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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUMToday’s Objectives:Students will be able to:1. Calculate the linear momentum

of a particle and linear impulseof a particle and linear impulse of a force.

2. Apply the principle of linear impulse and momentum.

In-Class Activities:• Check Homework• Reading Quiz• Applications• Linear Momentum and Impulse• Principle of Linear Impulse and p p

Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

APPLICATIONS

A dent in an automotive fender can be removed using an impulse tool, which delivers a force over a very short time interval. To do so the weight is gripped and jerked upwards, striking the stop ring.

How can we determine the magnitude of the linear impulse applied to the fender?

Could you analyze a carpenter’s hammer striking a nail in the same fashion? Sure!

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APPLICATIONS(continued)

When a stake is struck by a sledgehammer, a large impulse f i d li d t th t k dforce is delivered to the stake and drives it into the ground.

If we know the initial speed of the sledgehammer and the duration of impact, how can we determine the

it d f th i l i fmagnitude of the impulsive force delivered to the stake?

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(Section 15.1)

The next method we will consider for solving particle kinetics problems is obtained by integrating the equation of

The result is referred to as the principle of impulse and momentum. It can be applied to problems involving both linear and angular motion.

kinetics problems is obtained by integrating the equation of motion with respect to time.

This principle is useful for solving problems that involve force, velocity, and time. It can also be used to analyze the mechanics of impact (taken up in a later section).

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The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. The equation of motion can be written

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)

The equation of motion can be written ∑F = m a = m (dv/dt)

Separating variables and integrating between the limits v = v1at t = t1 and v = v2 at t = t2 results in

mv2 – mv1dvmF dtv2t2

== ∫∑ ∫

This equation represents the principle of linear impulse and momentum. It relates the particle’s final velocity (v2) and initial velocity (v1) and the forces acting on the particle as a function of time.

v1t1

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (continued)

Linear momentum: The vector mv is called the linear momentum, denoted as L. This vector has the same direction as v. The linear momentum vector has units of (kg·m)/s or (slug·ft)/s.

Linear impulse: The integral ∫F dt is the linear impulse, denoted I. It is a vector quantity measuring the effect of a force during its time interval of action. I acts in the same direction as F and has units of N·s or lb·s.

momentum vector has units of (kg m)/s or (slug ft)/s.

The impulse may be determined by di t i t ti G hi ll itdirect integration. Graphically, it can be represented by the area under the force versus time curve. If F is constant, then

I = F (t2 – t1) .

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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)

The principle of linear impulse and momentum in vector form is written as

F dtt2

∑∫mv + = mv

The two momentum diagrams indicate direction

The particle’s initial momentum plus the sum of all the impulses applied from t1 to t2 is equal to the particle’s final momentum.

F dtt1

∑∫mv1 + mv2

and magnitude of the particle’s initial and final momentum, mv1 and mv2. The impulse diagram is similar to a free body diagram, but includes the time duration of the forces acting on the particle.

IMPULSE AND MOMENTUM: SCALAR EQUATIONSSince the principle of linear impulse and momentum is a vector equation, it can be resolved into its x, y, z componentscalar equations:

m(v ) + ∑ F dt = m(v )

t2

∫m(vx)1 + ∑ Fx dt = m(vx)2

m(vy)1 + ∑ Fy dt = m(vy)2

m(v )1 + ∑ F dt = m(v )2

t1

∫t2

t1

∫t2

∫The scalar equations provide a convenient means for applying the principle of linear impulse and momentum once the velocity and force vectors have been resolved into x, y, z components.

m(vz)1 + ∑ Fz dt m(vz)2t1

∫

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PROBLEM SOLVING

• Establish the x, y, z coordinate system.

• Draw the particle’s free body diagram and establish the di ti f th ti l ’ i iti l d fi l l iti d i

• Resolve the force and velocity (or impulse and momentum) vectors into their x, y, z components, and apply the principle of linear impulse and momentum using its scalar form.

direction of the particle’s initial and final velocities, drawing the impulse and momentum diagrams for the particle. Show the linear momenta and force impulse vectors.

• Forces as functions of time must be integrated to obtain impulses. If a force is constant, its impulse is the product of the force’s magnitude and time interval over which it acts.

of linear impulse and momentum using its scalar form.

EXAMPLE

Given: A 0.5 kg ball strikes the rough ground and rebounds with the velocities shown. Neglect the ball’s weight during the time it impacts the ground.

Find: The magnitude of impulsive force exerted on the ball.

