11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse...

11 - 1

Impulse and MomentumSome more extremely useful relations are the concepts known as impulse and momentum. Conservation of momentum underlies several important physical laws in a variety of scientific disciplines from quantum electrodynamics to fluid mechanics. A derivation for this fundamental quantity is shown below:

1212

2

1

2

1

2

1

2

1

vmvmrmrmrmddtdt

rdmdtrmFdt

r

r

t

t

t

t

t

t

1212

2

1

2

1

PPvmvmvmdFdtIt

t

t

t

vmrmP

We now define a quantity known as momentum, as , and the change in momentum over a discrete time interval is defined as the impulse, :I

Note: In comparison with potential and kinetic energy, which are scalar quantities, impulse and momentum are vector quantities and must be added vectorially.

11 - 2

PdvmddtF

PddtF

PPdt

drmF

.0 constPPF

Starting from a generalized expression for momentum

We can take the derivative of both sides of the equation:

And we arrive at another expression for Newton’s Second law. The force is equal to the rate of change of momentum.

Thus, when the net force on an object is zero, then the momentum must be constant. This is the famous law of conservation of momentum.

Conservation of Linear Momentum

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11 - 4

30sinˆ30cosˆ00 yxvvvAi

CfCBfBAfACiCBiBAiA vmvmvmvmvmvm

oo

CfC

ooBfB

ooAfA

CfCBfBAfA

yxvm

yxvm

yxvm

vmvmvm

45sinˆ45cosˆ

3.49cosˆ3.49sinˆ

4.7cosˆ4.7sinˆ

Initial conditions:

Linear Momentum prior to impact Linear Momentum after impact

Conservation of Linear Momentum

After impact:

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CBAsm

Csm mmmvv ,1.2,0.40

osmo

Bfo

Afo vvvx 45cos23.49sin4.7sin30cos:ˆ 0

osmo

Bfo

Afo vvvy 45sin23.49cos4.7cos30cos:ˆ 0

sm

Af

sm

Bf

v

v

01.2

27.2

Given the following information:

Leads to 2 equations and 2 unknowns.

These can be solved simultaneously, leading to the following result:

11 - 6

Relation between Force and Impulse

PddtFI

vmPtFdtFI avg

As shown in previous slides, the impulse can be written as:

If the force is constant over a discrete time interval, then it can be taken out of the integral and the impulse is re-written as:

This equation is very useful for modeling collisions, because the force of impact is not usually known (or measurable), but the initial and final velocities are known and/or measurable.

11 - 7

11 - 8

if

s

s

s

s

s

s

s

s

s

s

vvmvmFdt

dtmgdtNmgdts

NtmgFdt

5

0

5

4

4

2

2

0

5

0

800400

stsmg

stsNmg

ststs

Nmg

tTmgF

54

42800

20400)(

mgssNsNss

NsmgFdt

s

s

52400280022005 25

0

s

m

s

mgssN

kgv f 1.265

200

2400

11 - 9

Angular Momentum

vmrHO

r

Like linear momentum, there is also an expression for angular momentum, which describes how much inertia an object carries in rotating about a coordinate center. While its derivation will be reserved for later in this course, the mathematical expression is given here as:

Path of particle

v

The cross product denotes that the angular momentum points in an entirely different direction from the instantaneous position vector, , and the instantaneous velocity vector . The direction can be found according to the right-hand rule. The O means that the angular momentum is taken with respect to the origin.

v r

11 - 10

Conservation of Angular Momentum

Frrmrrmrvmrvmrvmrdt

dHO

0

.

00

constH

rFrH

O

O

The rate of change of angular momentum can be derived as follows:

FrHM OO

(i.e. the cross product of a vector with itself is always zero (i.e. they point in the same direction)

We define this new vector, , which is the rate of change of angular moment, and it is called the moment of force about a pivot. Note: If the net force (or moment in this case) is zero, then it leads directly to the law of conservation of angular momentum.

M

11 - 11

Momentum according to Isaac AsimovIsaac Asimov wrote in "Understanding Physics": “This tendency for motion (or for rest) to maintain itself steadily unless made to do otherwise by some interfering force can be viewed as a kind of "laziness," a kind of unwillingness to make a change. And indeed, Newton's first law of motion is referred to as the principle of inertia, from a Latin word meaning "idleness" or "laziness."

