CEE 271: Applied Mechanics II, Dynamics– Lecture 14: Ch.15, Sec.1-3 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Thursday, October 4, 2012

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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM

Today’s objectives: Studentswill be able to

1 Calculate the linearmomentum of a particleand linear impulse of aforce.

2 Apply the principle of linearimpulse and momentum.

In-class activities:

• Reading Quiz• Applications• Linear Momentum and

Impulse• Principle of Linear Impulse

and Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

Figure: impulse! 2 / 20

READING QUIZ

1 The linear impulse and momentum equation is obtained byintegrating the with respect to time.(a) friction force(b) equation of motion(c) kinetic energy(d) potential energy

ANS: (b)

2 Which parameter is not involved in the linear impulse andmomentum equation?(a) Velocity(b) Displacement(c) Time(d) Force

ANS: (b)

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APPLICATIONS

• A dent in an automotive fender can beremoved using an impulse tool, whichdelivers a force over a very short timeinterval. To do so the weight is grippedand jerked upwards, striking the stopring.

• How can we determine the magnitudeof the linear impulse applied to thefender?

• Could you analyze a carpenter’shammer striking a nail in the samefashion?

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APPLICATIONS(continued)

• When a stake is struck by asledgehammer, a large impulseforce is delivered to the stake anddrives it into the ground.

• If we know the initial speed of thesledgehammer and the durationof impact, how can we determinethe magnitude of the impulsiveforce delivered to the stake?

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PRINCIPLE OF LINEAR IMPULSE ANDMOMENTUM (Section 15.1)

• The next method we will consider for solving particlekinetics problems is obtained by integrating theequation of motion with respect to time.

• The result is referred to as theprinciple of impulse and momentum. It can be applied toproblems involving both linear and angular motion.

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

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PRINCIPLE OF LINEAR IMPULSE ANDMOMENTUM (continued)

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

ΣF = ma = m(dv

dt)

• Separating variables and integrating between the limitsv = v1 at t = t1 and v = v2 at t = t2 results in

Σ

∫ t2

t1

F dt = m

∫ v2

v1

dv = mv2 −mv1

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

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PRINCIPLE OF LINEAR IMPULSE ANDMOMENTUM (continued)

• Linear momentum: The vector mv is called the linearmomentum, denoted as L. This vector has thesame direction as v. The linear momentum vector has unitsof (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 aforce during its time interval of action. I acts in the samedirection as F and has units of N·s or lb·s.

• The impulse may be determined bydirect integration. Graphically, it canbe represented by thearea under the force versus time curve.If F is constant, then I = F (t2 − t1).

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PRINCIPLE OF LINEAR IMPULSE ANDMOMENTUM (continued)

• The principle of linear impulse andmomentum in vector form is written as

mv1 + Σ

∫ t2

t1

F dt = mv2

• The particle’s initial momentum plus thesum of all the impulses applied from t1 to t2is equal to the particle’s final momentum.

• The two momentum diagrams indicatedirection and magnitude of the particle’sinitial and final momentum, mv1 and mv2.The impulse diagram is similar to a freebody diagram, but includes thetime duration of the forces acting on theparticle.

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IMPULSE AND MOMENTUM: SCALAR EQUATIONS

• Since the principle of linear impulse and momentum is avector equation, it can be resolved into its x, y, zcomponent scalar equations:

mv1x = Σ

∫ t2

t1

Fxdt = mv2x (1)

mv1y = Σ

∫ t2

t1

Fydt = mv2y (2)

mv1z = Σ

∫ t2

t1

Fzdt = mv2z (3)

• The scalar equations provide a convenient means forapplying the principle of linear impulse and momentumonce the velocity and force vectors have been resolved intox, y, z components.

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PROBLEM SOLVING

1 Establish the x, y, z coordinate system.2 Draw the particle’s free body diagram and establish the

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

3 Resolve the force and velocity (or impulse and momentum)vectors into their x, y, z components, and apply theprinciple of linear impulse and momentum using its scalarform.

