Real Life
Examples in
Dynamics
Lesson plans and solutions
Eann A Patterson, Editor
Real Life Examples in Dynamics
Lesson plans and solutions
First edition July 2009
Second [electronic] edition, 2011
Real Life Examples in Dynamics
Lesson plans and solutions
Copyright © 2009, 2011 Eann A Patterson (editor)
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(www.EngageEngineering.org), which is supported by the National Science Foundation
under Grant No. 0833076.
Real Life
Examples in
Dynamics
Lesson plans and solutions
Suggested exemplars within lesson plans for Junior 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
eann.patterson@liverpool.ac.uk
This work was developed during the NSF-supported project (#0431756) entitled:
“Enhancing Diversity in the Undergraduate Mechanical Engineering Population through
Curriculum Change”. Any opinions, findings, and conclusions or recommendations
expressed in this material are those of the authors and do not necessarily reflect the views
of the National Science Foundation. The following members of the project contributed to
the many discussions which led to this production of this volume:
Ilene Busch-Vishniac
McMaster University
(Project leader)
Patricia B. Campbell
Campbell-Kibler Associates Inc
Darrell Guillaume
California State University, Los Angeles
Eann Patterson
Michigan State University
Real Life Examples in Dynamics
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
3
INTRODUCTION
These real life examples and supporting materials are designed to enhance the teaching of
a junior level course in dynamics, increase the accessibility of the principles, and raise the
appeal of the subject to students from diverse backgrounds. The examples have been
embedded 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 the
Biological Sciences Curriculum Study1 in the 1980s from work by Atkin and Karplus2 in
1962. Today this approach is considered to form part of the constructivist learning theory
and 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 notes
and examples interesting, stimulating and hopefully illuminating, particularly when their
instructor 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 in
textbooks which this booklet is not intended to replace but rather to supplement and
enhance. Similarly, it is anticipated that these lessons plans can be used to generate
lectures/lessons that supplement those covering the fundamentals of each topic.
It is assumed that students have acquired a knowledge and understanding of topics
usually found in a Sophomore level course in Statics, including free-body diagrams and
efficiency.
This is the second in a series of booklets. The first in the series entitled ‘Real Life
Examples 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.
1
Engleman, Laura (ed.), The BSCS Story: A History of the Biological Sciences Curriculum Study. Colorado
Springs: BSCS, 2001.
2
Atkin, J. M. and Karplus, R. (1962). Discovery or invention? Science Teacher 29(5): 45.
3
e.g. Trowbridge, L.W., and Bybee, R.W., Becoming a secondary school science teacher. Merrill Pub. Co.
Inc., 1990.
Real Life Examples in Dynamics
KINEMATICS OF PARTICLES
1.
Topic: Rectilinear and curvilinear motion
Engage:
Stop by the office recycling box on your way to
class and pick-up enough sheets of paper to give
one to each member of the class and have some
left over. Pass them around the around the class
and invite the students to write on their sheet
three reasons why they chose to study
engineering; then ask them to throw it to a
member 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 a
paper airplane. If no one else does, you should demonstrate what happens if you try to
throw it as a sheet of paper.
Explore:
Discuss what is meant by the term ‘kinematics’, i.e. cases in which only the geometric
aspects of motion are considered and that a particle has mass but negligible size and
shape (no air resistance). For the sheets of paper to be considered particles then their
dimensions 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 be
neglected. So, the flat sheets don’t fly well due to the effect of air resistance on their
shape but the tight balls behave as particles and go where you aim them. The trajectory
of the paper airplanes is strongly dependent on their shape.
Explain:
In a heavy rainstorm large droplets fall from low
Nimbostratus cloud at 600 ft (183m). Let’s
assume that raindrops are approximately
spherical and of diameter, d3mm, so given the
density of water is 1000 kg/m3, their mass (m)
will be:
m   H 2 OV   H 2 O
6
4 r 3
3
 1000 
4  0.0015 3
 1.413  10 5 kg ( 0.014g)
3
Real Life Examples in Dynamics
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 raindrops
fall 183m we find using:
v y2183  v y20  2a y ( s y 183  s y 0 )
that
v y 183  2 gs y 183  2  9.81  183  3590 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. kinetics and
calculate a terminal velocity, vt which is approached when the drag force, Fd due to air
resistance equals the gravitational force, Fg (= mg), i.e. Fd  Fg
where Fd 
1
 air v 2 AC d
2
and the density of air, air 1.2 kg/m3, A is the frontal area and Cd is the drag coefficient
( 0.4 for a rough sphere).
Thus mg 
1
 air v 2 AC
2
d
and v yt 