Plan: 1) Draw the momentum and impulse diagrams of the ball as it hits the surface.ball as it hits the surface.

2) Apply the principle of impulse and momentum to determine the impulsive force.

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EXAMPLE (continued)

Solution:

1) The impulse and momentum diagrams can be drawn as:

+ =∫ W dt ≈ 0

∫ N dt ≈ 0∫ F dt

mv2

30°

mv1

45°

The impulse caused by the ball’s weight and the normal force N can be neglected because their magnitudes are very small as compared to the impulse from the ground.

2) The principle of impulse and momentum can be applied along the direction of motion:

mv1 + ∑ F dt = mv2

t2

∫

EXAMPLE (continued)

mv1 + ∑ F dt mv2t1∫

0.5 (25 cos 45° i − 25 sin 45° j) + ∑ F dt

= 0.5 (10 cos 30° i + 10 sin 30° j)

t2

t1

∫ ⇒

The impulsive force vector ist

I = ∑ F dt = (4.509 i + 11.34 j ) N⋅s

Magnitude: I = √ 4.5092 + 11.342 = 12.2 N⋅s

∫t1

t2

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CHECK YOUR UNDERSTANDING QUIZ

1. Calculate the impulse due to the force.

A) 20 kg·m/s B) 10 kg·m/s

F

10 N Force curve

A) 20 kg m/s B) 10 kg m/s

C) 5 N·s D) 15 N·s t2 s

2. A constant force F is applied for 2 s to change the particle’s velocity from v1 to v2. Determine the force F if the particle’s mass is 2 kg.

A) (17.3 j) N B) (–10 i +17.3 j) N

C) (20 i +17.3 j) N D) ( 10 i +17.3 j) Nv2=20 m/s

v1=10 m/s60°

GROUP PROBLEM SOLVING

Given: The 20 kg crate is resting on the floor. The motor M pulls on the cable with a f f F hi h hforce of F, which has a magnitude that varies as shown on the graph.

Find: The speed of the crate when t = 6 s.

Plan:1) Determine the force needed to begin lifting the crate, and

then the time needed for the motor to generate this force.2) After the crate starts moving, apply the principle of

impulse and momentum to determine the speed of the crate at t = 6 s.

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GROUP PROBLEM SOLVING (continued)Solution:

F = mg = (20) (9 81) = 196 2 N

1) The crate begins moving when the cable force F exceeds the crate weight. Solve for the force, then the time.

F = mg = (20) (9.81) = 196.2 NF = 196.2 N = 50 tt = 3.924 s

2) Apply the principle of impulse and momentum from the time the crate starts lifting at t1 = 3.924 s to t2 = 6 s.

Note that there are two external forces (cable force and weight) we need to consider.

A. The impulse due to cable force:

F dt = [0.5(250) 5 + (250) 1] – 0.5(196.2)3.924= 490.1 N⋅s ∫3.924

6+↑

GROUP PROBLEM SOLVING (continued)

B. The impulse due to weight:

+↑ (− mg) dt = − 196.2 (6 − 3.924) = − 407.3 N⋅s ∫3.924

6

Now, apply the principle of impulse and momentum

mv1 + ∑ F dt = mv2 where v1 = 0t2

t1

∫

0 + 490.1 − 407.3 = (20) v2

+↑

=> v2 = 4.14 m/s

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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM AND CONSERVATION OF LINEAR MOMENTUM FOR

SYSTEMS OF PARTICLESToday’s Objectives:Students will be able to: In-Class Activities:Students will be able to:1. Apply the principle of linear

impulse and momentum to a system of particles.

2. Understand the conditions for conservation of momentum.

• Check Homework• Reading Quiz• Applications• Linear Impulse and

Momentum for a System of Particles

• Conservation of Linear• Conservation of Linear Momentum

• Concept Quiz• Group Problem Solving• Attention Quiz

APPLICATIONS

As the wheels of this pitching machine rotate, they apply frictional impulses to the ball, thereby giving it linearthe ball, thereby giving it linear momentum in the direction of Fdt and F ’dt.The weight impulse, WΔt is very small since the time the ball is in contact with the wheels is very small.

Does the release velocity of the ball depend on the mass of the ball?

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APPLICATIONS (continued)

This large crane-mounted hammer is used to drive piles into the ground.

Conservation of momentum can be used to find the velocity of the pile just after impact, assuming the hammer does not rebound off the pile.

