Examples in Dynamics.pdf

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Real Life

Examples in

Lesson plans and solutions

Eann A Patterson, Edito


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Real Life Examples in Dynamics



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First edition July 2009

Second [electronic] edition, 2011



Copyright 2009, 2011 Eann A Patterson (editor)

This work is licensed under the Creative Commons Attribution-NonCommercial-

NoDerivs 3.0 Unported License. To view a copy of this license, visit

http://creativecommons.org/licenses/by-nc-nd/3.0/ or send a letter to Creative Commons,

444 Castro Street, Suite 900, Mountain View, California, 94041, USA.

The Editor has no responsibility for the persistence or accuracy of URLs for external

or third-party internet websites referred to in this publication, and does not guarantee

that any content on such websites is, or will remain, accurate or appropriate.

Bounded printed copies can be purchased on-line at www.engineeringexamples.org

This edition is distributed free of charge by the ENGAGE project(www.EngageEngineering.org), which is supported by the National Science Foundationunder Grant No. 0833076.


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Real LifeExamples in

Dynamics


Suggested exemplars within lesson plans for J unior level courses in Dynamics. Prepared

as part of the NSF-supported project (#0431756) entitled: Enhancing Diversity in the

Undergraduate Mechanical Engineering Population through Curriculum Change

Eann A Patterson, Editor

The University of Liverpool, England

[email protected]


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This work was developed during the NSF-supported project (#0431756) entitled:Enhancing Diversity in the Undergraduate Mechanical Engineering Population throughCurriculum Change. Any opinions, findings, and conclusions or recommendationsexpressed in this material are those of the authors and do not necessarily reflect the viewsof the National Science Foundation. The following members of the project contributed tothe many discussions which led to this production of this volume:Ilene Busch-Vishniac McMaster University (Project leader)Patricia B. Campbell Campbell-Kibler Associates IncDarrell Guillaume California State University, Los AngelesEann Patterson Michigan State University


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CONTENTS

page no.

Introduction 5

KINEMATICS OF PARTICLES

1. Rectilinear and curvilinear motion 6

Paper airplanes, raindrops, sneezing, iPod falling from bike

KINETICS OF PARTICLES

2. Force and acceleration 9

Hockey puck, dustpan and brush, car skidding

3. Work & energy 13

Sling shot, two-slice toaster, medieval longbow

4. Impulse and momentum 17

Tennis balls, kids on water slide

SYSTEMS OF PARTICLES

5. Steady streams of particles 20

Desk fan, hairdryer, wind turbine

KINEMATICS OF RIGID BODIES

6. Kinematic diagrams and angular velocity & acceleration 23

Yoyo, bicycle pedaling

PLANE MOTION OF RIGID BODIES

7. Forces and acceleration 26

Bicycle braking, roly-poly, pizza cutting

8. Work and energy 30

Marbles, yoyo, broken down car

9. Impulse and momentum methods 33

Basketball, skateboard, tennis topspin

THREE DIMENSIONAL RIGID BODY MOTION

10. Kinematics of rigid bodies in three dimensions 37

Violin-playing robot, industrial robot, teeth cleaning

11. Kinetics of rigid bodies in three dimensions 42

Spinning top, bicycle stability, equinox precession

MECHANICAL VIBRATIONS

12. Undamped and damped vibrations 47

Hula hooping, singing ruler, whip aerial, earthquake protection bearings


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INTRODUCTIONThese real life examples and supporting materials are designed to enhance the teaching ofa junior level course in dynamics, increase the accessibility of the principles, and raise theappeal of the subject to students from diverse backgrounds. The examples have beenembedded in skeletal lesson plans using the principle of the 5Es: Engage, Explore,Explain, Elaborate and Evaluate. The 5E outline is not original and was developed by theBiological Sciences Curriculum Study

1in the 1980s from work by Atkin and Karplus

2in

1962. Today this approach is considered to form part of the constructivist learning theoryand a number of websites provide easy-to-follow explanations of them3.

This booklet is intended to be used by instructors and is written in a style that addresses

the instructor, however this is not intended to exclude students who should find the notesand examples interesting, stimulating and hopefully illuminating, particularly when theirinstructor is not utilizing them. In the interest of brevity and clarity of presentation,standard derivations and definitions are not included since these are readily available intextbooks which this booklet is not intended to replace but rather to supplement andenhance. Similarly, it is anticipated that these lessons plans can be used to generatelectures/lessons that supplement those covering the fundamentals of each topic.

It is assumed that students have acquired a knowledge and understanding of topicsusually found in a Sophomore level course in Statics, including free-body diagrams andefficiency.

This is the second in a series of booklets. The first in the series entitled Real LifeExamples in Mechanics of Solids edited by Eann Patterson (ISBN: 978-0-615-20394-2)was produced in 2006 and is available on-line at www.engineeringexamples.org.

1Engleman, Laura (ed.), The BSCS Story: A History of the Biological Sciences Curriculum Study. ColoradoSprings: BSCS, 2001.2Atkin, J. M. and Karplus, R. (1962). Discovery or invention? Science Teacher 29(5): 45.3e.g. Trowbridge, L.W., and Bybee, R.W.,Becoming a secondary school science teacher. Merrill Pub. Co.Inc., 1990.


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1. Topic: Rectilinear and curvilinear motion

Engage:

Stop by the office recycling box on your way toclass and pick-up enough sheets of paper to giveone to each member of the class and have someleft over. Pass them around the around the classand invite the students to write on their sheetthree reasons why they chose to studyengineering; then ask them to throw it to amember of the class on the other side of the

room.Most of the students will crunch the sheet up into ball to throw it, and a few might make apaper airplane. If no one else does, you should demonstrate what happens if you try tothrow it as a sheet of paper.

Explore:

Discuss what is meant by the term kinematics, i.e. cases in which only the geometricaspects of motion are considered and that a particle has mass but negligible size andshape (no air resistance). For the sheets of paper to be considered particles then theirdimensions must have no influence on the analysis of their motion, i.e. their motion is

characterized by the motion of their center of mass and any rotation of the body can beneglected. So, the flat sheets dont fly well due to the effect of air resistance on theirshape but the tight balls behave as particles and go where you aim them. The trajectoryof the paper airplanes is strongly dependent on their shape.

Explain:

In a heavy rainstorm large droplets fall from lowNimbostratus cloud at 600 ft (183m). Letsassume that raindrops are approximately

spherical and of diameter, d3mm, so given thedensity of water is 1000 kg/m3, their mass (m)will be:

533

10413.13

0015.041000

3

4

22

r

Vm OHOH kg (0.014g)


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Assuming a constant acceleration,g = 9.81m/s2, and that the time at the start of the fall is

t = 0 when velocity, vy= vy=0= 0, and displacement, sy= sy=0 = 0 then when the raindropsfall 183m we find using:

)(2 01832

0

2

183 yyyyy ssavv

that 359018381.922 183183 yy gsv m/s

This is ten times the speed of sound (340 m/s at sea level) and is unrealistic.

In practice we need to include an analysis of the forces causing motion, i.e. kineticsandcalculate a terminal velocity, vtwhich is approached when the drag force, Fddue to air

resistanceequals the gravitational force, Fg(= mg), i.e. gd FF

where daird ACvF2

2

1

and the density of air,air1.2 kg/m3

,Ais the frontal area and Cdis the drag coefficient(0.4 for a rough sphere).

Thus dair ACvmg2

2

1 and

4.010342.1

81.910413.12223

5

dairyt

AC

mgv = 9 m/s

Elaborate:

The horizontal dispersion of the rain droplets can be estimated by considering thehorizontal component of the motion after calculating, in the vertical direction, the time

from release to impact with the ground, i.e. 22

1 atvts but at terminal velocity theacceleration is zero,

so tvs yty and 3.209

183

yt

y

v

st s

Plenty of time to dodge it if there was only one! Considering the horizontal direction

with a constant wind speed of 22m/s (50 mph a force 10 gale according the Beaufortscale) then

4473.2022221

0 tatvs xxx m

i.e. it travels further horizontally than it falls vertically which is what we often see in aviolent storm. If the wind drops to force 3, more likely with Nimbostratus, with a wind

speed of 10mph (4.5 m/s) then they fall much closer to the vertical, i.e.

4.913.205.4221

0 tatvs xxx m

It is assumed that the upwards force of buoyancy is negligible.


