4 - Kinetics of Particles - Impulse & Momentum

8/12/2019 4 - Kinetics of Particles - Impulse & Momentum

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Principle of Impulse and Momentum

Ken Youssefi MAE 1

The third method of solving engineering problems deals

with force, mass, velocity and time.

The method is particularly effective when dealing with

problems involving impulsive motion, force applied over a

short interval of time, and problems involving impact.


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Concept of Impulse and Momentum

Ken Youssefi MAE 2

Vector equation. Scalar equation.

The concept of workrelates

force to displacement.

The concept of impulse relates

force to time.

Greater force or greater

displacement is associated with

more work done.

Greater force or greater time of

action is associated with more

impulse applied.

More work done changes the

motion of a system to a greater

degree.

More impulse changes the

motion of a system to a greater

degree.

That which is changed is called

kinetic energy.

That which is changed is called

momentum.

Work & Energy Impulse & Momentum


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Principle of Linear Impulse and Momentum

Ken Youssefi MAE 3

Consider Newtons 2nd law.

mv (vector quantity) is called the linear

momentum of the particle

Linear impulse

Principle of linearImpulse & Momentum


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Components of Impulse

Ken Youssefi MAE 4

Rectangular coordinate system


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Principle of Impulse and Momentum

Ken Youssefi MAE 5

The final momentum of a particle is obtained by adding vectoriallyits initial

momentum and the impulse of the force F acting during the interval considered.


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Conservation of Linear Momentum

Ken Youssefi MAE 6

When the sum of the external impulses acting on a system of particle is

zero, the equation for the principle of linear impulse and momentum

reduces to the following:

Consider two boats initially at rest, which are pulled together


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Impulsive forces and Motion

Ken Youssefi MAE 7

Impulsive force is a force that acts on a particle during a very short

time interval and produces a definite change in momentum. The

resulting motion is called an impulsive motion. Baseball hitting a bat.

Non impulsive forces like weight of the body, the force exerted by spring, orany other force which is known to be small compared with the impulsive

force may be neglected.

In case of the impulsive motion of several particles, we can write:

No impulsive external

forces acting on the body


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Example

Ken Youssefi MAE 8

A car weighing 4000 lb is moving down a 5o incline at a

speed of 60 mi/h when the brakes are applied, causing

a constant braking force of 1500 lb (applied by the road

on the tires). Determine the time required for the car to

come to stop.

Apply the principle of impulse and momentum


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Example

Ken Youssefi MAE 9

A 4-oz baseball is pitched with a velocity of 80 ft/s. After

the ball is hit by the bat atB, it has a velocity of 120 ft/s in

the direction shown. If the bat and the ball are in contact

for .015 second, determine the average impulsive forceexerted on the ball during the impact.

Apply the principle of impulse and momentum to the ball, weight of the

ball can be neglected (weight is nonimpulsive force)

xx

t

t x mvmvdtF 12

2

1

yyt

t y mvmvdtF 12

2

1


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Example

Ken Youssefi MAE 10

x components

y components


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Example

Ken Youssefi MAE 12

x components

y components


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example

Ken Youssefi MAE 13

Initial energy

Final energy


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Direct Central Impact

Ken Youssefi MAE 15

Consider the impact of two particles

Before impact

The total momentum of the two particles is conserved

Scalar components

During impact

Same velocity

after impact

Restoration takes place


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Velocities after the Impact & the Coefficient of Restitution (e)

Ken Youssefi MAE 17

The ratio of the magnitude of the impulses corresponding to

the period of restitution and to the period of deformation is

called the coefficient of restitution, e is always between 0 and 1.

Same approach for particle B gives:

Substitute for the impulses

Eliminate u


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Oblique Central Impact

Ken Youssefi MAE 20

Velocities of the two colliding particles are not directed

along the line of impact.

Resolve the velocities along the n direction (along the line of impact)and the tdirection (along the common tangent, perpendicular to the

n direction).


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Oblique Central Impact

Ken Youssefi MAE 21

1. The tcomponents of the velocity of each particle

remains the same.

2. The component along the n axis of the total momentum

of the two particles is conserved.