If the hammer rebounds, does the pile velocity change from th h th h d ’t b d ? Wh ?the case when the hammer doesn’t rebound ? Why ?

In the impulse-momentum analysis, do we have to consider the impulses of the weights of the hammer and pile and the resistance force ? Why or why not ?

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES

(Section 15.2)

For the system of particles shown,

The linear impulse and momentum equation for this system

the internal forces fi between particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero.

The linear impulse and momentum equation for this system only includes the impulse of external forces.

mi(vi)2dtFimi(vi)1

t2

t1∑=∑+∑ ∫

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For a system of particles, we can define a “fictitious” center of mass of an aggregate particle of mass mtot, where mtot is

MOTION OF THE CENTER OF MASS

gg g p tot totthe sum (∑ mi) of all the particles. This system of particles then has an aggregate velocity of vG = (∑ mivi) / mtot.

The motion of this fictitious mass is based on motion of the center of mass for the system.

The position vector rG = (∑ miri) / mtot describes the motion of the center of mass.

CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES (Section 15.3)

When the sum of external impulses acting on a system of objects is zero, the linear impulse-momentum equation simplifies toequation simplifies to

∑ mi(vi)1 = ∑ mi(vi)2

This equation is referred to as the conservation of linear momentum. Conservation of linear momentum is often applied when particles collide or interact. When particles impact, only impulsive forces cause a change of linear momentum.g

The sledgehammer applies an impulsive force to the stake. The weight of the stake is considered negligible, or non-impulsive, as compared to the force of the sledgehammer. Also, provided the stake is driven into soft ground with little resistance, the impulse of the ground acting on the stake is considered non-impulsive.

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EXAMPLE I

Given: M = 100 kg, vi = 20j (m/s)mA = 20 kg, vA = 50i + 50j (m/s)mB = 30 kg, vB = -30i – 50k (m/s)y vB

vA

B g, B ( )An explosion has broken themass m into 3 smaller particles, a, b and c.

Fi d Th l it f f t C ft

x

y

Mvi vC

C

B

A

=

Find: The velocity of fragment C after the explosion.

Plan: Since the internal forces of the explosion cancel out, we can apply the conservation of linear momentum to the SYSTEM.

z

mvi = mAvA + mBvB + mCvC

EXAMPLE I (continued)

Solution:

100(20j) = 20(50i + 50j) + 30(-30i-50k) + 50(vcx i + vcy j + vcz k)

Equating the components on the left and right side yields:

0 = 1000 – 900 + 50(vcx) vcx = -2 m/s

2000 = 1000 + 50 (vcy) vcy = 20 m/s

0 = -1500 + 50 (vcz) vcz = 30 m/s

So vc = (-2i + 20j + 30k) m/s immediately after the explosion.

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EXAMPLE II

Given: Two rail cars with masses of mA = 20 Mg and mB = 15 Mg and velocities as h

Find: The speed of the car A after collision if the cars collide and rebound such that B moves to the right with a speed of 2 m/s. Also find the average impulsive force between the cars if the collision place in 0.5 s.

shown.

Plan: Use conservation of linear momentum to find the velocity of the car A after collision (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car.

EXAMPLE II (continued)

Conservation of linear momentum (x-dir):mA(vA1) + mB(vB1) = mA(vA2)+ mB(vB2)

Solution:

20,000 (3) + 15,000 (-1.5) = (20,000) vA2 + 15,000 (2)

vA2 = 0.375 m/s

Impulse and momentum on car A (x-dir):

mA (vA1)+ ∫ F dt = mA (vA2)

The average force is

∫ F dt = 52,500 N·s = Favg(0.5 sec); Favg = 105 kN

mA (vA1) ∫ F dt mA (vA2)

20,000 (3) - ∫ F dt = 20,000 (0.375)

∫ F dt = 52,500 N·s

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CONCEPT QUIZ1) Over the short time span of a tennis ball hitting the racket

during a player’s serve, the ball’s weight can be considered _____________A) nonimpulsive

2) A drill rod is used with a air hammer for making holes in hard rock so explosives can be placed in them How many

A) nonimpulsive.B) impulsive.C) not subject to Newton’s second law.D) Both A and C.

hard rock so explosives can be placed in them. How many impulsive forces act on the drill rod during the drilling?

A) None B) One

C) Two D) Three

GROUP PROBLEM SOLVING

Given:The free-rolling ramp has a weight of 120 lb. The 80 lb crate slides from rest at A, 15 ft down

Find: The ramp’s speed when the crate reaches B.

the ramp to B. Assume that the ramp is smooth, and neglect the mass of the wheels.

