me103 04 work energy

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MECH103Mechanisms & Dynamics of Machinery

Chapter 4Energy Methods

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Chapter 4 - Principle of Work & Energy• Principle of work & Energy:

– Newton’s second lawdtdm vF

Dot both sides with v and integrate them with respect to t

2 2 2 2 2

1 1 1 1 1

1 ( )2

t t

t t

dd dt m dt m d mddt

r v v

r v v

vF r F v v v v v v

1 [ ( )]2

d dm mdt dt

vF v v v v dt d F v F r

Work done to an object equals the change in the kinetic energy of the object

2

1

)(21)(

21 2

1221122

r

rvvvvrF vvmmdU

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Chapter 4 – Work Done

• Work done by F from r1 to r2

Evaluation of the work

2

1

r

1 2 rU U d F r

s1 s2

Ft

s

2

1

s

s tdsFU

x

y

r1

r2

dr(r1)

z

r2

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Evaluation of the work

Ft

s1 s2s

2

1

s

s tdsFU

Ft

s1 s2

sNegative work: Motion is in the opposite direction of force (e.g. work done by friction)

Ft

s1 s2s

Constant Force: U = Ft(s2 –s1)

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Work done by weight

F = - mgj

d dx dy dzr i j k

)( 122

1

2

1

yymgmgdydUy

y

r

rrF

eF rE

rmgRIf 2

2

12

2 11rr

mgRU E

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Work done by spring

F = k(r – ro)er

21or ( )2 oU k r r

( )o od k x x dx dy dz k x x dxor F r i i j k

F F

ro or xo

F Fr or x

F = k(x – xo)i or

( )o r r od k r r dr rd k r r dr F r e e e

2)(21

oxxkU

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Chapter 4 - Example

• Example 1 – Potential energy of weight

kjiF mg 00

kjir dzdydxd dV = - F dr = mgdz

V2 – V1 = mg(z2 – z1) (= mgh)

2 2 2

1 1 1

V r z

V r zdV d mgdz F r

j

k

i

g

dVd -rF

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Chapter 4 - Example

• Example 2 – Potential energy in gravitational field

rE

rmgR eF 2

2

eer rddrd r

22- E

drdV d mgRr

F r

1

21

11rr

mgRVV E- -1 1

22

V r

EV r

drdV mgRr

y

x

r

mdr

drer

rde

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PowerRate at which work is done

Unit: Joule/sec or Watt, hp (=746 Watt)=550 ft-lb/s

2

21 mv

dtd

dtdUP

Average Power:

2

1

2 22 1

2 1 2 1

1 12

t

av t

mP Pdt v vt t t t

http://www.sizes.com/units/horsepower_british.htm

Dynamometer: Engine horsepower measurements, transmission fatigue testing, and driveline parasitic horsepower loss measurements.

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dVd -rF If there exists a function V of position only such that

2

1

r 2 21 2 1 2 2 1r

1 ( )2

U d V V m v v F r

222

211 2

121 mvVmvV 21i.e. Constant

2V mv

V(x,y,z) is called the potential energy.

Principle of Conservation of Energy The sum of the kinetic energy and potential

energy (or strain energy) is constant. True only if F is conservative!!

: kinetic energy

2211 VTVT

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Chapter 4 – Example - Pendulum

• Example 3 – Conservation of energy : kinetic energyV : potential energy

21At point A, 2Ah L v gL v

1 1 2 2T V T V

2 21 1 2 2

1 12 2

mv mgy mv mgy

2 22 1 2 1 12 ( ) [ 2 ]v g y y v gh v

v1mg

y

x

mg

FL

v2

LF

v1 L

v3

mg

F

y1

y3

mg

FL

v2

y2

h

Amg

FL

v2

y2

sinLh

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Example 3-6: Pendulum

The mass m is released from rest with the string horizontal. By using Newton’s 2nd law in terms of polar coordinates, determine the velocity magnitude v and the tension T in the string as functions of

cos2

2

mgdtdmr

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Ex 3-6: Pendulum

Solution Draw free body diagram Express the force in polar coordinates

Weight: ( sin cos )rmg W e e

2

2

Acceleration: 2

0

r

r

r r r r

L L r, r

a e e

e e

Tension : rTT e

m T W a

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2 2sinrF T mg mL mL

cosF mg mL

21 sin , (for 0 at 0) or2

2 sin

gL

v L gL

cosd gd L

2sinrF T mg mL sin3sinsin2 mgmgmgT

Ex 3-6: Pendulum

How to solve this problem using Energy Method?

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Chapter 4 – EnergyPotential energy (strain energy) of linear spring

or F = k(r – ro)er

21or ( )2 oU k r r

( ) o o

dV dU dk x x dx dy dz k x x dx or

F ri i j k

F F

ro or xo

F Fr or x

F = k(x – xo)i

( )o r r od k r r dr rd k r r dr F r e e e

2

Strain energy1 ( )2 oU k x x

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Example 4-1 – Potential energy (strain energy) : spring and friction

A spring is used to stop a 60-kg package which is

sliding on a surface. The spring has a constant

k=20kN/m and is held by cables so that it is initially

compressed 120mm. Knowing that the package has

v= 2.5m/s shown and that the max additional spring

deflection is 40 mm, determine (a) the coefficient of

kinetic friction (k) between the package and the

surface, (b) the vel. of the package as it passes

again through the position shown

S.E. = Strain Energy: one kind of potential energy

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Ex 4-1 – Example-package stopped by spring