He added a footnote: "In Aristotle's time the earth was considered a motionless body fixed at the center of the universe; the notion of 'rest' therefore had a literal meaning. What we ordinarily consider 'rest' nowadays is a state of being motionless with respect to the surface of the earth. But we know (and Newton did, too) that the earth itself is in motion about the sun and about its own axis. A body resting on the surface of the earth is therefore not really in a state of rest at all."

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Central Forces

ˆ2ˆˆˆ, 2

rrmrrrmFrFamrF r

2r

rrmF 20

In many forces, such as gravitational, electrical forces, etc., the magnitude of the force only depends on the distance away from the source (i.e. ), and the direction is always towards or away from the source . r̂

2

2

2

.

021

mr

H

Hconstmr

rrrdt

d

r

Using these assumptions, we can derive an expression for the rotation frequency:

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2mr

H

32

22

22

rm

Hrm

mr

HrrmrrmFr

03

2

rfrm

Hrm

Using this result, we can arrive at an expression for radial force written purely in terms of the radial distance

32

2

rm

Hrmrf

Now the question becomes, what function f(r) satisfies the solution to this differential form?

Or alternatively, what is the solution to this equation:

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22

22

2

2222222

2 ˆˆˆˆ

rm

Hr

mr

Hrrrrv

rrrrrrrrvvv

03

2

rfrrrm

Hrrm

rrm

Hrrv

dt

d

rm

Hr

dt

dvv

dt

dv

dt

d

32

22

22

222

22

For example, consider the dot product of the velocity vector with itself.

What is the time derivative of this quantity?

The trick is to write it in differential form:

11 - 15

03

2

rfrrrm

Hrrm

If you now multiply this function by ½ m, you will find that:

02

1 2

rfrmvdt

d

r

rm

Hrrmmv

dt

d3

22

2

1

The result is that the equation for radial motion is now reduced to:

The next step is to write f(r) in a form that allows for solution of this differential equation.

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02

1

2

1

2

1 222

rUmvdt

d

dt

rdUmv

dt

drfrmv

dt

d

rdt

rdU

dr

dt

dt

dr

dr

rdU

dr

rdUrf

1

rfrdt

rdU

Consider for the moment that the function f(r) is actually derivative of another function.

After some manipulation, this formula becomes:

So, now we can re-write the equation on the previous slide as:

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.2

1 2 ConstrUmv

02

1 2

rUmv

dt

d

22

222

rm

Hrv

This equation is easily solvable. It is simply a constant.

.2

1

2

12

22 ConstrU

rm

Hrm

But we also know that the expression for the square of velocity.

This equation then becomes upon multiplication by ½ m…

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.2

1

2

12

22 ConstrU

rm

Hrm

.

2

1

2 2

22

ConstrUrm

H

m

P

vmP

222 vmP

This is the kinetic energy stored in the linear velocity

This is the kinetic energy stored in the angular momentum

This is the potential energy term

When we make the substitution…

The above equation becomes:

Linear momentum Energy term

Angular momentum Energy term

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Problem 1

A 25-g steel-jacket bullet is firedhorizontally with a velocity of600 m/s and ricochets off a steelplate along the path CD with avelocity of 400 m/s. Knowing thatthe bullet leaves a 10-mm scratchon the plate and assuming that itsaverage speed is 500 m/s while itis in contact with the plate,determine the magnitude anddirection of the average impulsiveforce exerted by the bullet on theplate.

10 mm

15o

20o

B C

D

A

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Problem 110 mm

15o

20o

B C

D

AA 25-g steel-jacket bullet is firedhorizontally with a velocity of600 m/s and ricochets off a steelplate along the path CD with a

velocity of 400 m/s. Knowing that the bullet leaves a 10-mmscratch on the plate and assuming that its average speed is500 m/s while it is in contact with the plate, determine themagnitude and direction of the average impulsive force exertedby the bullet on the plate.

1. Draw a momentum impulse diagram: The diagramshows the particle, its momentum at t1 and at t2, and the impulsesof the forces exerted on the particle during the time interval t1 to t2.

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Problem 110 mm

15o

20o

B C

D

AA 25-g steel-jacket bullet is firedhorizontally with a velocity of600 m/s and ricochets off a steelplate along the path CD with a

velocity of 400 m/s. Knowing that the bullet leaves a 10-mmscratch on the plate and assuming that its average speed is500 m/s while it is in contact with the plate, determine themagnitude and direction of the average impulsive force exertedby the bullet on the plate.