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Evaluate:

Invite students to attempt the following examples:

Example 1.1

An individual, who is 1.7m (5ft 6in) tall, tilts her head back to sneeze so that particlesleave her nostrils horizontally with a velocity of 40m/s (90mph), how far away from theindividual will the particles land?

Solution:

Time to fall vertically: 221

0 tatvs yyy so 589.081.9

7.122

y

y

a

st s given vy=0 = 0

Distance travelled horizontally: 5.230589.040221

0 tatvs xx m

Example 1.2

While riding your bike across a pedestrian bridge at about 50km/hr (30mph) your iPoddrops out of your pocket and, missing the bridge, falls to the ground 6m below. Estimatethe velocity at which your iPod will hit the ground below the bridge and the horizontaldistance that it will travel before impact. Comment on the factors influencing whetheryour iPod will survive the fall.

Solution:

Time to fall vertically: 221

0 tatvs yyy so 11.181.9

622

y

y

a

st s given vy=0 = 0

Distance travelled horizontally:

4.1511.189.13011.16060

100050221

0

tatvs xx m

Vertical component of velocity on impact:

tavv yyy 0 so 9.1011.181.9 tav yy m/s

Magnitude of velocity on impact, 7.178.109.13 2222 yx vvv m/s

Direction of impact, 389.139.10tan 1

So your iPod will hit the ground at 18m/s (40mph) at an angle of 38to the ground.


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2. Topic: Force and acceleration

Engage:

Take an ice hockey stick and puck intoclass and push the puck around on thefloor. If you know nothing about icehockey shots then try typing ice hockeysnapshot in www.youtube.comfor someshort coaching videos

4. Alternatively use

a dustpan and broom with a jar lid (e.g. ajam jar lid). Sweep the jar lid into the

dustpan.

4http://www.youtube.com/watch?v=TYEE7tZhRtk

photo by Hkan Dahlstrmon Flickr

Explore:

Discuss the meaning of kinetics which involves the analysis of forces causing motion.Remind the class about how consideration of the drag force gave a more realistic answerfor the velocity of the raindrops in the previous lesson. Discuss the forces acting on the

puck (jar lid) when it is pushed across the floor: driving force from the stick (broom);friction; drag (probably negligible), gravity and the reaction from the floor.

Explain:

Sketch the free body diagrams below explaining Newtons laws of motion as the situationbecomes progressively more complicated:

Puck resting on ice (just before being push)

Fg=mg

N=Fg

y

x

Fg=mg

N=Fg

y

x

y

x

Newtons Third Law:for every action, there is an equal and opposite reaction. So themass of the puck acts on the floor and there is an equal and opposite reaction from thefloor,N.


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Puck being pushed across the floor by stick

Fg=mg

N=Fg

y

x

S

Ff=uN

ax

Fg=mg

N=Fg

y

x

y

x

S

Ff=uN

ax

Newtons Second Law: the acceleration of an object is proportional to the force actingon it and inversely proportional to its mass. So when a force, Sis applied with the stick,the puck will accelerate in the direction of this force, such that:

xxmaF

There will be also a frictional force resisting motion at the interface between the floor andthe bottom of the puck:

Ff= N

where is the coefficient of friction.

Puck sliding across the ice on its own

Fg=mg

N=Fg

yx Ff=uN

ax

Fg=mg

N=Fg

yx

yx Ff=uN

ax

Newtons First Law: a body continues to maintain its state of rest or of uniform motionunless acted upon by an external force. This also applied to the first diagram in whichthe puck is in a state of rest. In this case if the floor is highly polished or the puck istravelling on ice then the coefficient of friction could be assumed to be almost zero andthe acceleration will be approximately zero; otherwise the puck will suffer a decelerationdue to the friction force from the floor.

Elaborate:

An NHL ice hockey puck has a maximum mass of 170g. So if a player applied a force of35N and the coefficient of friction between the ice and the puck (vulcanized rubber) is

approximately 0.15 then applying Newtons Second Law:

xx maF and

204170.0

81.9170.015.035

m

NS

m

Fa

x

x

m/s

2

and if the player pushes it for 1.5m then the velocity as the puck leaves the stick will be:

asvv xx 22

0

2 , i.e. 8.245.120422 asvx m/s (55 mph)


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Alternatively, for a jar lid of mass 15g sliding on a wooden floor (= 0.2) assuming aforce of only 1N is applied with the broom then:

65

015.0

81.9015.02.01

m

NSa

x m/s2

And if it is pushed by the broom for 0.05m (2 inches) then its velocity will be:

54.205.06522 asvx m/s (5.7 mph)

Evaluate:

Ask students to attempt the following examples:

Example 2.1

Sometimes during half-time at a hockey game, a car is driven into the ice-ring (perhaps to

show it off as a lottery prize). If the driver applies the brakes so that they lock, calculatethe maximum speed for which the skid will be no more than 2m if the car has a mass of

2000kg. Compare this with a skid on a wet road (=0.45)

Solution:

Free-body diagram:

Ff=uNN=Fg

Fg=mg

ax

yx Ff=uNN=Fg

Fg=mg

ax

yx

yx

applying Newtons Second Law for car on ice:

xx maF and

47.12000

81.9200015.0

m

N

m

Fa

f

x

m/s

2

Using kinematics: asvv xx 22

0

2 , i.e. 4.2247.1220 asvx m/s (5.4 mph)

Again applying Newtons Second Law for car on wet concrete:

xx maF and 41.42000 81.9200045.0 mNax m/s2

Using kinematics: asvv xx 22

0

2 , i.e. 2.4241.4220 asvx m/s (9.4 mph)

This is why cars are fitted with Antilock Brake Systems (ABS) !


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Example 2.2

Ask the students to continue exploring the lesson example by calculating the maximumdistance from the goal-keeper that the shot can be taken in order to beat the reactions ofthe goal-keeper, assuming a typical visual reflex time is 180 milliseconds.

Solution:

For the puck moving across the ice,

xxmaF and

47.1

170.0

81.9170.015.0

m

N

m

Fa

f

x

m/s2

The distance that the puck will travel in 180 milliseconds is:

4.418.047.15.0180.08.24 2221 tatvs xx m

So the shot should be taken within 4m of the goal-keeper.


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3. Topic: Work and Energy

Engage:

Bring a slingshot and a handful of rubberballs into class, pull the elastic band back andrelease a few rubber balls into the class.

Explore:

Remind students about the basic meaning of conservation of energy. Ask students towork in pairs and to identify the conservation of energy during the loading, firing andtrajectory of the balls. Invite some pairs to talk through to the class their understandingof the energy conversions. Discuss how strain energy stored in the sling material istransferred as work to the ball and becomes kinetic energy. Ask them in their pairs toreconsider conservation of energy during loading, firing and flight of projectile.

Explain:

Applying the principle of work and energy to the rubber ball during firing:

2211 KEUKE

where KE1and KE2are the initial and final kinetic energies of the ball respectively andU1-2is the work done by all the external and internal forces acting on the ball betweenthe initial and final state.

By definition, kinetic energy, 2

2

1mvKE , and KE1 = 0 (because v1=0).

The work done on the ball during firing is the strain energy released from the sling shotwhich was stored in the material as it was stretched, uVU where u is the specificstrain energy at yield and V is the volume of material and assuming the sling shot isstretched to the limit of its elastic behavior.

During flight potential energy maybe gained or lost with height but this is negligiblecompared to the kinetic energy at launch. Some kinetic energy will be lost due to dragbut again it is small so that the ball will arrive at its target with a substantial portion of thelaunch kinetic energy.


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Elaborate:

The strain energy stored at yield in an elastic band is u= 3.6 MJ/m3; if it is of length,

150mm with a cross-section of 16mm2 and volume, V of 2400mm

3, then the energy

stored when it is pulled to yield is (see chapter 6 in Real Life Examples in Mechanics of

Solids):

6.8102400106.3 96 uVU J

so for a yellow dot squash ball of mass 24g with initial kinetic energyKE1 =0:

2

221

2211 mvKEUKE and 271024

6.82232

mU

v ms-1

If the ball is launched parallel to the ground from a height of 1.5m, then its range can becalculated using kinematics (neglecting drag) since in the vertical direction: sy= 1.5m,vy=0 = 0, and ay= g= 9.81m/s

2,

so savv yyy 2 02 and 4.55.181.922 yyy sav m/s

and tavv yyy 20 so 28.081.92

42.5

2

y

y

a

vt s

In the horizontal direction (neglecting drag): 45.7028.027221 tatvs xx m

so the slingshot has a range of almost 7.5m.