3. The coefficient of restitution equation can be used inthe n direction


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Example

Ken Youssefi MAE 24

The magnitude and direction of the velocities

of two balls before they collide are shown in

the figure. If the coefficient of restitution is 0.9,

determine the magnitude and direction of thevelocity of each ball after the impact (neglect

friction)

Resolve the velocities into components along the line of impact and

along the common tangent to the surface in contact

BallA

BallB


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Example

Ken Youssefi MAE 25

Motion along the common tangent

Motion along the line of impact


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Using Conservation of energy and Momentum

Ken Youssefi MAE 27

Consider a pendulumA released with no velocity from positionA1.

PendulumA hits pendulumB which is initially at rest. After the impact

pendulumB swings through angle . Determine the angle .


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Using Conservation of energy and Momentum

Ken Youssefi MAE 28

2. PendulumA hits pendulumB. Use conservation of momentum and theequation for coefficient of restitution to solve for velocities of the two

pendulums after the impact, (vA)3 and (vB)3

1. PendulumA swings fromA1 toA2. Use conservation of energy to find the

velocity of pendulumA, (vA)2 , just before it collides with pendulumB.

3. PendulumB swings from B3 toB4. Use conservation of energy for

pendulum B to determine the heighty4. Use trigonometry to find .


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Example

Ken Youssefi MAE 29

A 30 kg block is dropped from a height of 2 m onto the 10 kg pan of a spring

scale. If the impact is perfectly plastic, determine the maximum deflection of

the pan. The spring stiffness is k= 20 kN/m.


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Example

Ken Youssefi MAE 30


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Example

Ken Youssefi MAE 31


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Example

Ken Youssefi MAE 32

Conservation of energy

Initial deflection of the spring due to the scale pan (WB)


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l


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Angular Momentum

Ken Youssefi MAE 35

The linear momentum was defined as mv. The moment of the

vector mvabout point O is called the angular moment or the

moment of momentum of the particle about point O

Ho = (d)(mv)

Consider a 2D motion, particle

moving along a path onxy plane.

The magnitude of the angularmoment is;

l


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Angular Momentum

Ken Youssefi MAE 36

Ho

is a vector perpendicular to the

plane containing vectors r and mv

Plane containing

vector r and

vector mv

A l M


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Angular Momentum

Ken Youssefi MAE 37

r = rx i + ry j + rzk , v = vx i + vy j + vzk

Position and velocity vectors in rectangular coordinate system


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A l M t


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Angular Momentum

Ken Youssefi MAE 39

Relationship between moment of a force and angular momentum.

Equation of motion

Moments of the forces about point O

Same result for

linear momentum

P i i l f A l I l d M t


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Principle of Angular Impulse and Momentum

Ken Youssefi MAE 40

Modt= dHo

At time t= t1 Ho = (Ho)1 , At time t= t2Ho = (Ho)2

Principle of

angular impulse

and momentum

Angular Impulse

Linear Impulse


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P i i l f Li & A l I l d M t


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Principle of Linear & Angular Impulse and Momentum

Ken Youssefi MAE 42

Linear and Angular impulse & momentum

3D vector

equations

2D scalarequations

Linear

Angular

Conser ation of Ang lar Moment m


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Conservation of Angular Momentum

Ken Youssefi MAE 43

If the angular impulse acting on a particle is

zero the principle equation reduces to the

following:

Conservation ofAngular Momentum

If force F is directed toward

point O the moment is zero

so the angular momentum is

conserved. Force F is knownas a Central Force.


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Example


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Example

Ken Youssefi MAE 46

A 2 kg bob is given a horizontal speed of 1.5 m/s.

It rotates on a circular pathA. If the force F on

the cord is increased, the bob rises and rotates

around the horizontal circular pathB. Determine

the speed of the bob around pathB. Also, find

the work done by force F.

There are three forces involved in this problem, W,F,

and the cord force. WandFproduce no moment

about thezaxis. Also, the cord force is along the cord

so it does not produce a moment about thezaxis.

Therefore, the conservation of momentum applies.

r1mv1 = r2mv2

r1 = l1 sin1 r2 = l2 sin2


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Example


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Example

Ken Youssefi MAE 48

Position 2

l= l2 = .3 m , v = v2 and = 2

Eliminate v2

(1)

r1mv1 = r2mv2

r1 = l1 sin1 r2 = l2 sin2

Conservation of momentum

(2)

Example


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Example

Principle of Work and Energy

When changes to 1 to 2 , Wdisplaces vertically upward.

Therefore, Wdoes negative work

h = .6 cos (34.21) - .3 cos (57.866) = .3366 mh

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4 - Kinetics of Particles - Impulse & Momentum

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