2mg
2  1.413  10 5  9.81

= 9 m/s
2
 air ACd
1.2   4  3  10 3  0.4


Elaborate:
The horizontal dispersion of the rain droplets can be estimated by considering the
horizontal component of the motion after calculating, in the vertical direction, the time
from release to impact with the ground, i.e. s  vt  12 at 2 but at terminal velocity the
acceleration is zero,
so
s y  v yt t and t 
sy
v yt

183
 20.3 s
9
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 Beaufort
scale) then
s x  v x 0 t  12 a x t 2  22  20.3  447 m
i.e. it travels further horizontally than it falls vertically which is what we often see in a
violent 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.
s x  v x 0 t  12 a x t 2  4.5  20.3  91.4 m

It is assumed that the upwards force of buoyancy is negligible.
7
Real Life Examples in Dynamics
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 particles
leave her nostrils horizontally with a velocity of 40m/s ( 90mph), how far away from the
individual will the particles land?
Solution:
Time to fall vertically: s y  v y 0 t  12 a y t 2 so t 
2s y
ay

2  1.7
 0.589 s given vy=0 = 0
9.81
Distance travelled horizontally: s  v x 0 t  12 a x t 2  40  0.589  0  23.5 m
Example 1.2
While riding your bike across a pedestrian bridge at about 50km/hr (30mph) your iPod
drops out of your pocket and, missing the bridge, falls to the ground 6m below. Estimate
the velocity at which your iPod will hit the ground below the bridge and the horizontal
distance that it will travel before impact. Comment on the factors influencing whether
your iPod will survive the fall.
Solution:
Time to fall vertically: s y  v y 0 t  12 a y t 2 so t 
2s y
ay

26
 1.11 s given vy=0 = 0
9.81
Distance travelled horizontally:
 50  1000

s  v x 0 t  12 a x t 2  
 1.11  0  13.89  1.11  15.4 m
 60  60

Vertical component of velocity on impact:
v y  v y 0  a y t so v y  a y t  9.81  1.11  10.9 m/s
Magnitude of velocity on impact, v  v x2  v y2  13.9 2  10.8 2  17.7 m/s
Direction of impact,   tan 1
10.9
 38 
13.9
So your iPod will hit the ground at 18m/s (40mph) at an angle of 38 to the ground.
8
Real Life Examples in Dynamics
KINETICS OF PARTICLES
2.
Topic: Force and acceleration
Engage:
Take an ice hockey stick and puck into
class and push the puck around on the
floor. If you know nothing about ice
hockey shots then try typing ‘ice hockey
snapshot’ in www.youtube.com for some
short coaching videos4. Alternatively use
a dustpan and broom with a jar lid (e.g. a
jam jar lid). Sweep the jar lid into the
dustpan.
photo by Håkan Dahlström on Flickr
4
http://www.youtube.com/watch?v=TYEE7tZhRtk
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 answer
for 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 Newton’s laws of motion as the situation
becomes progressively more complicated:
Puck resting on ice (just before being push)
Fg=mg
y
x
N=Fg
Newton’s Third Law: for every action, there is an equal and opposite reaction. So the
mass of the puck acts on the floor and there is an equal and opposite reaction from the
floor, N.
9
Real Life Examples in Dynamics
Puck being pushed across the floor by stick
ax
S
Fg=mg
y
N=Fg
x
Ff=uN
Newton’s Second Law: the acceleration of an object is proportional to the force acting
on it and inversely proportional to its mass. So when a force, S is applied with the stick,
the puck will accelerate in the direction of this force, such that:
 Fx  ma x
There will be also a frictional force resisting motion at the interface between the floor and
the bottom of the puck:
Ff = N
where  is the coefficient of friction.
Puck sliding across the ice on its own
ax
Fg=mg
y
N=Fg
x
Ff=uN
Newton’s First Law: a body continues to maintain its state of rest or of uniform motion
unless acted upon by an external force. This also applied to the first diagram in which
the puck is in a state of rest. In this case if the floor is highly polished or the puck is
travelling on ice then the coefficient of friction could be assumed to be almost zero and
the acceleration will be approximately zero; otherwise the puck will suffer a deceleration
due 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 of
35N and the coefficient of friction between the ice and the puck (vulcanized rubber) is
approximately 0.15 then applying Newton’s Second Law:
 Fx  ma x and a x 
F
x
m