(Note: potential energy gained in flight is (mgh= 0.0249.811.5 = ) 0.35J or 4% of thekinetic energy at launch).

Evaluate:

Ask students to do the following examples:

Example 3.1

A two-slice toaster is switched on by depressing a slider which causes the slices of breadto be drawn downwards into the toaster between the heating elements and also extends aspring at each end of the toaster. When the toast is ready a pair of triggers releases bothsprings simultaneously which in turn provide an impulse to the toast so that it pops up. A

typical slice of bread is 125mm 125mm and has a mass of 40g, and in the off position

a slice sits with two-thirds of its length inside the toaster.(a) Calculate the force which must be applied to the slider so that toast pops up but justdoes not come out of the toaster when it is ready, neglecting the weight of the mechanismand assuming there are no losses in it.

(b) If the toaster is redesigned with each slot rotated outwards so that the plane of the

slice is at 60to the horizontal table top and two additional identical springs are fitted;then how far away from their respective slots will the slices land if the effect of the shapeof slice is neglected.


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Solution

(a) Toast must not jump higher than minimum depth of slotlength of a slice of bread

At top of jump, vertical velocity,vy=vy2= 0 thus using yyyy savv 22122

57.1125.081.9221 yyy sav ms-1

Hence the kinetic energy received by each slice from the springs at launch is:

049.02

57.104.0 2212

1

ymvKE J

To store this much energy in a spring the work done (W) on the springs is

FdWKE 21 where Fis the force applied and moved through a distance d. Since in

the resting position the bread is two-thirds within the toaster the available travel for the

slider, 0833.0125.032 d m.

So 18.10833.0

049.022

d

WF N

For the two-slice toaster the force that needs to be applied is 2.35N (= 1.182)

(b) With the additional springs the energy stored will be doubled, i.e. the kinetic energy

delivered to each slice will be 0.098J (= 0.0492).

Thus, the exit velocity will be 2.204.0

098.022

mKE

v ms-1

from2

21

mvKE

The vertical component will be 91.160sin2.260sin1 vvy ms-1

and the horizontal component, 10.160cos2.260cos1 vvx ms-1

For time to top of flight: tavv yyy 12 and vy2= 0 so 19.081.9

91.1t s

The time for the rise and fall to the top of the flight is 0.39 seconds (=20.19), anddistance travelled horizontally is

43.0039.010.12

21 tatvs xxx mi.e. each slice will land 430mm away from the toaster.


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Example 3.2

Estimate the range of a longbow of the type used in medieval England (e.g. by RobinHood). Such bows are about 2m long, made of yew with a draw of about 0.76m andwere used with arrows having a mass of about 50g. The draw force has been estimated tobe about 265N. To see a longbow in action just type longbow inwww.youtube.com(http://www.youtube.com/watch?v=YtTyOf8OCKg).

Solution:

The incremental work done, W by the archer as the bow is pull back (drawn) over aincremental distance, d is given by dFW , so integrating this over the total drawgives the total work done, 10176.02652

121 FdW J assuming the draw force is

linearly related to the draw, i.e. the bow has a constant stiffness.

This work is stored as strain energy in the bow and, when the arrow is released, it istransferred to the arrow as kinetic energy, i.e.

222

12211 mvKEUKE and 6.63

1050

101223

m

Uv ms

-1

If the arrow is launched at an angle, to the ground then the vertical component of

velocity is sin0 vvy and horizontal component is cos0 vvx . The time of flight

can be calculated as the twice the time required to reach its maximum height when itsvertical velocity, vy= 0, i.e.

tavv yyy 0 andg

vavt

y sin0

The distance travelled horizontally during time, 2twill be

221 tatvs xx thus sincos

22cos

2

g

vtvs

For maximum range

222

sincos2

0 g

v

d

ds i.e. 1

cos

sin

and =45

Hence substituting in 4122

1

81.9

6.632sincos

2222

g

vs m

Replica longbows have been shown to have a range of about 300m so this estimate is alittle long which is unsurprising since drag and energy losses were neglected.


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4. Topic: Impulse and momentum methods

Engage part I:

Bring atennis racket and a ball (or two) into class plus ameter rule. Drop a tennis ball on the floor and let itbounce freely. Do this a number of times and ask theclass to count the number of bounces before the ball rollsacross the floor.

Explore part I:

Ask the class why the ball does not bounce back to the same height, i.e. why the height ofrebound decays. Explain that the ball loses a fraction of its mechanical energy with eachimpact with the floor.

Since the conservation of energy must apply, ask the class to identify what happens to thelost energy. Explain that it is dissipated in acoustical energy (they can hear the bounce)and heat. Those that have played squash will know that the ball warms up sufficientlyduring a game to feel the temperature difference.

Note that the sound of the bounce changes with successive bounces, with the impactsurface and with the height of the initial drop. Use the meter rule to drop it from differentheights in order to illustrate this effect.

Explain part I:

A ball of mass, mon released from a height, hhas a potential energy, PE = mgh; afterhitting the floor it bounces to height, dwhere it has a potential energy, PE = mgd. Thecoefficient of restitution, e= d/hand is equal to the fraction of mechanical energy lost.

This fraction is the same for successive bounces, i.e.

3423121 ddddddhde

where d1, d2, d3, d4are the heights of successive bounces.

The coefficient of restitution of a tennis ball bouncing on a concrete floor is about 0.71 soif the ball is dropped from 1m the height of successive bounces will be

d1= eh =0.711 = 0.71m and d2= 0.5m, d3= 0.36m, d4= 0.25m

Engage part II:

Place a second tennis ball on a bench and roll the first one into it with sufficientmomentum to cause them both to move after the impact.


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Explore part II:

Explain how conservation of energycauses kinetic energy from the moving ball, beforeimpact to be converted to strain energy of deformation for both balls whilst they are incontact and then restituted as kinetic energy in both balls after impact. The coefficient of

restitution can also be defined in terms of velocities, so for two balls (particles), A and B:

beforeBbeforeA

afterAafterB

vv

vve

Elaborate:

In a tennis machine (for an example goto www.youtube.comand type in tennis machine5)

tennis ball A rolls down a chute from a vertical height of 0.25m and impacts tennis ball Bat the bottom of the chute, we can calculate the velocity at the instant before impact byconservation of energy. The kinetic energy of the ball, A at the bottom of the chute

equals its potential energy at the top, i.e. ghmvm AbeforeAA 2

21 so

2.225.081.922 ghvbeforeA

m/s

Now considering the coefficient of restitution in terms of velocities, assuming ball B isstationary before the impact:

02.2

afterAafterB

beforeBbeforeA

afterAafterBvv

vv

vve so

afterAafterBvv 6.1

Now applying the principle of conservation of momentum:

afterBBafterAAbeforeBBbeforeAA

vmvmvmvm and mA=mB

so afterBafterAbeforeBbeforeA

vvvv

and afterAafterA

vv 6.102.2 or 3.02

6.12.2

afterAv m/s

and 9.13.06.122.2 afterAafterB

vv m/s

In other words the stationary ball moves off at 1.9m/s and the impacting ball at 0.3m/s.

Evaluate


Example 4.1

A 4 year-old boy weighing 15kg is sitting on the bottom edge of a waterslide (with hislegs dangling in the swimming pool) when his 7 year-old sister weighing 22kg comes

5http://www.youtube.com/watch?v=EWNq0t6ap-8


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down the slide from a height of 3m. Assuming the end of the slide is horizontal andneglecting friction, calculate the velocity of the boy after impact if (a) his sister grabs himand they move off together and (b) if they separate after contact with a coefficient ofrestitution of 0.8.

Solution:

(a) For the girl (subscript G), the kinetic energy before impact is equal to her potentialenergy at top of slide:

So ghmvm GbeforeGG 2

21

and 7.7381.922 ghvbeforeG

m/s

So in case (a) applying conservation of momentum using subscript B for the boy:

afterGGafterBBbeforeGGbeforeBB

vmvmvmvm

with afterafterGafterB vvv

afterv 22157.722015

and 56.42215

7.722

afterv m/s

i.e. they will shoot off the slide together at a velocity of 4.6m/s (10 mph) horizontally.

