Impulse & Momentum

Objectives

• To develop the principle of linear impulse and

momentum for a particle.

• To study the conservation of linear momentum for

particles.particles.

• To analyze the mechanics of impact.

• To introduce the concept of angular impulse and

momentum.

• To solve problems involving steady fluid streams and

propulsion with variable mass.

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Principle of linear impulse and momentum

∫ ∫=Σ

==Σ

2 2t

t

v

vvdmdtF

dt

vdmamF

vv

vvv

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∫ ∫=Σ1 1t v

vdmdtF

∫ −=Σ2

112

t

tvmvmdtFvvv

Linear Impulse Linear Momentum

Principle of linear impulse and momentum

)()(2t

vmdtFvm =+∑∫

21

2

1

vFvvvvmdtm

t

t=+∑∫

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21

21

21

)()(

)()(

)()(

2

1

2

1

2

1

z

t

tzz

y

t

tyy

x

t

txx

vmdtFvm

vmdtFvm

vmdtFvm

=+

=+

=+

∑∫

∑∫

∑∫

Procedure for Analysis

• Select the inertial coordinate system

• Draw FBD to account for all forces

• Establish direction and sense of acceleration

• Solve for unknowns• Solve for unknowns

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Problem 15-10

A man kicks the 200-g ball such that it leaves the

ground at an angle of 30⁰ with the horizontal and

strikes the ground at the same elevation a distance

of 15 m away. Determine the impulse of his foot F on the ball. the ball.

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Problem 15-11

The particle P is acted upon by its weight of 30 N and

forces F1 and F2, where t is in seconds. If the particle

originally has a velocity of v1 = (3i + 1j + 6k) m/s, determine its speed after 2 s.

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Problem 15-27

Block A weighs 100 N and block B weighs 30 N. If B is

moving downward with a velocity (vB)1 = 1 m/s at

t=0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth.

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Principle of Linear Impulse and Momentum for a System of Particles

∑∑ =d

m iii

vF

vv

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∑∑ =dt

miiF

( ) ( )21

2

1∑∑∫∑ =+ ii

t

tiii mdtm vFv

vvv

Conservation of Linear Momentum for a System of Particles

• When the sum of the external impulses acting

on a system of particles is zero, the equation is

( ) ( )∑∑ = mm vvvv

• This equation is referred to as the conservation of linear momentum.

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( ) ( )∑∑ =21 iiii mm vv

Example 1

The 15-Mg boxcar A is coasting at 1.5 m/s on the

horizontal track when it encounters a 12-Mg tank B

coasting at 0.75 m/s toward it. If the cars meet and

couple together, determine (a) the speed of both

cars just after the coupling, and (b) the average force cars just after the coupling, and (b) the average force

between them if the coupling takes place in 0.8 s.

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Problem 15-52

The boy B jumps off the canoe at A with a velocity of

5 m/s relative to the canoe as shown. If he lands in

the second canoe C, determine the final speed of

both canoes after the motion. Each canoe has a mass

of 40 kg. The boy’s mass is 30 kg, and the girl has a of 40 kg. The boy’s mass is 30 kg, and the girl has a mass of 25 kg. Both canoes are originally at rest.

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Impact

• Central Impact

– Velocities are colinear (a)

– Deforms during collision (b)

– Restore after deformation (b)– Restore after deformation (b)

– Coefficient of restitution

= Irestitution/Ideform

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Coefficient of Restitution

• Consider m1

01

10

101

011

)]([

)]([0

vv

vv

vvm

vvm

dtF

dtF

et

d

t

t

r

−

′−=

−−−

−−′−==

∫

∫

• Consider m2

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0t

d∫

20

02

202

022

][

][

0

0

vv

vv

vvm

vvm

dtF

dtF

et

t

d

t

t

r

−

−′=

−

−′==

∫

∫

21

12

vv

vve

−

′−′=

Impact

• Oblique Impact

vv )()( ′−′

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nn

nn

vv

vve

)()(

)()(

21

12

−

′−′=

Problem 15-64

• If the girl throws the ball with a horizontal

velocity of 3 m/s, determine the distance d so

that the ball bounces once on the smooth

surface and then lands in the cup at C. Take surface and then lands in the cup at C. Take

e=0.8.

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Angular Momentum

• If the particle is moving along a space curve,

the vector cross product can be used to determine the angular momentum about O.

vrHvvvm×=

kjir zyx rrr ++=

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vrHvvvmO ×=

kjir zyx rrr ++=

kjiv zyx vvv ++=

θsin0 rvmH =

Angular Momentum

• RecallLvm

dt

dF &vv

==Σ )(

Hvmrd

Fr &vvvv=×=×Σ )()(

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OHvmrdt

Fr &=×=×Σ )()(

OO HM &v

=Σ

The moment about the fixed point O of all forces acting on m equals to the rate of change of angular momentum.

Principle of Angular Momentum

dt

HdHM OOO

v

&vv

==Σ

OO HddtMvv

=Σ

2t vvv

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12

2

1

OO

t

t

O HHdtMvvv

−=Σ∫

2

2

1

1 O

t

t

OO HdtMHvvv

=Σ+ ∫2

2

1

1 O

t

t

OO HdtMHvvv

Σ=Σ+Σ ∫

Conservation of Angular Momentum

• Finally, if all angular impulse is zero.

21 OO HHvv

=

• For a system of particles

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21 OO HHvv

Σ=Σ

Example 2

The 2 kg disk rests on a smooth horizontal surface

and is attached to an elastic cord that has a stiffness

kc = 20 N/m and is initially unstretched. If the disk is

given a velocity (vD)1 = 1.5 m/s, perpendicular to the

cord, determine the rate at which the cord is being D

cord, determine the rate at which the cord is being

stretched and the speed of the disk at the instant the

cord is stretched 0.2 m.

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Meriam 3-226

The small spheres, which have the masses and initial

velocities shown in the figure, strike and become

attached to the spiked ends of the rod, which is

freely pivoted at O and is initially at rest. Determine

the angular velocity ω of the assembly after impact. the angular velocity ω of the assembly after impact. Neglect the mass of the rod.

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Meriam 3-230

The 6-kg sphere and 4-kg block are secured to the arm of negligible mass which rotates in the vertical plane about a horizontal axis at O. The 2-kg plug is released from rest at A and falls into the recess in the block when the into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity ω0 = 2 rad/s. Determine the angular velocity ω of the arm immediately after the plug has wedged itself in the block.

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