S  N 35  0.15  0.170  9.81

 204 m/s2
0.170
m
and if the player pushes it for 1.5m then the velocity as the puck leaves the stick will be:
vx2  vx20  2as , i.e. v x  2as  2  204  1.5  24.8 m/s (55 mph)
10
Real Life Examples in Dynamics
Alternatively, for a jar lid of mass 15g sliding on a wooden floor ( = 0.2) assuming a
force of only 1N is applied with the broom then:
ax 
S  N 1  0.2  0.015  9.81

 65 m/s2
m
0.015
And if it is pushed by the broom for 0.05m ( 2 inches) then its velocity will be:
v x  2as  2  65  0.05  2.54 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, calculate
the 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:
ax
Fg=mg
y
x
N=Fg
Ff=uN
applying Newton’s Second Law for car on ice:
 Fx  ma x and a x 
Ff
m

N
m

0.15  2000  9.81  1.47 m/s2
2000
Using kinematics: v x  v x20  2as , i.e. v x 0  2as  2  1.47  2  2.4 m/s (5.4 mph)
2
Again applying Newton’s Second Law for car on wet concrete:
 Fx  ma x and a x 
N
m

0.45  2000  9.81  4.41 m/s2
2000
Using kinematics: v x  v x20  2as , i.e. v x 0  2as  2  4.41  2  4.2 m/s (9.4 mph)
2
This is why cars are fitted with Antilock Brake Systems (ABS) !
11
Real Life Examples in Dynamics
Example 2.2
Ask the students to continue exploring the lesson example by calculating the maximum
distance from the goal-keeper that the shot can be taken in order to beat the reactions of
the goal-keeper, assuming a typical visual reflex time is 180 milliseconds.
Solution:
For the puck moving across the ice,
 Fx  ma x and a x 
Ff
m

N
m

0.15  0.170  9.81  1.47
0.170
The distance that the puck will travel in 180 milliseconds is:
m/s2
s  v x t  12 a x t 2  24.8  0.180   0.5   1.47   0.18 2   4.4 m
So the shot should be taken within 4m of the goal-keeper.
12
Real Life Examples in Dynamics
KINEMATICS OF PARTICLES
3.
Topic: Work and Energy
Engage:
Bring a slingshot and a handful of rubber
balls into class, pull the elastic band back and
release a few rubber balls into the class.
Explore:
Remind students about the basic meaning of conservation of energy. Ask students to
work in pairs and to identify the conservation of energy during the loading, firing and
trajectory of the balls. Invite some pairs to talk through to the class their understanding
of the energy conversions. Discuss how strain energy stored in the sling material is
transferred as work to the ball and becomes kinetic energy. Ask them in their pairs to
reconsider conservation of energy during loading, firing and flight of projectile.
Explain:
Applying the principle of work and energy to the rubber ball during firing:
 KE1   U 1 2   KE 2
where KE1 and KE2 are the initial and final kinetic energies of the ball respectively and
U1-2 is the work done by all the external and internal forces acting on the ball between
the initial and final state.
1
By definition, kinetic energy, KE  mv 2 , and KE1 = 0 (because v1=0).
2
The work done on the ball during firing is the strain energy released from the sling shot
which was stored in the material as it was stretched, U  uV where u is the specific
strain energy at yield and V is the volume of material and assuming the sling shot is
stretched to the limit of its elastic behavior.
During flight potential energy maybe gained or lost with height but this is negligible
compared to the kinetic energy at launch. Some kinetic energy will be lost due to drag
but again it is small so that the ball will arrive at its target with a substantial portion of the
launch kinetic energy.
13
Real Life Examples in Dynamics
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 2400mm3, then the energy
stored when it is pulled to yield is (see chapter 6 in Real Life Examples in Mechanics of
Solids):