2. Apply the principle of impulse and momentum: The finalmomentum mv2 of the particle is obtained by adding its initialmomentum mv1 and the impulse of the forces F acting on theparticle during the time interval considered.

mv1 +F t = mv2

F is sum of the impulsive forces (the forces that are largeenough to produce a definite change in momentum).

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Problem 1 Solution10 mm

15o

20o

B C

D

A

Draw a momentum impulsediagram.

+ =x

y

Fx t

Fy t

m v1 xy

15o

x

y

m v2

20o

Since the bullet leaves a 10-mm scratch and its average speedis 500 m/s, the time of contact t is:

t = (0.010 m) / (500 m/s) = 2x10-5 s

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Problem 1 Solution

+ =x

y

Fx t

Fy t

m v1 xy

15o

x

y

m v2

20o

Apply the principle of impulseand momentum.

mv1 +F t = mv2

(0.025 kg)(600 m/s)cos15o+Fx2x10-5s= (0.025 kg)(400 m/s)cos20o

Fx = - 254.6 kN

-(0.025 kg)(600 m/s)sin15o+Fy2x10-5s=(0.025 kg)(400 m/s) sin20o

Fy = 365.1 kN

+ x components:

+ y components:

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Problem 1 Solution

+ =x

y

Fx t

Fy t

m v1 xy

15o

x

y

m v2

20o

Fx = - 254.6 kN, Fy = 365.1 kN

F = ( -254.6 kN )2 + ( 365.12 kN )2 = 445 kN

F = 445 kN 40.1o

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140 kg

650 kg

1.2 m

Problem 2

The 650-kg hammer of a drop-hammerpile driver falls from a height of 1.2 monto the top of a 140-kg pile, drivingit 110 mm into the ground. Assumingperfectly plastic impact (e = 0 ),determine the average resistance ofthe ground to penetration.

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Problem 2

140 kg

650 kg

1.2 mThe 650-kg hammer of a drop-hammerpile driver falls from a height of 1.2 monto the top of a 140-kg pile, drivingit 110 mm into the ground. Assumingperfectly plastic impact (e = 0 ),determine the average resistance ofthe ground to penetration.

1. Apply conservation of energy principle : When a particlemoves under the action of a conservative force, the sum of thekinetic and potential energies of the particle remains constant.

T1 + V1 = T2 + V2

where 1 and 2 are two positions of the particle.1a. Kinetic energy: The kinetic energy at each point on the pathis given by:

T = m v212

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Problem 2

140 kg

650 kg

1.2 m

1b. Potential energy: The potential energy of a weight W closeto the surface of the earth at a height y above a given datum isgiven by: Vg = W y


11 - 28

Problem 2

140 kg

650 kg

1.2 m

2. Apply conservation of momentum principle: During an impactof two bodies A and B, the total momentum of A and B isconserved if no impulsive external force is applied.

mA vA + mB vB = mA v’A + mB v’B

where vA and vB denote the velocities of the bodies before the

impact and v’A and v’B denote their velocities after the impact.

For perfectly plastic impact (e = 0), v’A = v’B = v’ and

mA vA + mB vB = (mA + mB) v’


11 - 29

Problem 2

140 kg

650 kg

1.2 m

3. Apply principle of work and energy: When a particle movesfrom position 1 to position 2 under the action of a force F, thework of the force F is equal to to the change in the kinetic energyof the particle.

T1 + U1 2 = T2

where

T1 = m v12 , T2 = m v2

2 and U 1 2 = F . dr12

12


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Problem 2 Solution

650 kg

1.2 my

Apply conservation of energy principle.

Motion of the hammer during the drop just before impact.

Position 1

T1 + V1 = T2 + V2

0 + mg (1.2 m) = m v22 + 0

v22 = 2 ( 9.81 m/s2 )(1.2 m)

v2 = 4.85 m/s

12

v1 = 0 650 kg

1.2 mPosition 2

v2

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Problem 2 Solution

Impact process:

Apply conservation ofmomentum principle.

mH vH

mP vP = 0

mH v’

mP v’

Beforeimpact:

Afterimpact:

mH vH + mP vP = ( mH + mP ) v’

(650 kg)( 4.85 m/s) + 0 = (650 kg + 140 kg) v’

v’ = 3.99 m/s

11 - 32

Problem 2 Solution

v’ v = 0W

R

110 mm

Apply principle of work and energy.