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

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EXAMPLE

• Given: A 0.5 kg ball strikes therough ground and rebounds withthe velocities shown. Neglect theball’s weight during the time itimpacts the ground.

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

1 Draw the momentum and impulse diagrams of the ball as ithits the surface.

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

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EXAMPLE (solution)

1 The impulse and momentum diagrams can be drawn as:

!"#$%$&$!'"#$!

("#$%$&$!

)*+$,-.$

/&+$

,-0$

2 The impulse caused by the ball’s weight and thenormal force N can be neglected because their magnitudesare very small as compared to the impulse from theground.

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EXAMPLE(solution)

1 The principle of impulse and momentum can be appliedalong the direction of motion:

mv1 + Σ

∫ t2

t1

F dt = mv2 (4)

0.5(25 cos 45◦i− 25 sin 45◦j) +

∫ t2

t1

ΣF dt (5)

= 0.5(10 cos 30◦i + 10 sin 30◦j) (6)

2 The impulsive force vector is

I =

∫ t2

t1

ΣF dt = (4.509i + 11.34j)N · s

3 Magnitude: I =√

4.5092 + 11.342 = 12.2N · s

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CHECK YOUR UNDERSTANDING QUIZ

1 Calculate the impulse due to the force.

!"#

$%&#

'#

(#

)!#*#!%+,"#

)$*#$%+,"#-%.#

(/012##13042#

(a) 20 kg·m/s

(c) 5 kg·m/s

(b) 10 N·s

(d) 15 N·s

ANS: (b)

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

!"#

$%&#

'#

(#

)!#*#!%+,"#

)$*#$%+,"#-%.#

(/012##13042# (a) (17.3j) N

(c) (20i + 17.3j) N

(b) (−10i+ 17.3j) N

(d) ( 10i + 17.3j) N

ANS: (c)

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GROUP PROBLEM SOLVING

• Given: The 20 kg crate is resting on thefloor. The motor M pulls on the cable witha force of F , which has a magnitude thatvaries 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 att = 6 s.

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GROUP PROBLEM SOLVING (solution)

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

F = mg = (20)(9.81) = 196.2N

F = 196.2N = 50 t

t = 3.924s

2 Apply the principle of impulse and momentum from thetime the crate starts lifting at t1 = 3.924 s to t2 = 6 s. Notethat there are two external forces (cable force and weight)we need to consider.

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GROUP PROBLEM SOLVING (solution)

A. The impulse due to cable force: F1 = 50 t and F2 = 250

(+ ↑)∫ 6

3.924Fdt =

∫ 5

3.924F1dt +

∫ 6

5F2dt

= [0.5(250)5− 0.5(196.2)3.924] + (250)(6− 5)

= 490.1N · s (7)

B. The impulse due to weight:

+ ↑∫ 6

3.924(−mg)dt = −196.2(6− 3.924) = −407.3N · s

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GROUP PROBLEM SOLVING (solution)

• Now, apply the principle of impulse and momentum

(+ ↑)mv1 + Σ

∫ t2

t1

Fdt = mv2 where v1 = 0 (8)

0 + 490.1− 407.3 = (20)v2 (9)v2 = 4.14m/s (10)

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ATTENTION QUIZ

1 Jet engines on the 100 Mg VTOL aircraft exert a constantvertical force of 981 kN as it hovers. Determine the netimpulse on the aircraft over t = 10 s.

(a) -981 kN·s

(c) 981 kN·s

(b) 0 kN·s

(d) 9810 kN·s

ANS: (b)2 A 100 lb cabinet is placed on a smooth surface. If a force

of a 100 lb is applied for 2 s, determine the net impulse onthe cabinet during this time interval.

(a) 0 lb·s

(c) 200 lb·s

(b) 100 lb·s

(d) 300 lb·s

ANS: (b)20 / 20