

(HZ)2 = 4× (5/32.2)(0.6)v2 = 0.3727v2

• Angular impulse:∫ t2

t1

Mdt =

∫ t2

t1

(0.5t+ 0.8)dt

=[(0.5/2)t2 + 0.8t

]40

(10)= 7.2 lb · ft · s (11)

• Apply the principle of angular impulse and momentum.

0 + 7.2 = 0.3727v2

v2 = 719.4 ft/s

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ATTENTION QUIZ1 A ball is traveling on a smooth surface in a 3 ft radius circle

with a speed of 6 ft/s. If the attached cord is pulled downwith a constant speed of 2 ft/s, which of the followingprinciples can be applied to solve for the velocity of the ballwhen r = 2 ft?

(a) Conservation of energy(b) Conservation of angular momentum(c) Conservation of linear momentum(d) Conservation of mass

ANS: (b)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)34 / 34