 U  uV  3.6  10 6
 2400  10   8.6 J
9
so for a yellow dot squash ball of mass 24g with initial kinetic energy KE1 =0:
 KE1   U12   KE2  12 mv22 and v2 
2U

m
2  8.6
 27 ms-1
3
24  10
If the ball is launched parallel to the ground from a height of 1.5m, then its range can be
calculated using kinematics (neglecting drag) since in the vertical direction: sy = 1.5m,
vy=0 = 0, and ay = g = 9.81m/s2,
so
v y2  v y20  a y s
and
v y  2a y s y  2  9.81  1.5  5.4 m/s
and
v y  v y 0  2a y t
so
t
vy
2a y

5.42
 0.28 s
2  9.81
In the horizontal direction (neglecting drag): s  vx t  12 ax t 2  27  0.28  0  7.45 m
so the slingshot has a range of almost 7.5m.
(Note: potential energy gained in flight is (mgh = 0.0249.811.5 = ) 0.35J or 4% of the
kinetic 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 bread
to be drawn downwards into the toaster between the heating elements and also extends a
spring at each end of the toaster. When the toast is ready a pair of triggers releases both
springs 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 just
does not come out of the toaster when it is ready, neglecting the weight of the mechanism
and 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 60 to 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 shape
of slice is neglected.
14
Real Life Examples in Dynamics
Solution
(a) Toast must not jump higher than minimum depth of slot  length of a slice of bread
At top of jump, vertical velocity, vy = vy2 = 0 thus using
v y22  v y21  2a y s y
v y1  2a y s y  2  9.81  0.125  1.57 ms-1
Hence the kinetic energy received by each slice from the springs at launch is:
KE  12 mv y21 
0.04  1.57 2
 0.049 J
2
To store this much energy in a spring the work done (W) on the springs is
KE  W  12 Fd where F is 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, d  23  0.125  0.0833 m.
So
F
2W 2  0.049

 1.18 N
0.0833
d
For the two-slice toaster the force that needs to be applied is 2.35N (= 1.18  2)
(b) With the additional springs the energy stored will be doubled, i.e. the kinetic energy
delivered to each slice will be 0.098J (= 0.049  2).
Thus, the exit velocity will be v 
2  KE

m
2  0.098
 2.2 ms-1 from KE  12 mv 2
0.04
The vertical component will be v y1  v sin 60  2.2 sin 60  1.91 ms-1
and the horizontal component, vx1  v cos 60  2.2 cos 60  1.10 ms-1
For time to top of flight:
v y 2  v y1  a y t and vy2 = 0 so t 
1.91
 0.19 s
9.81
The time for the rise and fall to the top of the flight is 0.39 seconds (=20.19), and
distance travelled horizontally is
s x  vxt  12 a xt 2  1.10  0.39  0  0.43 m
i.e. each slice will land 430mm away from the toaster.
15
Real Life Examples in Dynamics
Example 3.2
Estimate the range of a longbow of the type used in medieval England (e.g. by Robin
Hood). Such bows are about 2m long, made of yew with a draw of about 0.76m and
were used with arrows having a mass of about 50g. The draw force has been estimated to
be about 265N. To see a longbow in action just type ‘longbow’ in www.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 a
incremental distance, d is given by W  Fd , so integrating this over the total draw
gives the total work done, W  12 Fd  12  265  0.76  101 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 is
transferred to the arrow as kinetic energy, i.e.
 KE1   U 12   KE2  12 mv22 and v 
2U
2  101

 63.6 ms-1
3
m
50  10
If the arrow is launched at an angle,  to the ground then the vertical component of
velocity is v y 0  v sin  and horizontal component is v x 0  v cos  . The time of flight
can be calculated as the twice the time required to reach its maximum height when its
vertical velocity, vy = 0, i.e.
v y  v y 0  a y t and t 
 vy0
a

v sin 
g
The distance travelled horizontally during time, 2t will be
“ s  vxt  12 axt 2 ” thus s  v cos    2t 
For maximum range
2v 2
cos  sin 
g
ds
2v 2
0
cos2   sin 2  
d
g
i.e.
sin 
 1 and  =45
cos 
2
2v 2
2  63.6 2  1 
Hence substituting in s 
cos  sin  

