emm 222 - dynamics mechanism - chapter 4

8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4

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Kinetics of a Particle

Impulse and Momentum

Chapter Four


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In this chapter, you will learn how to: -

derive the principle of impulse and

momentum for a particle from

Newtons second law.

express the conservation oftotal

momentum of the particles whenimpulse of the external forces is zero.


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Principle of Linear Impulse

and Momentum


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Principle of Linear Impulse and

Momentum

21

12

MomentumImpulseLinear

mvFdtmv

mvmvFdt

dvmFdt

mdvdtF

dt

dvmF

maF


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Resolving into x,y,z components

21

21

21

2

1

2

1

2

1

z

t

t

zz

y

t

t

yy

x

t

t

xx

vmdtFvm

vmdtFvm

vmdtFvm


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Example 4.1:

A 20 N block slides down a 30 inclined

plane with an initial velocity of 2 m/s.

Determine the velocity of the block in 3 s

if the coefficient between the block andthe plane is k = 0.25.

30

2 m/s


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Solution 4.1:

FBD

N

Ff= 0.25 N

W=20 N

y

x


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Solution 4.1: (continued)

___

030cos20

0

___

030cos200

3

0

21

2

1

N

N

maF

or

NN

dtN

vmdtFvm

yy

y

t

t

yy+ y

17.32 N


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smv

vdtN

vmdtFvm x

t

t

xx

/___

)(81.9

2025.030sin20)2(

81.9

20

3

0

21

2

1

+ x

10.3


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Example 4.2:

A 20 N ball is thrown in the direction shown

with an initial speed vA=6 m/s. Determine

the time needed for it to reach its highest

point B and the speed which it is travelling

at B. Use the principle of impulse and

momentum for the solution.


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Solution 4.2:

st

dt

vmdtFvm

t

y

t

t

yy

___

020)30sin6(81.9

20

0

21

2

1

+

0.306


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smv

v

vmdtFvm

B

B

x

t

t

xx

/___

)(81.9

200)30cos6(

81.9

20

21

2

1

+

5.2


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Example 4.3:

A hammer head having a

weight of1 N is movingvertically downward at 10 m/s

when it strikes the head of a

nail of negligible mass and

drives into a block of wood.

Find the impulse on the nail if

it is assumed that the grip at A

is loose, the handle has anegligible mass, and the

hammer stays in contact with

the nail while it comes to rest.

Neglect impulse that

caused by hammer

head.


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Solution 4.3:

NsFdt

Fdt

vmdtFvm

t

y

t

t

yy

___

0)10(81.9

1

0

21

2

1

+

1.02


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Conservation of

Linear Momentum


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When the sum of the external impulses acting

on a system of particles iszero, the equation is

21

2

1

0

iiii

t

t

mm

Fdt

vv

This equation is referred to as the conservation

of linear momentum.

It states that the total momentum for a system of

particles remains constant during the time period

t1 to t2.


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Example 4.4:

The car A has a weight of22.5 kN and is

travelling to the right at 1 m/s. Meanwhile a

15 kN car B is travelling at 2 m/s to the left.

If the cars crash, determine their commonvelocity just after the collision. Assume

brakes are not applied during collision.


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Solution 4.4:

smv

v

vmmvmvm BABBAA

/___

81.9

1500022500)2(81.9

15000)1(81.9

22500

2

2

211

- 0.2


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Example 4.5:

Given:Two rail cars with masses ofmA = 15

Mg and mB = 12 Mg and velocities as

shown.

Find:The speed of the cars after they meet

and connect. Also find the averageimpulsive force between the cars if the

coupling takes place in 0.8 s.


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Plan:

1. Use conservation of linear momentum

to find the velocity of the two cars after

connection (all internal impulses

cancel).

2. Then use the principle of impulse and

momentum to find the impulsive forceby looking at only one car.


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Solution 4.5:Conservation of linear momentum (x-dir):

mA(vA)1 + mB(vB)1 = (mA + mB) v2

15,000(1.5) + 12,000(- 0.75) = (27,000)v2

v2 = 0.5 m/s

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

mA(vA)1 + F dt = mA(v2)

15,000(1.5) - F dt = 15,000(0.5)

F dt = 15,000 Ns

The average force is

F dt = 15,000 Ns = Favg(0.8 sec); Favg = 18,750 N

Solution:


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Impact


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Impactoccurs when two bodies collide

with each other during a very shortperiod

of time, causing relatively large (impulsive)

forces to be exerted between the bodies.