Hammer and pile move against ground resistance.

Position 1 Position 2Work

T1 + U1 2 = T2

( mH + mP ) v’2 + (WH + WP - R) y = 0

(650 + 140) (3.99 m/s)2 + [(650 + 140)(9.81) - R](0.110) = 0

R = 65,000 N

1212

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Problem 3

A small sphere B of mass m is attachedto an inextensible cord of length 2a,which passes around the fixed peg A and is attached to a fixed support at O.The sphere is held close to the supportat O and released with no initial velocity.It drops freely to point C, where the cordbecomes taut, and swings in a verticalplane, first about A and then about O.Determine the vertical distance fromline OD to the highest point C’’ that thesphere will reach.

45o

OB

C C’

C’’

D

Aa

11 - 34

Problem 3

A small sphere B of mass m is attachedto an inextensible cord of length 2a,which passes around the fixed peg A and is attached to a fixed support at O.The sphere is held close to the support

at O and released with no initial velocity. It drops freely to point C,where the cord becomes taut, and swings in a vertical plane, firstabout A and then about O. Determine the vertical distance fromline OD to the highest point C’’ that the sphere will reach.

1. Apply conservation of energy principle : When a particlemoves under the action of a conservative force, the sum of thekinetic and potential energies of the particle remains constant.

T1 + V1 = T2 + V2

where 1 and 2 are two positions of the particle.

45o

OB

C C’

C’’

D

Aa

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1a. Kinetic energy: The kinetic energy at each end of the pathis given by:

T = m v2

Problem 3



12

45o

OB

C C’

C’’

D

Aa

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Problem 3




45o

OB

C C’

C’’

D

Aa

11 - 37

Problem 3



45o

OB

C C’

C’’

D

Aa

2. Apply the principle of impulse and momentum: The finalmomentum mv2 of the particle is obtained by adding its initialmomentum mv1 and the impulse of the forces F acting on theparticle during the time interval considered.

mv1 +F t = mv2

F is sum of the impulsive forces (the forces that are largeenough to produce a definite change in momentum).

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Problem 3 SolutionApply conservation of energy principle.

Motion of the sphere from point B to point C (just before thecord is taut).

45o

OB

D

AavB = 0 Position 1

45o

OB

C

D

Aa

vC

Position 2

y

T1 + V1 = T2 + V2

0 + 0 = m vC2 - m g (2 a sin 45o )

vC = 1.682 g a

12

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Problem 3 Solution

Consider the sphere at point C as the cord becomes taut and thevelocity of the sphere changes to be in direction normal to the cord.

Apply impulse and momentum principle.

45o

OB

C

D

Aa

vC

45o

OB

C

D

Aa

v’C45o

t t

Momentum is conservedin the tangential directionsince the external impulse(the cord on the sphere)is in the normal direction.

m vC cos 45o = m v’C

v’C = vC cos 45o = 1.682 g a cos 45o

v’C = 1.1892 g a

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Problem 3 Solution

Motion of the sphere from point C to point C’’.

45o

OB

C

D

Aa

v’C = 1.1892 g a

Position 2

45o

OB

C’’

D

Aa

y

d2 a sin45o

Position 3

T2 + V2 = T3 + V3

m (v’C)2 - m g (2 a sin 45o ) = 0 - m g d

m (1.1892)2 g a - m g (2 a sin 45o ) = 0 - m g d

d = 0.707 a

vC’’= 0

1212

Apply conservation of energy principle.

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A

B

lA

lB

A

B

C

D

Problem 4

A small sphere A attached to a cordAC is released from rest in theposition shown and hits an identicalsphere B hanging from a verticalcord BD. If the maximum angle B

formed by cord BD with the verticalin the subsequent motion of sphereB is to be equal to the angle A ,determine the required value of theratio lB / lA of the lengths of the twocords in terms of the coefficient ofrestitution e between the twospheres.

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Problem 4

A

B

lA

lB

A

B

C

D A small sphere A attached to a cordAC is released from rest in theposition shown and hits an identicalsphere B hanging from a verticalcord BD. If the maximum angle B

formed by cord BD with the vertical in the subsequent motion ofsphere B is to be equal to the angle A , determine the requiredvalue of the ratio lB / lA of the lengths of the two cords in terms ofthe coefficient of restitution e between the two spheres.