  412 m
g
9.81
 2
Replica longbows have been shown to have a range of about 300m so this estimate is a
little long which is unsurprising since drag and energy losses were neglected.
16
Real Life Examples in Dynamics
KINETICS OF PARTICLES
4.
Topic: Impulse and momentum methods
Engage – part I:
Bring a tennis racket and a ball (or two) into class plus a
meter rule. Drop a tennis ball on the floor and let it
bounce freely. Do this a number of times and ask the
class to count the number of bounces before the ball rolls
across the floor.
Explore – part I:
Ask the class why the ball does not bounce back to the same height, i.e. why the height of
rebound decays. Explain that the ball loses a fraction of its mechanical energy with each
impact with the floor.
Since the conservation of energy must apply, ask the class to identify what happens to the
lost 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 sufficiently
during a game to feel the temperature difference.
Note that the sound of the bounce changes with successive bounces, with the impact
surface and with the height of the initial drop. Use the meter rule to drop it from different
heights in order to illustrate this effect.
Explain – part I:
A ball of mass, m on released from a height, h has a potential energy, PE = mgh; after
hitting the floor it bounces to height, d where it has a potential energy, PE = mgd. The
coefficient of restitution, e = d/h and is equal to the fraction of mechanical energy lost.
This fraction is the same for successive bounces, i.e.
e  d1 h  d 2 d1  d 3 d 2  d 4 d 3
where d1, d2, d3, d4 are the heights of successive bounces.
The coefficient of restitution of a tennis ball bouncing on a concrete floor is about 0.71 so
if the ball is dropped from 1m the height of successive bounces will be
d1 = eh = 0.711 = 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 sufficient
momentum to cause them both to move after the impact.
17
Real Life Examples in Dynamics
Explore – part II:
Explain how conservation of energy causes kinetic energy from the moving ball, before
impact to be converted to strain energy of deformation for both balls whilst they are in
contact 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:
e
vB after  v A after
v A before  vB before
Elaborate:
In a tennis machine (for an example goto www.youtube.com and type in tennis machine5)
tennis ball A rolls down a chute from a vertical height of 0.25m and impacts tennis ball B
at the bottom of the chute, we can calculate the velocity at the instant before impact by
conservation of energy. The kinetic energy of the ball, A at the bottom of the chute
2
equals its potential energy at the top, i.e. 1 2 mA v A before  m A gh so
v A before 
2 gh  2  9.81 0.25  2.2 m/s
Now considering the coefficient of restitution in terms of velocities, assuming ball B is
stationary before the impact:
e
vB after  v A after vB after  v A after

v A before  vB before
2.2  0
so vB after  1.6  v A after
Now applying the principle of conservation of momentum:
mA v A before  mB vB before  mA v A after  mB vB after and
so
v A before  v B before  v A after  v B after
and
2.2  0  v A after  1.6  v A after
and


or
mA=mB
v A after  2.2  1.6  0.3 m/s
2
vB after  2.22  v A after  1.6  0.3  1.9 m/s
In other words the stationary ball moves off at 1.9m/s and the impacting ball at 0.3m/s.
Evaluate
Ask students to attempt the following examples:
Example 4.1
A 4 year-old boy weighing 15kg is sitting on the bottom edge of a waterslide (with his
legs dangling in the swimming pool) when his 7 year-old sister weighing 22kg comes
5
http://www.youtube.com/watch?v=EWNq0t6ap-8
18
Real Life Examples in Dynamics
down the slide from a height of 3m. Assuming the end of the slide is horizontal and
neglecting friction, calculate the velocity of the boy after impact if (a) his sister grabs him
and they move off together and (b) if they separate after contact with a coefficient of
restitution of 0.8.
Solution:
(a) For the girl (subscript ‘G’), the kinetic energy before impact is equal to her potential
energy at top of slide:
mG vG before  mG gh
2
So
1
and
vG before 
2
2 gh  2  9.81  3  7.7 m/s
So in case (a) applying conservation of momentum using subscript ‘B’ for the boy:
m B v B before  mG vG before  m B v B after  mG vG after
with
v B after  vG after
 v after
15  0  22  7.7   15  22 vafter
and
vafter 
22  7.7
 4.56 m/s
15  22
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
e
and
vB after  vG after
vG before  vB before
vB after