(b) for a coefficient of restitution of e = 0.8

067.7

afterGafterB

beforeBbeforeG

afterGafterBvv

vv

vve

and afterGafterB

vev 67.7

now applying conservation of momentum:

afterGGafterBBbeforeGGbeforeBB

vmvmvmvm

thus afterGafterG

vve 227.71567.722015

and

1.2

2215

8.07.71567.722

afterG

v m/s

so 3.81.22.67.7 afterGafterB vev m/s

i.e. the boy slides off the slide at 8 m/s (18 mph) and the girl follows at 2m/s (4 mph).

Example 4.2

Ask students to identify other events with which they are familiar that can be analyzed inthis way and then to select one and perform a similar analysis to the one above.


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5. Topic: Systems of Particles

Engage:

Bring a desk fan into class, place it on the table andswitch it on. It would be good to attach someribbons or strips of tissue paper to the protectivecage around the fan to illustrate the flow.

Explore:

Discuss the action of the fan blades on the air. The air on one side of the fan is

essentially at rest. As it passes over the blades its momentum is increased. The bladesmust exert a horizontal force on the air flow in order to generate the momentum increase.Packets or bundles of air can be considered and the principle of linear impulse andmomentum applied. Since the flow is steady, the applied force will be constant during atime interval.

Explain:

Explain the concept of a control volume as a box around the fan containing a mass ofair, mwith an average velocity, v;and a small mass of air, dmabout to enter the box witha velocity vAat time, tthen subsequently at time, (t + dt) a corresponding mass dmleaves

the other side of the box having been accelerated tovBby the fan.

The principle of linear impulse and momentum:

beforeafter

t

t

mvmvFdt

after

before

can be applied to the air stream in the box:

mvvdmmvvdmFdt AB

i.e. AB vvdt

dmF

the term dtdm is known as the mass flow. Continuity of mass requires:

BBBAAA AvAvdt

dm


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21

Elaborate:

Fans are often powered by shaded pole AC induction motors which have a very lowefficiency of the order of 20%; so a 50W fan will only deliver about 10W to the blades.Power, Pis the rate of work, or for a fan the rate of kinetic energy change delivered to the

airstream, i.e.

dt

vvm

dt

dKEP AB

222

1

And if the airstream is initially at rest, vB = 0 then utilizing continuity of mass:

AvPA

32

1 and rearranging 32

A

PvA

So, if the fan has a radius of 0.1m and the density of air is 1.2 kg/m3then the velocity of

the air leaving from the fan will be:

1.81.02.1

10223

23

A

PvA m/s

The force on the air is 47.21.81.02.1 222 AAB Avvvdt

dmF N

Now simple statics can be employed to calculate the minimum mass of the fan to preventit sliding along the desk:

Free-body diagram:

F

mg

N N

yx

F

mg

N N

yx

0 NFFx and 0 mgNFy

So mgF and 42.081.96.0

47.2

g

Fm

kg

Taking the coefficient of friction between the rubber base of the fan

and plastic coated table or floor to be= 0.6.

If the fan weighs less than 420g then it will move across the floorwhen it is switched on.

Evaluate:


Example 5.1

When the outlet nozzle of a 1875W hair-dryer is placed vertically above and just nottouching the stainless steel plate of a set of kitchen scales, the scales register 46g whenthe hair-dryer is switched on. If the nozzle is an ellipse with a major diameter of 75mmand minor diameter of 22mm and the fan has a diameter of 60mm, calculate thepercentage of the power used by the fan, neglecting losses.


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22

Solution

Nomenclature: subscripts A at the intake, B at the outlet and C just after impact withthe plate.

The force exerted by the hairdryer can be obtained using the principle of linear impulseand momentum, and by assuming the air comes to rest (in the direction perpendicular tothe plate) after impact with the plate, i.e. vC= 0

2BnozBC vAvvdt

dmF

and the area of an elliptical nozzle is given by:

31029.12

022.0

2

075.0 NJnoz rrA m

where rJand rNare the major and minor semi-axes lengths, so for air of density,

=1.2 kg/m

3

:

9.53

1029.12.1

81.946.03

noz

BA

Fv

m/s

For the fan, the power is the rate of delivery of kinetic energy, i.e.

dt

vvm

dt

KEdP AB

222

1

and if the airstream at the hairdryer intake is initially at rest, vA = 0 then utilizingcontinuity of mass:

2654

06.09.532.123

213

21 fanB AvP W

So the percentage of the power used by the fan is:

%141001875

265

The remaining 86% is used in the heating elements!

Example 5.2

Ask the students to consider the forces acting on a supporting structure suitable for asmall wind turbine capable of providing power for their home.


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23

KINEMATICS OF RIGID BODIES

6. Topic: Angular velocity & acceleration

Engage:

Take a yoyo into class and also an empty cable spool with a lengthof string wound on to it. You should be able to get cable spoolsfrom your Department of Electrical Engineering, alternatively askyour campus maintenance staff if their electrical team can collectthem for you.

Practice with the yoyo while the class assembles and settles down.

Explore:

Place the spool on the bench with the string partially unwound (orientated as shown inthe figure below) and ask the students to predict which way the spool will move whenyou pull the string and whether the string will be wound or unwound. Collect the optionsand take a class vote on it before pulling the string carefully so that the spool does notslip. This also works with the yoyo but is less easy to see in a large class.

O

ABC

O

ABC

Explain:

Draw the kinematic diagram:

The spool does not slip so, at contact between spool and table,

the velocity is zero, i.e. 0Cv

Velocity is constant all along the unwound string (assume the

string is pulled at 150 mm/s) so 15.0Bv m/s

(Angular velocity of spool, 31050

15.03

BO

B

r

v rad/s

i.e. spool moves in the direction it is being pulled!

O

BC

vB

100mm

150mm

O

BC

vB

100mm

150mm

Velocity of center of spool, 45.0310150 3 OCO rv m/s

So since BO vv the string will be wound up at a rate equal to the difference, i.e. 0.3 m/s.


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24

Elaborate:

Consider a yoyo falling under gravity when the unwindingof the string causes it to rotate giving it gyroscopicstability, i.e. it will resist forces attempting to move its axis

of rotation in the same way as a spinning top (see lesson#11). If the string is attached securely to the axis of theyoyo then, when it is fully unwound, the rotational energyin the yoyo will cause it to start winding back up the stringagain. The yoyoist will have to pull up on the stringslightly in order to replace the energy loss due to friction.In some designs of yoyo, the string is not firmly attached tothe axle, in order to allow the yoyo to spin freely or sleepwhen the string is full unwound, with these designs theyoyoist needs to give a tug to start it winding up again. Thetug has the same effect as pulling the string around the

spool.

O

A

B

mg

aOvO

O

A

B

mg

aOvO

The center of mass of a yoyo will fall at a constant acceleration due to gravity: 81.9Oa m/s

2so at the end of its 1.15m string it will have achieved a velocity given by:

asvv 2212

2 and 75.415.181.9222 asvO m/s

The diameter of the axle is typically 10mm so the rate of rotation is given by,

950105

75.43

BO

O

r

v rad/s

This ignores any losses due to friction or air resistance and also ignores the resistance to

rotation, i.e. inertia of the yoyo (see example 8.1).

Evaluate:

Ask the students to calculate the horizontal velocity of their foot when riding their bicycle

at an increasing velocity of 15mph ( 6.7m/s) using the middle gear of the rangeavailable, at the instant when the pedal is at the top of its rotation.

2rCO2rTO 2rPD

rPF

vO

vPvC

2rCO2rTO 2rPD

rPF

vO

vPvC


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25

Solution:

First each student will need to make some measurements on their bike. Typicalmeasurements are given below.

Radius of wheel including tire, rTO: 340 mm

Radius of middle gear at the wheel, rCO: 32 mm

Radius of middle gear at the pedals,rPD: 75 mm

Radius of pedal crank rPF: 175 mm

Consider rotational speed of rear wheel: 7.19340.0

7.6

TO

O

wheelr

v rad/s

Velocity of the chain and the outer radius of the gear at the rear wheel,

63.07.19032.0 wheelCOC rv m/s

The chain will have the same velocity at the pedal gear, so the rotational speed of the

pedals will be 4.8075.0

63.0

PD

Cpedal

r

v rad/s

And the velocity of the pedal will be 5.14.8175.0 pedalPFP rv m/s

So your foot would be moving forward with an instantaneous velocity of 1.5m/s (3.3mph) which is approximately walking speed!