• Solution Free Body Diagram

k=0

W

Nf = kN

F= kx

Energy in each state

1

v1x0 +x1

x1 : Initial compression

21

21

1. .21. .2

K E mv

S E k x

State 1Initial Condition

2s

v2 = 0

x0 + x2

22

22

1. . 021. .2

K E mv

S E k x

State 2Spring is fully compressed

x2 : Final compressions : total distance traveled

2 2 21 1 2

1 1 12 2 2

mv k x k x

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Conservation of energyInitial K.E. of the package – |Work done by friction|+ initial S.E. stored in spring = Zero K.E. + S.E. stored in spring

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21

1. .21. .2

K E mv

S E k x

State 1Initial Condition

22

22

1. . 021. .2

K E mv

S E k x


2 2 21 1 2

1 1 12 2 2kmv mg s k x k x k is unknown

1

v1 v2 = 0

2s

2 1 0.04 0.16x x m m 1 0.12x m

2 10.6 .ks x x can be determined 1 2.5 /v m s


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2v2 = 0

22

22

1. . 021. .2

K E mv

S E k x


3

v3

23

21

1. .21. .2

K E mv

S E k x

State 3Rebounce

s2

v2 = 0

W

N f = kN

F=kx

Conservation of energyFinal K.E. of the package (State 3) + Initial S.E. stored in spring (State 3 = State1) = S.E. stored in spring (State 2)– Work done by friction

2 2 23 1 2

1 1 12 2 2 kmv k x k x mg s v3 is unknown

2 1 0.04x x m 1 0.12x m 2 10.6s x x

3 .v can be determined


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Example 4-2: Crate on inclined surfaceThe two crates in the figure are released from rest. Their masses are mA=40 kg and mB = 30 kg, and the kinetic coefficient of friction between crate A and the inclined surface is k = 0.15. What is their velocity when they have moved 400mm?

v

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• Solution Free Body Diagram mA g

Nf = kNT = 20

T

mB g

Crate A Crate B

Method 1:Consider Crates A & B separately

BAivvmdsΣF i

s

s t ,),(21 2

122

2

1

)0(

21)20cos(20sin 24.0

0 vmdsgmgmT AAkA

Crate A

Crate B0.4 2

0

1( ) ( 0)2B Bm g T ds m v 2 eqs, 2 unknowns

BA

kBA

mmmm

T)sincos1(

BA

kAB

mmmmsg

v

)cos(sin(22 v = 2.07 m/s

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Conservation of energy

0

21

21

)20cos(20sin

22

4.0

0

4.0

0

vmvm

gdsmdsgmgm

BA

BAkA

.determinedbecanv

Example 4-2: Crate on inclined surface

Method 2:Consider Crates A & B together

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Ex 4-3: Car on the track

A 2000 lb car starts from rest at point 1 and

moves w/o friction down the track shown. (a)

Determine the force exerted by the track on

the car at point 2, where the radius of

curvature of the track is 20 ft. (b) Determine

the minimum safe value of the radius of

curvature at point 3.

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Ex 4-3: at point 2

• Solution

W

N

man = mv22/

State 2

Conservation of energy2 21 1 2 2

1 12 2

mv mgh mv mgh

and can be determinednN a .

22

nvN mg ma m

1 1 2

2

0, 40 .

v h h ftv can be determined.

Centripetal acceleration

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Ex 4-3: at point 3

W

man = mv32/N

State 3

Conservation of energy

2 21 1 3 3

1 12 2

mv mgh mv mgh

23

nvN mg ma m

Min. value of can be determined by setting 0N .

1 1 3

3

0, 25 .

v h h ftv can be determined.

Centripetal acceleration

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Ex 4-3: Skier on the ramp

Textbook Ex 15.4

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222

211 2

121 mvmgymvmgy

Textbook Ex 15.4

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Solution On the top of ramp (State 1); leaving the ramp (State 2)

2 21 1 2 2

1 12 2

mv mgh mv mgh

Add (jump) vertical velocity of 3 m/s at point 2

22

22 3 vv

1 1 2

2

0, 20 (given)the horizontal component can be determined

v h h meterv .


v2 = 24.8 m/s

v’2 = 25.0 m/s

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Leaving the ramp (State 2) to the highest point (State 3)

2 22 2 3 3

1 12 2

m v mgh mv mgh

2 2 22 2

3 2

3

3 .At the highest point of jump:

can be determined

v vv v

h .


h3 = 0.459 m

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Example 4.4 Net Work by Internal Forces

• QuestionCrates A and B in Fig are released from rest. The coefficient of kinetic friction between A and B is μk, and the friction between B and the inclined surface can be neglected. What is the velocity of the crates when they have moved a distance b?

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• StrategyBy applying the principle of work and energy to each crate, we can obtain two equations in terms of the tension in the cable and the velocity.

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• SolutionWe draw the free-body diagrams of the crates in Figs a & b. The acceleration of A normal to the inclined surface is zero, so N = mAgcos θ. The magnitudes of the velocities of A and B are equal (Fig c). The work done on A equals the change in its kinetic energy.

1

21cossin

21

220

21

2212

vmdsgmgmT

vvmU

Ab

AkA

A

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The work done on B equals the change in its kinetic energy.

221cossin

21

220

21

2212

vmdsgmgmT

vvmU

Bb

AkB

B

Summing Eqns (1) & (2) to eliminate T to solvefor v2, we obtain

BAAkAB mmmmmgbv cos2sin22

me103 04 work energy

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