Plan: Use the energy conservation equation as well as conservation of linear momentum and the relative velocity equation (you thought you could safely forget it?) to find the velocity of the ramp.

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GROUP PROBLEM SOLVING(continued)

Energy conservation equation:0 + 80 (3/5) (15) = 0 5 (80/32 2)(v )2 + 0 5 (120/32 2)(v )2

Solution:

= 0.5 (80/32.2)(vB) + 0.5 (120/32.2)(vr)

Since vB = vr + vB/r ⇒ -vBx i + vBy j = vr i + vB/r (−4/5 i −3/5 j)

+

To find the relations between vB and vr, useconservation of linear momentum:→ 0 = (120/32.2) vr − (80/32.2) vBx

⇒ vBx =1.5 vr (1)

Since vB vr vB/r ⇒ vBx i vBy j vr i vB/r ( 4/5 i 3/5 j)⇒ -vBx = vr − (4/5) vB/r (2)

vBy = − (3/5) vB/r (3)

Eliminating vB/r from Eqs. (2) and (3) and Substituting Eq. (1) results in vBy =1.875 vr

GROUP PROBLEM SOLVING(continued)

Then, energy conservation equation can be rewritten ;

0 + 80 (3/5) (15) = 0.5 (80/32.2)(vB)2 + 0.5 (120/32.2)(vr)2

0 + 80 (3/5) (15) = 0.5 (80/32.2) [(1.5 vr)2 +(1.875 vr)2]+ 0.5 (120/32.2) (vr)2

720 = 9.023 (vr)2

8 93 ft/vr = 8.93 ft/s

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IMPACTToday’s Objectives:Students will be able to:1. Understand and analyze the

mechanics of impact.In Class Activities:

2. Analyze the motion of bodies undergoing a collision, in both central and oblique cases of impact.

In-Class Activities:• Check Homework• Reading Quiz• Applications• Central Impact• Coefficient of Restitution• Oblique Impact• Concept Quiz• Group Problem Solving• Attention Quiz

APPLICATIONS

The quality of a tennis ball is measured by the height of its bounce. This can be quantified by the coefficient of

tit ti f th b llrestitution 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?

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APPLICATIONS(continued)

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?

IMPACT (Section 15.4)

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.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.

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CENTRAL IMPACT

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

vA vB

Line of impact

Once the particles contact, they may deform if they are non-rigid. In any

i t f d b t th

There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation.

case, energy is transferred between the two particles.

CENTRAL IMPACT (continued)

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

(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2

This provides one equation but there are usually two unknownsThis provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. The principle of impulse and momentum is used to develop this equation, which involves the coefficient of restitution, or e.

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The coefficient of restitution, e, is the ratio of the particles’ relative separation velocity after impact, (vB)2 – (vA)2, to the

CENTRAL IMPACT (continued)

p y p ( B)2 ( A)2particles’ relative approach velocity before impact, (vA)1 – (vB)1. The coefficient of restitution is also an indicator of the energy lost during the impact.

The equation defining the coefficient of restitution, e, is

(vB)2 – (vA)2

(vA)1 - (vB)1

( B)2 ( A)2e =

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

COEFFICIENT OF RESTITUTION

El i i ( 1) I f l l i lli i

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

• Plastic impact (e = 0): In a plastic impact, the relative separation velocity is zero. The particles stick together and

• Elastic impact (e = 1): In a perfectly elastic collision, no energy is lost and the relative separation velocity equals the relative approach velocity of the particles. In practical situations, this condition cannot be achieved.

p y p gmove 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 collision have been determined, the energy loss during the collision can be calculated on the basis of the difference in the

During a collision, some of the particles’ initial kinetic energy will be lost in the form of heat, sound, or due to localized deformation

can be calculated on the basis of the difference in the particles’ kinetic energy. The energy loss is

∑ U1-2 = ∑ T2 − ∑ T1 where Ti = 0.5mi (vi)2

In a plastic collision (e = 0), the energy lost is a maximum, although it does not necessarily go to zero. Why?

localized deformation.

OBLIQUE IMPACT

In an oblique impact, one or both of the particles’ motion is at an angle to the line of impact. Typically, there will be four unknowns: the magnitudes and directions ofunknowns: the magnitudes and directions of the final velocities.