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26


7. Topic: Forces & accelerations

Engage:

Ride your bike into class, apply the front brakes too hard so that the rear wheel justbegins to lift off the ground. Wear a helmet!

Explore:

Dismount from your bike and standing on the floor,demonstrate how applying the front brakes whengripping the handle-bars causes the rear wheel of the

bike to lift off the floor. Use a mountaineeringkarabiner or other similar clip to suspend a backpack ofbooks on the bike saddle. Invite members of the class,standing on the floor, to repeat the demonstration withdifferent numbers of books in the backpack and withbackpack suspended at different locations. Talk aboutthe moment of inertia as a measure of the resistance ofa body (the bike and backpack in this case) to angularacceleration (tendency for the rear wheel to lift in thiscase) in the same way that mass is a measure of bodysresistance to acceleration (i.e. Newtons Second Law).

Explain:

When the front brake is locked so that there is no motion of the front wheel relative to thebike, then the bike (including the front wheel) and backpack rotate about the point on thefront wheel that is in contact with the road, i.e.A

The moment of inertia of a body of mass, mabout an axis is: m

dmrI 2

G

1.2m

A

G

1.2m

A


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27

where r is the perpendicular distance from the axis to an arbitrary element dm. Themoment of inertia for a solid cuboid of heighth, width, wdepth dand mass mis:

2212

dwm

Ih

whereIhis the moment of inertia about the axis through the center of mass parallel to theheight dimension. So, for a pile of textbooks in a backpack of mass 3.8 kg and of

dimensions 0.120.210.26m:

035.026.021.012

8.3

12

2222 dwm

Ih kg.m2

When the moment of inertia, IG about the bodys center of mass is known then themoment of inertia,Iabout any other parallel axis is given by the parallel axis theorem:

2mbII G

where bis the perpendicular distance between the axes. So if we assume that the center ofmass of the backpack is at the location, G in the diagram above then, its moment ofinertia aboutA, IAis given by:

51.547.5035.0)2.18.3(035.0 22 mbII GA kg.m2

We neglected the moment of inertia of the backpack but this is inconsequential becausethe second term in the expression above dominates and location of the backpack and themass of the books has a huge effect on the resistance to rotational acceleration.

Elaborate:

The equation of rotational motion is given by: GG IM where the momentsMGare considered about an axis through the center of mass andisthe angular acceleration of the body.

When a child rolls down a steep grassy bank (roly-poly) we can calculate theiracceleration by using the above expression and a few simplifying assumptions. Let usassume that the child has uniform radius of 0.15m, a mass of 40kg and a moment of

inertia of 0.8kg.m2; that the grassy bank has a 45 degree slope (=45) and that we can

represent the child by a cylinder or imagine them in a plastic pipe.

Then, drawing a free body diagram:

mg

C

N F

r

C

ma

mg

C

N F

r

C

ma


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28

Considering motion about C, the equation of rotational motion is:

IrmamgrMc )(sinNow, the linear acceleration is, a= rand the radius of gyration, kis defined byI = mk2,

so 2)(sin mkrmrmgr

rearranging22

sin

kr

gr

and

67.3408.015.0

45sin81.915.0sin2

2

22

2

kr

gra

ms

-2

If the child starts from rest and the bank is 5m high then their velocity at the bottom canbe calculated from:

asuv 222

i.e. 1.6567.322 asv m/s

so the child will have a speed of 6.1 m/s (13.6 mph) at the bottom of the grassy bank.

Evaluate:

Ask students to attempt the following example:

Example 7.1

A pizza cutter is pushed through a thick crust pizza using a handle held such that thedriving force is at 45 degrees to the cutting board. The cutting wheel and handle have atotal mass of 500g. The blade is of radius 50mm with an average thickness 0.5mm andthe cylindrical handle is 180mm long with a radius of 15mm. The coefficient of frictionbetween the pizza and the blade is 0.01 and it can be assumed that the pizza exactlyopposes the magnitude of the driving force with a force acting towards the axis of thecutting wheel at 10mm above the cutting board. Calculate the constant driving forcerequired to cut through a pizza 30cm wide exiting at a velocity of 1.5m/s from a standingstart is this viable for a 12 year old child?

Solution:

Volume of pizza cutter wheel,

622

1093.30005.005.0

trV m3

So for stainless steel of density, = 8000 kg.m

-3,

the mass of the pizza cutter wheel is

031.01093.38000 6 Vm kg

and the moment of inertia of the wheel about itsaxis of rotation is given by:


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29

522

1093.32

05.0031.0

2

mr

I kg.m2

The mass of the handle is 0.469kg (= 0.5-0.031) and its moment of inertia about an axisparallel to the axis of rotation of the wheel and though its center of mass is

00129.018.0015.0312

469.03

12

2222 hrm

I kg.m2

And using the parallel axis theorem to shift this to be about the center of rotation of thecutter wheel:

0025.005.0469.000129.0 22 mbII G kg.m2

so the moment of inertia of the wheel is negligible compared to the heavy handle.

For an exit velocity of 1.5 m/s after crossing a pizza of diameter 0.3m and assuming anentry velocity of zero, using:

asuv 222 we have 75.36.0

5.1

2

222

s

uva ms

-2

and for the cutter wheel ra so 7505.0

75.3

r

a rad/s

2

Drawing a free body diagram for the cutter wheel:

For motion about C, consider the equation of rotational motion:

IrmIrmaFFPPDM ppc 2)(cos3.0sin1.0sin3.0cos1.0cos5.0now,D = P = (Fp/) so re-arranging we can obtain:

)cos3.0sin1.0()sin3.0cos1.0(cos5.0

2

ImrD

and substituting values for the parameters, to obtain:

1.3

09.0

281.0

)50

12

50

3(01.0)

50

9

50

4(

2

1

2

1

0025.005.05.075 2

D N

So, the driving force required is 3.1N which is very small.

mg

CP

FP

0.5

C

ma

N

FN

D

0.4

0.1

mg

CP

FP

0.5

C

ma

N

FN

D

0.4

0.1


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8. Topic: Work and energy

Engage:

Take a bag of marbles into class. Empty it ontothe bench so that marbles scatter everywhereand fall off the bench. Invite students to collectthem and put them on their tables.

Explore:

Ask the students, working in pairs, to identify the types of energy possessed by a marble

as it rolls along the bench, falls off the end and hits the floor. Invite some pairs to talkthrough to the class their understanding of the energy conversions. Remind them about

translational kinetic energy ( 22mvKE ) and tell them about the concept of rotational

kinetic energy ( 22oIKE ) about center of mass of a marble.

Explain:

Consider the moment of inertia about the center of mass for a marble of mass 5g & radius15mm:

722

105.4015.0005.05

2

5

2

mrI kgm2

The kinetic energy as it rolls along the bench at 0.5 m/s:

2222222

10

7

5

2

2

1

2

1

2

1 mrmrmrImvKE

where rv so 3.331015

5.03

r

v rad/s

thus, the kinetic energy of a rolling marble is given by:

42222

1075.83.33015.0005.010

7

10

7 mrKE J

Elaborate:

Now consider a marble starting from a stationary position on the edge of the bench androlling over the edge.

Applying conservation of energy between two positions: 2211 PEKEPEKE


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31

Position 1: with the center of the marble vertically above the edge (corner) of bench.

Position 2: when the marble just ceases to make contact with the bench, i.e. it has rolled

over the edge to an angle.

Rearranging the conservation of energy expression, so that the gain in kinetic energy

equals loss of potential energy, i.e. 1221 PEKE and substituting:

)cos1(10

7 21

2

2

2 mgrmr

and the marble is initially stationary so 01 andr

g

7

)cos1(1022

Free body diagram at point of losing contact, i.e. position 2:

So resolving forces normal to the contact force at the corner, noting that when contact islost, 0 NFN

0nF so 0cos nmamg and nmamg cos

since 22

2

2 rr

va n then

2

2cos rg andr

g

cos22

so equating with expression above from conservation of energy:

r

g

r

g

7

)cos1(10cos and

17

10cos i.e. 54

The marble loses contact with the bench when it has rolled over by 54 degrees and at thisinstant it has a velocity:

294.017

10015.081.9cos22 grrv m/s

at 54to the horizontal so that it will land at some distance from the bench, depending onthe height of the bench.

mg

0.12

N

FN

r sin

n

man

mat

r(1-cos)

mg

0.12

N

FN

r sin

n

man

mat

r(1-cos)


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32

Evaluate:


Example 8.1

A yoyo of diameter 6cm and mass 50g has an axle of diameter 10mm around which astring of length 1.15m is wound. Use the principle of conservation of energy to calculateits angular velocity of the yoyo when it is allowed to fall and unwind under gravity.