1. Apply principle of conservation of energy: When a particlemoves under the action of a conservative force, the sum of thekinetic and potential energies of the particle remains constant.

T1 + V1 = T2 + V2

where 1 and 2 are two positions of the particle.

11 - 43

Problem 4

A

B

lA

lB

A

B

C

DA small sphere A attached to a cordAC is released from rest in theposition shown and hits an identicalsphere B hanging from a verticalcord BD. If the maximum angle B

formed by cord BD with the vertical in the subsequent motion ofsphere B is to be equal to the angle A , determine the requiredvalue of the ratio lB / lA of the lengths of the two cords in terms ofthe coefficient of restitution e between the two spheres.1a. Kinetic energy: The kinetic energy at each point on the pathis given by:

T = m v2


12

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Problem 4

A

B

lA

lB

A

B

C


formed by cord BD with the vertical in the subsequent motion ofsphere B is to be equal to the angle A , determine the requiredvalue of the ratio lB / lA of the lengths of the two cords in terms ofthe coefficient of restitution e between the two spheres.2. Apply conservation of momentum principle: During an impactof two bodies A and B, the total momentum of A and B isconserved if no impulsive external force is applied.

mA vA + mB vB = mA v’A + mB v’B

where vA and vB denote the velocities of the bodies before the

impact and v’A and v’B denote their velocities after the impact.

11 - 45

Problem 4

A

B

lA

lB

A

B

C


formed by cord BD with the vertical in the subsequent motion ofsphere B is to be equal to the angle A , determine the requiredvalue of the ratio lB / lA of the lengths of the two cords in terms ofthe coefficient of restitution e between the two spheres.3. Apply the relationship for the coefficient of restitution: Forimpact of two particles A and B:

v’B - v’A = e (vA - vB)

where v’B - v’A and vA - vB are the relative velocities, normal to the

impact plane, after and before the impact, respectively, and e isthe coefficient of restitution.

11 - 46

Problem 4 Solution

A

B

lA

lB

A

B

C

D Motion of sphere A from itsrelease until it hits sphere B.

Position 1

Position 2

A

lA

A

C

vA2

vA1= 0lA( 1 - cos A )

y

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Problem 4 Solution

Apply principle of conservationof energy.

Position 1

Position 2

A

lA

A

C

vA2

vA1= 0lA( 1 - cos A )

y

T1 + V1 = T2 + V2

0 + m g lA( 1 - cos A ) = m (vA2)2 + 0

vA2 = 2 g lA( 1 - cos A )

12

Motion of sphere A from itsrelease until it hits sphere B.

11 - 48

Problem 4 Solution

A

B

lA

lB

A

B

C

D Collision of balls A and B.

B

C

D

vA2 vB2 = 0

Before impact

B

C

D

v’A2 v’B2

After impact

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Problem 4 Solution

B

C

D

vA2 vB2 = 0

Before impact

B

C

D

v’A2 v’B2

After impact

Apply conservation ofmomentum principle.

m vA2 = m v’A2 + m v’B2

vA2 = v’A2 + v’B2 (1)

Apply the relationshipfor the coefficient ofrestitution.

( v’B2 - v’A2 ) = e ( vA2 ) (2)

Eliminating v’A2 from equations (1) and (2) gives:

v’B2 = ( 1+ e )2

vA2

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Problem 4 Solution

A

B

lA

lB

A

B

C

DMotion of sphere B followingthe collision.

B

lB

B

D

lB( 1 - cos B )

Position 2v’B2

vB3

Position 3

11 - 51

Problem 4 Solution

Motion of sphere Bfollowing the collision.

B

lB

B

DlB( 1 - cos B )

Position 2v’B2

vB3

Position 3y

Apply principle ofconservation of energy.

T2 + V2 = T3 + V3

m ( v’B2 )2 + 0 = 0 + m g lB( 1 - cos B )

12

Substituting v’B2 = ( 1+ e )2

vA2 and vA2 = 2 g lA( 1 - cos A )

and B = A gives:

lBlA

= ( )1 + e2

2

11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse...

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Transcript of 11 - 1 Impulse and Momentum Some more extremely useful relations are the concepts known as impulse...