There are two types of impact.


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1.Central impactoccurs when the direction of

motion of the mass centers of the two

colliding particles is along theline of impact.

2.Oblique impactoccurs when one or both of

the particles is at an angle with the line of

impact.

The line of impact passes through the masscenters of the particles.


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In most problems, the initial velocities of the particles,

(vA)1 and (vB)1, are known, and it is necessary to

determine the final velocities, (vA)2 and (vB)2. So the firstequation 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

unknowns, (vA)2 and (vB)2. So another equation isneeded. 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 velocity after impact,(vB)2 (vA)2, to the particles relative velocity

before impact, (vA)1 (vB)1.

The equation defining the coefficient of

restitution, e, is

(vA)1(vB)1(vB)2

(vA)2e =


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In general, e has a value between zero and one.

Elastic impact (e = 1): In a perfectly elasticcollision, 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.

Plastic impact (e = 0): In a plastic impact, the

relative separation velocity is zero. The particlesstick together and move with a common velocity

after the impact.


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IMPACT: ENERGY LOSSES

During a collision, some of the particles initialkinetic energy will be lost in the form of heat,

sound, or due to localized deformation.

Once the particles velocities before and afterthe collision have been determined, the energy

loss during the collision 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


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Example 4.6:

Disk A has a mass of2 kg and is sliding forward

on the smooth surface with a velocity vA = 5m/s when it strikes the 4 kg disk B, which is

sliding towards A at vB = 2 m/s, with direct

central impact. If the coefficient of restitutionbetween the disk is e = 0.4, compute the

velocity of A and B just after collision.


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Solution 4.6:

)1(42)2(452

MomentumofonConservati

22

2211

BA

BBAABBAA

vv

vmvmvmvm

)2(

)2(5

)()(4.0

)()(

)()(

nRestitutiooftCoefficien

22

11

22

AB

BA

AB

vv

vv

vve


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Solution 4.6:(continued)

smv

smsmv

B

A

/27.1

/53.1/53.1

(2)and(1)equationsSolving

2

2


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Example 4.7:

Two smooth disks A and B, having mass of 1 kg

and 2 kg respectively, collide with the velocitiesshown. If the coefficient of restitution for the disksis e= 0.75, determine the xand ycomponents ofthe final velocity of each disk just after collision.


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Solution 4.7:

Resolving each of the initial velocities into x and ycomponents, we have

smv

smv

smv

smv

By

Bx

Ay

Ax

/707.045sin1)(

/707.045cos1)(

/50.130sin3)(

/60.230cos3)(

1

1

1

1


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The four unknown velocity components aftercollision are assumed to act in the positivedirections. Since the impact occurs only in the x

direction (line of impact), the conservation ofmomentum for bothdisks can be applied in thisdirection.


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Conservation of x Momentum.

18.1)(2)(

)(2)(1)707.0(2)60.2(1

)()()()(

22

22

2211

BxAx

BxAx

BxBAxABxBAxA

vv

vv

vmvmvmvm( )


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Conservation of (x) Restitution. Both disks areassumedto have components of velocity in the +xdirection after collision,

48.2)()(

)07.0(60.2

)()(75.0

)()(

)()(

22

22

1122

AxBx

AxBx

BxAxAxBx

vv

vv

vv

vve( )


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Solving the two simultaneous equations,

smv

smsmv

Bx

Ax

/22.1)(

/26.1/26.1)(

2

2


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Conservation of y Momentum.The momentum of each diskis conservedin the ydirection (plane of contact), since the disks aresmooth and therefore noexternal impulse acts in

this direction.

smsmv

vmvmsmv

vmvm

By

ByBByB

Ay

AyAAyA

/707.0/707.0)(

)()(/5.1)(

)()(

2

21

2

21( )

( )


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Tutorial:Problems (12th Edition)

15-2, 15-38, 15-67

emm 222 - dynamics mechanism - chapter 4

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