Solution

The moment of inertia of the yoyo is: 522

1025.22

03.005.0

2

mR

I kg.m2

Conservation of energy, i.e. from position 1 (start of fall) to position 2 (fully unwound):

2211 PEKEPEKE

222

221

21

21

21

21

21

21 mghImvmghImv

or 2222

221

2

1

2

1 Imrhhmg

so

1051025.205.005.0

15.181.905.022522

21

Imr

hhmg rad/s

which is about 17 revolutions per second.

Example 8.2

A broken down car of mass 1500kg is being pushed up a small incline by a student whocan exert a constant force of 600N. After pushing for 5m, the student is joined by asecond student who exerts a constant force or 900N and after another 2m a third studentjoins them and exerts a constant force of 700N but the second student slips and stopspushing only a 1m after the third joins and 2m later the car crests the incline andeveryone stops pushing. If the height gained while the car was being pushed was 0.7m,calculate the velocity of the car as it crests the hill.

Solution

Conservation of energy between when the first student starts to push with the car at rest(v1= 0) and when everyone stops pushing as the car crests the hill:

1212 PEKE Work Done

i.e. 22

112

22

Fdmghmv or

m

mghFdv

122

46.2

1500

7.081.91500307003900106002

v m/s

The car will crest the hill at 2.5m/s or 5.5 mph).


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33


9. Topic: Impulse and momentum methods

Engage:

Take your bike, a basketball and a short block of wood of

approximate cross-section 9cm 2 cm.

You will need some assistance from members of the class.Place the wood block on the floor with a student standingon each end of it. On your bike freewheel very gentlytowards the block of wood. Repeat the experiment untilyour front wheel just rolls over the wood block.

Explore:

Discuss the impulse received by your front wheel when it hits the wooden block; notethat the impulse force is of unknown magnitude and direction. Highlight that angularmomentum just before and just after impact must be conserved. Discuss the conservationof energy during the event and that kinetic energy at the impact must at least equal thepotential energy when the wheel is on top of the wooden block in order for the wheel toroll over the block.

Actually this example is easier if you consider a basketball rolling over the block ofwood, as in the photo above because then the effect of the rear wheel and the inertia of

the rider does not have to be handled. However, it is less engaging and of less practicalinterest, so explain this to the students and repeat the demonstration for the basketball.Ask them to draw the free-body diagrams for a basketball just before and just afterimpact with a wooden block.

Explain:

Free body diagrams:

+

Fdt

A

m(vA )1A A

mgt

B

d

A

m(vA )2

A

B

=

Momentum just before impact

Impulses during impact

Momentum just after impact

r

B b

+

Fdt

A

m(vA )1A A

mgt

B

d

A

m(vA )2

A

B

=

Momentum just before impact

Impulses during impact

Momentum just after impact

r

B b


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34

Lets assume that the basketball does not slip on the wooden block or rebound from it butessentially rolls over it by rotating about the corner B. Then there are two impulses orforces acting on the basketball: (a) its weight and (b) the impulse from the wooden blockat B. Since the duration of impact is very small it is reasonable to assume that the weight

generates a negligible impulse compared to the force at the wooden block. We do notknow the magnitude or direction of the force or impulse at the wooden block.

Elaborate:

Conservation of angular momentum about B from just before (1) to just after (2):

[System Momentum]1+ [External Impulses]12= [System Momentum]2

2211 AAAA IrvmIbrvm

For simplicity, consider the basketball as a hollow sphere with mass, m = 0.6kg andradius, r=0.012m:

3

2 2mrIA and

r

v ,

so

3

2

3

22

2

1

1

a

A

a

A

vrrv

vrbrv

Re-arranging*:

r

bvv

AA5

31

12

Conservation of energy from just after impact (2) to when the basketball is on top of the

wooden black (3), assuming the basketball can just mount the baton, i.e. 03Av :

3322 PEKEPEKE

`

mgbIvm A

00

22

2

2

2

2

Substituting forIand rv : 5

622

mgbvA

Thus combining with conservation of angular momentum*:

rb

vmgb

A321

56 1

hence 422.0

12.03

02.021

502.081.96.06

3

21

56

1

r

b

mgb

vA

m/s

Note that as the radius of the rolling object gets larger the required velocity becomessmaller, so it easier to get a larger diameter wheel up a curb than a small one.


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Evaluate:


Example 9.1

A girl of mass 50 kg is setting off along a level sidewalk on her skateboard. If she canpush off with one foot with a constant force of 150N for 0.5 seconds, calculate her speedafter this push. If, almost immediately she can push again for another 0.5 seconds, thencalculate her speed after this second push assuming the force applied is sinusoidal with amaximum of 150N.

Solution:

Change in Momentum = Linear Impulse

or 2

1

12

t

t

Fdtmvmv

so, in the first push with a constant force:

5.150

5.01502

m

Ftv m/s (3.36 mph)

and in the second push:

50

5.150sin1501

5.02

dtt

v

so

5.250

755.0cos1cos1502

v m/s (5.6mph)

Example 9.2

Calculate the topspin (in terms of revolutions per second) required to make a tennis ball,

that strikes a grass surface at 80 mph and at 25to the horizontal, bounce at an angle of15, if it has a diameter of 66mm and mass of 57g.

Solution

+

m(vA)1

m(vA)2

Fdt

Ndt

=

Momentum just before impact Momentum just after impact

External impulse during impact

+

m(vA)1

m(vA)2

Fdt

Ndt

=

Momentum just before impact Momentum just after impact

External impulse during impact


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36

Conservation of angular momentum about B:

2211

15cos025cos AAAA IvmIvm

assume a perfectly elastic impact, so

21 AA vv and 21 AA

and that the tennis ball is a hollow sphere, i.e.3

2 2mrI note that 80mph 35.8m/s

then 25cos15cos3

411

2

AA vmrmr

and

5.48033.04

25cos15cos8.353

4

25cos15cos31

1

r

vAA rads/s

or about 463 rpm!


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37


10. Topic: Kinematics of rigid bodies in three-dimensions

Engage:

If you play the violin, then bring it to classand play an excerpt from your favorite piece;or if you dont play, maybe you could ask astudent in your class or a colleague in themusic school. And, or search inwww.YouTube.com for Janine JansenMozart and select Mozart Violin Concerto 5(2of 5) for a good close-up of the violin beingplayed6. Play it a second time and ask thestudents play along on the air violin.

6http://www.youtube.com/watch?v=5QlyMNMLlfI

Explore:

Again in www.YouTube.com search under Robot violinistand show the video of the ToyotaViolin playing robot7.Discuss the need to calculate displacements, rotations,velocities, accelerations, forces and moments in order to be

able to program the robot to perform such a complex task. Itwould be relevant to introduce examples of engineeringapplications of robots, e.g. welding robots

8; again search in

www.YouTube.com for suitable videos (also see Robot ofthe Year 2007

9).

Ask students to explore in pairs the degrees of freedom andaxes of rotation in their own arms. Ask them to identify thedegrees of freedom used in playing the air violin.

7http://www.youtube.com/watch?v=EzjkBwZtxp48http://www.youtube.com/watch?v=362vMN7Ra4w9

http://www.youtube.com/watch?v=Q8WnAN9jmEc

Explain

You have seven degrees of freedom in your arm: at the shoulder, elbow and wrist, asshown in the picture. Use your own arm, or ask a student to help you. If you havebrought a violin to class then talk about the motions involved in moving the bow to theinstrument.


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38

Elaborate

A simple robotic arm for a violin playing robot is shown in the figure. To engage thebow with the instrument, the arm will rotate about the vertical axis shown with the elbowat E fixed. To avoid damage to the bow and strings we need to calculate the angular

velocity and acceleration of the bow.