Conservation of momentum and the coefficient of restitution equation are applied along the line of impact (x-axis):

mA(vA )1 + mB(vB )1 = mA(vA )2 + mB(vB )2

The four equations required to solve for the unknowns are:

Momentum of each particle is conserved in the direction perpendicular to the line of impact (y-axis):

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

mA(vAx)1 + mB(vBx)1 mA(vAx)2 + mB(vBx)2

e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]

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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.

• For both central and oblique impact problems, the following equations apply along the line of impact (x-dir.):∑ ( ) ∑ ( ) d [( ) ( ) ]/[( ) ( ) ]

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

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

∑ m(vx)1 = ∑ m(vx)2 and e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]

EXAMPLE

Given: The ball strikes the smooth wall with a velocity (vb)1 = 20 m/s. The coefficient of restitution between the ball and the wall is e = 0 75the ball and the wall is e = 0.75.

Find: The velocity of the ball just after the impact.

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.

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Solve the impact problem by using x-y axes defined along and perpendicular to the line of impact, respectively:

The momentum of the ball is conserved in

EXAMPLE (continued)Solution:

The momentum of the ball is conserved in the y-dir:

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

The coefficient of restitution applies in the x-dir:e = [ 0 – (vbx)2 ] / [ (vbx)1 – 0 ]

⇒ 0 75 = [ 0 – (-vb)2 cos θ ] / [ 20 cos 30° – 0]⇒ 0.75 [ 0 ( vb)2 cos θ ] / [ 20 cos 30 0]⇒ (vb)2 cos θ = 12.99 m/s (2)

Using Eqs. (1) and (2) and solving for the velocity and θ yields:(vb)2 = (12.992+102)0.5 = 16.4 m/sθ = tan-1(10/12.99)=37.6°

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?

2. Under what condition is the energy lost during a collision i ?

A) Yes B) No

C) Impossible to tell D) Don’t pick this one!

maximum?

A) e = 1.0 B) e = 0.0

C) e = -1.0 D) Collision is non-elastic.

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GROUP PROBLEM SOLVING

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

ffi i f i i bcoefficient of restitution between B and P is e = 0.6, and the spring stiffness is k = 30 N/m.

Find: The velocity of crate B just after the collision.

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.

GROUP PROBLEM SOLVING (continued)

Solution:

1) Determine the speed of block B just before impact by using ti f ( h ?) D fi th it ti l

0 0 0 5(2)( )2 (2)(9 81)( 0 5)

conservation of energy (why?). Define the gravitational datum at the initial position of the block (h1 = 0) and note the block is released from rest (v1 = 0):

T1 + V1 = T2 + V2

0.5m(v1)2 + mgh1 = 0.5m(v2)2 + mgh2

0 + 0 = 0.5(2)(v2)2 + (2)(9.81)(-0.5)

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

v2 = 3.132 m/s

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Apply conservation of momentum to the

2) Analyze the collision as a central impact problem.(vB)2 (vB)1 = 3.132 m/s

GROUP PROBLEM SOLVING (continued)

pp ysystem 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]

(vP)2 (vP)1 = 0

BP

=> 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.

ATTENTION QUIZ

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

2. A particle strikes the smooth surface with a velocity of 30 m/s. If e = 0.8, (vx) 2 is y

right, (vB)2 is

A) 2 m/s B) 7 m/s

C) 7 m/s D) 2 m/sB A

vB=10 m/s

y ( x) 2_____ after the collision.

A) zero B) equal to (vx) 1

C) less than (vx) 1 D) greater than (vx) 1

θ

v30 m/s x

y30°

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ANGULAR MOMENTUM, MOMENT OF A FORCE AND PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUMToday’s Objectives:Students will be able to:1. Determine the angular

momentum of a particle andIn-Class Activities:• Check Homeworkmomentum of a particle and

apply the principle of angular impulse & momentum.

2. Use conservation of angular momentum to solve problems.

• Check Homework• Reading Quiz• Applications• Angular Momentum• Angular Impulse and

Momentum Principle• Conservation of Angular

Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

APPLICATIONS

Planets and most satellites move in elliptical orbits. This motion is caused by gravitational attraction forces. Since these forces act in pairs, the sum of the moments of the forcesthese forces act in pairs, the sum of the moments of the forces acting on the system will be zero. This means that angular momentum is conserved.If the angular momentum is constant, does it mean the linear momentum is also constant? Why or why not?

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APPLICATIONS (continued)

The passengers on the amusement-park ride experience conservation of angular momentum about the axis of rotation (the z-axis). As shown on the free body diagram, the line of action of the normal force, N, passes through the z-axis and the weight’s line of action is parallel to it. Therefore, the sum of moments of these two forces about the z-axis is zero.