We can establish a coordinate system which is fixed relative to the arm with thex-axis inthe plane SEW and they-axis co-incident with the axis of rotation, i.e with an origin at O.

So the angular velocity of the arm, and of the coordinate system, are equal andj0arm . Thus, we can consider the position vector of W relative to O:

jir )sinsin(coscos/ ufufOW

now, BABA /rvv

so

0sinsincoscos

000 0/

ufuf

OWarmOW

kji

rvv

and kv coscos0 ufW

arm

b

f

u

P

S

E W

x

y

O

arm

b

f

u

P

S

E W

x

y

O

arm

S

E

O

relrel

W

P

arm

S

E

O

relrel

W

P


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39

The coordinate system is fixed with respect to the arm so the angular velocity of the bowwith respect to the coordinate system is:

ji sincos relrelrel

and the position of P, halfway along the bow, relative to W is: jir )cos(sin/ bbWP

so, the velocity of the point, P is given by:

0cos2

sin2

0sincoscoscos 00/

bb

uf relrelwPbowWP

kji

krvv

kv

sin

2sincos

2coscoscos 00

bbuf

relrelP

For a bow of length 75cm held by an arm with upper and forearm lengths of 25cm and

30cm respectively at angles of= 15and = 20, when the arm rotates at 0.75 rad/s(=o) and the wrist rotates the bow at 0.5rad/s (=rel) the velocity of point, P is given by:

15sin375.020sin5.075.020cos375.020cos5.015cos25.020cos3.075.0 Pv 99.0482.0117.0393.0 Pv m/s

The acceleration of the bow, is given by:

kji dt

d

dt

d

dt

dzyx

though the components ofare constant, the acceleration is not zero but

0sincos

00

relarmrel

armbowbow

kji

cosrelarm k

35.020cos5.075.0 rad/s about thez-axis.


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40

Evaluate

Example 10.1

An industrial robot uses its arm in the positionshown to roll out adhesive tape on the underside ofa structure as shown in the figure. If the arm CBrotates about the vertical axis through C at aconstant angular velocity of 3 rad/s find the velocity

at which the tape is being laid out when= 35.

Solution

Establish a coordinate system that is fixed

relative to the arm with the x-axis coincidentwith CB and they-axis co-incident with the axisof rotation, i.e with an origin at C. Then

j0arm

and we can consider the position vector of Arelative to C:

jir )sin(cos/ cbcCA

now

0sincos

000 0/

cbc

CAarmCA

kji

rvv

and kv cos0 bcA

The coordinate system is fixed with respect to the arm so the angular velocity of the discwith respect to the coordinate system is:

ji sincos relrelrel

For the point of contact, P there is no motion due to the adhesive on the tape so 0Pv .Now, the position of P relative to A is:

jir )cos2

(sin2

/ dd

AP

So, the velocity of the point, P is given by:

d=0.3mb=0.2

m

P

BC

A

c=0.18m

d=0.3mb=0.2

m

P

BC

A

c=0.18m

db

P

B

C

A

cx

y

relrel

db

P

B

C

A

cx

y

relrel


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41

0

0cos

2

sin

2

0sincoscos 00/

dd

bcrelrelAPdiskAP

kji

krvv

hence,

sincos22

0d

b

d

crel

and ji sincos 0 relrelreldisk

finally, the tape is used at

0cos2

sin2

0sincos 0/

dd

relrelAPdiscT

kji

rv

03.135cos2.018.03cos0 kv bcT m/s

So the tape is used at approximately 1 meter per second.

Example 10.2

What degrees of freedom are involved in cleaning your teeth using a tooth brush?Calculate the approximate velocities of the parts of your arm during a brushing stroke.

________________________________________________________________________

Note:

The determinant of a 3x3 matrix,

ihg

fed

cba

A

is given by cegcdhbdibfgafhaei Adet

and can remember using the diagonals as follows:

ihg

fed

cba

ihg

fed

cba+ + + - - -

ihg

fed

cba

ihg

fed

cba+ + + - - -


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42


11. Topic: Kinetics of rigid bodies in three-dimensions

Engage:

Take a childs spinning top into class. In fact wood ones arecheap enough for you to take a handful into class so that you canpass them around in case the students have forgotten what theyare like. Encourage the students to spin the tops. Also take thefront wheel of your bicycle.

Explore:

Spin your top sufficiently fast that its spin axis remains vertical this is known assleeping. Of course, friction will reduce the spin rate, , so that the top starts to lean

over which is known as nutation, and the axis about which it is spinning will rotate aboutthe vertical axis which is known as precession.

In practice, the spin rate of the top will continue to decrease due to friction but in steadyprecession it is assumed that the rate of spin is constant.

Explain:

Discuss the concept of two co-ordinate systems: one

fixed to the bench,XYZ; and one fixed to the top,xyz; but with a common origin at the point of thetop.

Re-iterate that three angles are required to definethe three-dimensional motion of a rigid bodyrelative to an axis: one angle to define the rotationof the body about the axis and two angles to definethe orientation of the axis. We just defined these

for the top, i.e. the spin angle,, plus the nutation,

and precession angles, . These are known asEulers angles and the equations of motions can be

expressed in terms of them, i.e.

sincossin2 zzxxzzxxx IIIIM

coscos2sin zzxxy IIM

sincos zzz IM

X

Y

Z

x

y

z

mg

h.

X

Y

Z

x

y

z

mg

h.

.


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43

Generally these equations have to be solved numerically; however when the precessionrate is constant these equations reduce to:

sincossin2 zzxxzzx IIIM , 0yM and 0zM .

Now, the mass of the top exerts a moment about the origin of: sinmghMx and hence,

sincossinsin 2 zzxxzz IIImgh

or

sin

cossinsin 2

zz

xxzz

I

IImgh

So, if you estimate the nutation (lean),and the precession rate, which is usually

fairly slow then you can calculate the spinning rate, .

For the cone in the diagram above, if it is of radius and length 6cm and made of teak(density,= 700 kg/m3) then:

0095.03

06.0700

3

43

hR

Vm kg

and 52222

1057.220

06.03

5

06.030095.0

20

3

5

3

RhmII yyxx kg.m

2

522

1003.110

06.00095.03

10

3

mR

Izz kg.m2

Now, estimating the nutation to be about 15and the precession rate to be about tenrevolution per minute (= 1.05 rads/s):

712

15sin05.1101

15cos15sin05.11056.203.115sin06.081.90095.05

25

34

rad/s

Or about 113 revolutions/second which 6800 r.p.m.


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44

Elaborate:

Engage the students by spinning your bicyclewheel and holding by the axle stubs in front ofyou so the axle is horizontal. Tilt the axle byraising one end of the axle and lower the otherend so that you experience gyroscopic momentsrestoring the axle to its original position. Explainto the students about the forces you areexperiencing. Perhaps have a student come to thefront of the class and hold the wheel while youspin it for them.

The angular velocity of the wheel is the vector sum of rate of spin about the local z-axis,

;nutation about the global Y-axis, and precession about the global Z-axis K j k

wherei,j, andk are unit vectors along the local axes andK is the unit vector along thefixed Z axis, such that sinK i cos k

hence, the angular velocity is given by: sin i j cos k

The local axes of rotation,xyzare also the axes of inertia, so the angular momentum,Howill be the sum of the moments of inertia multiplied by the appropriate component of theangular velocity, i.e.

sin0 IH i

I j cos I kwhereIand I are the moments of inertia about the spin axis and fixed axes respectively.Newtons second law of motion in terms of moments is:

OO HM and it can be shown that:

OxyzOO HHM

where xyzO

H is the rate of change of HOwith respect to rotating axesxyzand is the

angular velocity of the rotating axesxyz. Now, these axes nutate and precess but do notspin thus,

sin i cos k

Z

Y

X

z

x y

zk

isin

Z

Y

X

z

x y

zk

isin isin


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45

Now consider the case where, the nutation,, rate of

precession, and rate of spin, are all constant then

the angular momentum, HOis also constant, i.e.

0xyzO

H then

OO HM

sincoscos IIO M jThis is the couple that must be applied to maintain the

steady precession, i.e. 0 assuming the thatcenter of gravity of the wheel is fixed in space (F=0).

Now, for the situation in the demonstration the precision

and spin axes are at right angles, i.e. = 90so

IO M jWhen this is applied about an axis perpendicular to theaxis of spin the wheel will precess about an axisperpendicular to both the axis of spin and the couple, i.e.the globalZ-axis.

This behavior makes it easier to balance on your bicycleat higher speeds due to the stabilizing effects of thespinning wheels. It is also known as gyroscope motion.

Evaluate:


Example 11.1

The Earth can be thought of as a spinning top with a spinaxis through the North and South Poles which slowlyrotates, or precesses so that the North Pole draws out acircle in space. This precession is very slow (one degreeevery 71.6 years) and is known as the Precession of theEquinoxes. It was probably observed first by Hipparchus

in around 137BC and results in a slow change in theposition of the sun with respect to the stars at an equinox.This is important for calendars and their leap year rules.

The Precession of the Equinoxes is caused by a (force) couple acting on the earth due tothe gravitational attraction of the sun and moon. Assuming that the earth is an oblate

spheroid of average radius, 3960 miles and mass 5.97421024kg, and that the earthsnutation is constant at 23.45, calculate the couple.

N

MO

O

S

N

MO

O

S

Y

Z

X

z

x

K

k

MO

y

Y

Z

X

z

x

K

k

MO

y

Z

y

x

z

MO

K

k

Z

y

x

z

MO

KK

kk


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46

Solution:

Rate of precession, 121073.7360024365

1

360

2

6.71

1

rad/s

Rate of spin, 51027.7360024

2 rad/s

Assume a moment of inertia for a sphere,

37

2

24

1059.95

5

80003960109742.52

5

2

smRI kg.m2

And sincoscos IIOO HM j

sincos IIIO

M j

assuming II then:

sinIOM 45.23sin1073.71027.71059.9 12537

and finally:

221014.2 OM Nm

Example 11.2

Ask the students to design a gyroscope stabilizer for an experienced rider on a

motorcycle.


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47

MECHANICAL VIBRATIONS

12. Topic: Free and forced vibrations

Engage:

Take a hula hoop and a wooden rulerinto class. Place the ruler overhangingthe bench, lean on the end on the benchand flick the free end so that the rulervibrates. Repeat this a few times andslide the ruler onto the bench, as itvibrates, so that the pitch of the noisechanges the frequency will go up.

Suggest that the students do the experiment for themselves.This is noisy experiment so good for engaging the students but the mathematics is quitehard so tell them this and put it to one side. Pick up your hula hoop and, if you can, dosome hooping (see www.hooping.orgfor instructions on hula hooping and links on howto make your own).

Explore:

Allow the hoop to oscillate on your finger, in simple harmonicmotion, and discuss the characteristics of the motion.

The moment of inertia of a thin ring or hoop about an axisthrough its center is 2mrIC

and using the parallel axis system the moment of inertia about a

point,Eon the hoop is 22 2mrmrII CE

;

The kinetic energy of a hoop oscillating on your finger is the sum of its translational(=zero) and rotational kinetic energy:

2222

22

mr

ImvKE nEE

The potential energy is the sum of the elastic (=zero) and gravitational potential energies.For small oscillations the center of gravity of the hoop is raised by cos1r and for

small angles we can use the series expansion of 21cos 2 so:

2

2mgrmghPE

Now, the total energy of the system is constant:


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48

2

222 mgrmrPEKE = constant

And its derivative yields the equation of motion of the hoop, i.e.

022

mgrmr

or 02 grmr

and is not always zero, hence:

02

r

g

This is the equation of motion for the undamped free vibration of a simple block andspring system which could be used to represent the oscillating hoop. It is known assimple harmonic motion in which the angular acceleration is proportional to the angulardisplacement of the hoop. The natural frequency of oscillation is

r

gfn

22

1

If you oscillate the hoop on your finger at this frequency you will generate the largest

oscillations. For a 1m diameter hoop this frequency is about Hz 281.9 .The solution to the equation of motion is:

tr

gBt

r

gAtx

2cos

2sin

whereAandBmust determined from known boundary conditions. The amplitude of the

oscillation is 22 BA .

Explain:

Now, return to the ruler and repeat theanalysis. The kinetic energy stored inthe ruler is

"2

"2mv

KE

and for an element this is

2

2dmyKEd

So to obtain the total strain energy in theruler we need to integrate over its length,i.e.

dm

d

R

dx

y

d

M+dM

M dm

d

R

dx

y

d

M+dM

M


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49

l

dmyKE0

2

2

1

And in one period, any point on the ruler moves through y so

yy

y

and

l

dxym

KE0

22

2

where mis the mass per unit length. Now considering the strain energy stored in terms ofthe bending moment,Mand the curvature of the beam:

MdU 21

and for small deflections:

dx

dy and

2

21

dx

yd

dx

d

R

also from the theory of beams:

EI

M

R

1

thus

l

dxdx

ydEIMdU

0

2

2

2

2

1

2

1

Neglecting the gravitational potential energy, the total energy of the ruler is:

ll

dxdx

ydEIdxy

mUKE

0

2

2

2

0

22

2

1

2

=constant

Elaborate:

from Mechanics of Solids, the deflection of a cantilever can be described as:

323

36 l

x

l

x

EI

Ply

so the total energy can be demonstrated to be:

3

22

3

2

3

140

33

23 l

EIml

EI

Pl = constant


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50

At the natural frequency we can equate the energies, so:

3

2

2

3

140

33

2 l

EIml

and

411

140

ml

EI

Thus as the ruler gets shorter the frequency goes up. For a wooden mahogany ruler(E=9,200 MPa) of cross-section 28mm3mm and mass per unit length of 0.0459 kg/m(=1410-3/0.305):

1133

103.612

003.0028.0

12

bh

I m4

At l=22cm 63.26122.00459.011

103.6102.9140

11

1404

119

4

ml

EI Hz i.e. middle C.

Whereas a plastic ruler (polycarbonate, E = 2.2 GPa) of cross-section 452mm(I =310-11 m4) and mass per unit length 0.1076 kg/m requires a length of 10.3cm toproduce middle C.

Evaluate:

Ask students to complete the following examples.

Example 12.1

Find the natural frequency and the equation of motion for a ball of mass 12g stuck on theend of a whip antenna of length 0.7m.

Solution

For the ball, apply Newtons Second Law in the direction ofmotion, i.e. tangential to the support,F from the aerial:

tt maF so tmamg sin

using kinematics, xdt

xdat 2

2

and from the geometry, for small angles,

lx

so we can rewrite the equation of motion as

mlmg sin or 0 l

g

assuming for small angles sin and so the naturalfrequency is

74.37.0

81.9

l

gn Hz

l

s

l

s

mg

F

at

an

mg

F

at

an


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Example 12.2

The natural period of a building or structure can be lengthened to alleviate the highestearthquake forces using isolating bearings that consist of a ball carrying the structuralweight and rolling on a concave surface (e.g. www.earthquakeprotection.com). For abridge, sixteen such bearings are to be used in which the high strength steel ball has aradius of 10cm and the concave bearing surface a radius of curvature of 1.5m, find thenatural frequency of a bearing.

If the bearing is shaken by an earthquake with a seismic frequency of 40Hz that producessinusoidal motion of the ground of 150mm what will be the form of the equation ofmotion of the bearing.

Solution:

Kinetic energy at bottom of bearing:

22

22 ImvKE

andrR

r

rR

v

also

2

2mrI

so substituting for v, andI:

4

3 22rRm

KE

The potential energy gained when the ball is displaced from bottom of the bearing is:

cos1 rRmgmghPE

now 22

sin2cos12

2 for small angles, so 2

2rRmgPE

Thus the energy equation is:

24

3 222

rRgm

rRmPEKE

And differentiating with respect to time:

0

2

32

rRgrR

or 023 grRrR

Since is not always zero then: 03

2

rRg

And the natural frequency is:

16.21.05.13

81.92

3

2

rR

gn Hz

The earthquake will generate a forcing term of the form tF sin0 so that the equation of

motion will be

tFrR

g 80sin

3

20

.

R

R-r (R-r)cos

h

r

R

R

R-r (R-r)cos

h

r

R


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Junio r Dyna m ic s C ourse : Sug g e ste d e xem p la rs w ithin lesso n p la ns

NOTES FOR INSTRUCTORS ON EXAMPLE APPLICATIONS

Prepared as part of the NSF-supported project ( #0431756) entitled:

Edited by Eann A Patterson, University of Liverpool

Population through Curriculum Change

Examples in Dynamics.pdf

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