If the passenger moves away from the z-axis, will his speed increase or decrease? Why?

ANGULAR MOMENTUM (Section 15.5)

The angular momentum of a particle about point O is defined as the “moment” of the particle’s linear momentumdefined as the moment of the particle s linear momentum about O.

i j kHo = r × mv = rx ry rz

mvx mvy mvz

The magnitude of Ho is (Ho)z = mv d

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RELATIONSHIP BETWEEN MOMENT OF A FORCE AND ANGULAR MOMENTUM

(Section 15.6)The resultant force acting on the particle is equal to the time rate of change of the particle’s linear momentum. Showing the time derivative using the familiar “dot” notation results in the equation

∑F = L = mv••

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

∑Mo = r × F = Ho•

PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM (Section 15.7)

Considering the relationship between moment and time rate of change of angular momentum

∑Mo = Ho = dHo/dt•

By integrating between the time interval t1 to t2

∑∫ −=2

1

1)(2)(

t

t

HoHodtMo 1)(Ho 2)(Ho∑∫

2

1

t

t

dtMo+ =or

This equation is referred to as the principle of angular impulse and momentum. The second term on the left side, ∑∫ Mo dt, is called the angular impulse. In cases of 2D motion, it can be applied as a scalar equation using components about the z-axis.

1 1

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CONSERVATION OF ANGULAR MOMENTUM

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

(HO)1 = (HO)2

An example of this condition occurs when a particle is subjected only to a central force. In the figure, the f F i l di t d t dforce F is always directed toward point O. Thus, the angular impulse of F about O is always zero, and angular momentum of the particle about O is conserved.

EXAMPLE

Given:A satellite has an elliptical orbit about earth.msatellite = 700 kgm h = 5 976 × 1024 kgmearth 5.976 × 10 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.B,

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

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EXAMPLE (continued)

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

ms vA2 – = ms vB

2 –1 G ms me 1 G ms mes A s B

where G = 66.73×10-12 m3/(kg·s2). Dividing through by ms and

substituting values yields:

2 rA 2 rB

106x151024)×-12(5.97610×73.662)000,10(5.0 −

rB

1024)×-12(5.97610×73.66v5.0

106x 152B −=

23.4 × 106 = 0.5 (vB)2 – (3.99 × 1014)/rBor

EXAMPLE (continued)

Now use Conservation of Angular Momentum.

(rA m vA) sin φA = r m v

Solution:

Solving the two equations for rB and vB yields

r = 13 8 × 106 m v = 10 2 km/s

(rA ms vA) sin φA rB ms vB

(15 × 106)(10,000) sin 70° = rB vB or rB = (140.95 × 109)/vB

rB = 13.8 × 106 m vB = 10.2 km/s

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CONCEPT QUIZ

1. If a particle moves in the x - y plane, its angular momentum vector is in the

A) x direction. B) y direction.) ) y

C) z direction. D) x - y direction.

2. If there are no external impulses acting on a particle

A) only linear momentum is conserved.B) only angular momentum is conserved.) y gC) both linear momentum and angular momentum are

conserved.D) neither linear momentum nor angular momentum are

conserved.

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 gfrom rest.

Plan:Apply the principle of angular impulse and momentum about the axis of rotation (z-axis).

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GROUP PROBLEM SOLVING (continued)

Angular momentum: HZ = r × mv reduces to a scalar equation.

(H ) = 0 and (H ) = 4×{(5/32 2) (0 6) v } = 0 3727 v

Solution:

(HZ)1 = 0 and (HZ)2 = 4×{(5/32.2) (0.6) v2} = 0.3727 v2

Angular impulse:

M dt = (0.5t + 0.8) dt = [(0.5/2) t2 + 0.8 t] = 7.2 lb·ft·s ∫2

1

t

t∫

2

1

t

t 0

4

Apply the principle of angular impulse and momentum.

0 + 7.2 = 0.3727 v2 ⇒ v2 = 19.4 ft/s

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 down with a constant speed of 2 ft/s, which of the following principles can be applied to solve for the velocity of the ball when r = 2 ft?A) Conservation of energyB) Conservation of angular momentumC) Conservation of linear momentumD) Conservation of mass

2. If a particle moves in the z - y plane, its angular momentum vector is in theA) x direction. B) y direction.C) z direction. D) z - y direction.