vB after  vG after
7.67  0
 7.67e  vG after
now applying conservation of momentum:
m B v B before  mG vG before  m B v B after  mG vG after
thus
and
so
15  0  22  7.67   15  7.7e  vG after   22vG after
vG after  22  7.67   15  7.7  0.8  2.1 m/s
15  22
vB after  7.7e  vG after  6.2  2.1  8.3 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 in
this way and then to select one and perform a similar analysis to the one above.
19
Real Life Examples in Dynamics
KINETICS OF PARTICLES
5.
Topic: Systems of Particles
Engage:
Bring a desk fan into class, place it on the table and
switch it on. It would be good to attach some
ribbons or strips of tissue paper to the protective
cage 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 blades
must 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 and
momentum applied. Since the flow is steady, the applied force will be constant during a
time interval.
Explain:
Explain the concept of a control volume as a ‘box’ around the fan containing a mass of
air, m with an average velocity, v; and a small mass of air, dm about to enter the box with
a velocity vA at time, t then subsequently at time, (t + dt) a corresponding mass dm leaves
the other side of the box having been accelerated to vB by the fan.
The principle of linear impulse and momentum:
tafter
  Fdt  mv 
after
 mv before
tbefore
can be applied to the air stream in the box:
 Fdt  dm  v   mv   dm  v   mv 
B
i.e.
F 
A
dm
v B  v A 
dt
the term dm dt is known as the mass flow. Continuity of mass requires:
dm
  A v A AA   B v B AB
dt
20
Real Life Examples in Dynamics
Elaborate:
Fans are often powered by shaded pole AC induction motors which have a very low
efficiency of the order of 20%; so a 50W fan will only deliver about 10W to the blades.
Power, P is the rate of work, or for a fan the rate of kinetic energy change delivered to the
airstream, i.e.
P
dKE

dt
1
2

m v B2  v A2
dt

And if the airstream is initially at rest, vB = 0 then utilizing continuity of mass:
P
1
2
 v 3A A
and rearranging
vA  3
2P
A
So, if the fan has a radius of 0.1m and the density of air is 1.2 kg/m3 then the velocity of
the air leaving from the fan will be:
vA  3
2 P 3 2  10

 8.1 m/s
A
1.2   0.12
The force on the air is
F 


dm
vB  v A    Av A2  1.2   0.12  8.12  2.47 N
dt
Now simple statics can be employed to calculate the minimum mass of the fan to prevent
it sliding along the desk:
Free-body diagram:
y
x
 Fx  F  N  0
So
F
mg
N
N
and
F  mg
 Fy  N  mg  0
and
m
F
2.47

 0.42 kg
g 0.6  9.81
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 floor
when it is switched on.
Evaluate:
Ask students to attempt the following examples:
Example 5.1
When the outlet nozzle of a 1875W hair-dryer is placed vertically above and just not
touching the stainless steel plate of a set of kitchen scales, the scales register 46g when
the hair-dryer is switched on. If the nozzle is an ellipse with a major diameter of 75mm
and minor diameter of 22mm and the fan has a diameter of 60mm, calculate the
percentage of the power used by the fan, neglecting losses.
21
Real Life Examples in Dynamics
Solution
Nomenclature: subscripts ‘A’ at the intake, ‘B’ at the outlet and ‘C’ just after impact with
the plate.
The force exerted by the hairdryer can be obtained using the principle of linear impulse
and momentum, and by assuming the air comes to rest (in the direction perpendicular to
the plate) after impact with the plate, i.e. vC = 0
F 
dm
vC  v B   Anoz v B2
dt
and the area of an elliptical nozzle is given by:
0.075 0.022

 1.29 10 3 m
2
2
Anoz  rJ rN   
where rJ and rN are the major and minor semi-axes lengths, so for air of density,
=1.2 kg/m3:
0.46  9.81
F

 53.9 m/s
Anoz
1.2  1.29  10 3
vB 


For the fan, the power is the rate of delivery of kinetic energy, i.e.
P
d KE 

dt
1
2

m vB2  v A2
dt

and if the airstream at the hairdryer intake is initially at rest, vA = 0 then utilizing
continuity of mass:
P
1
2
 vB3 A fan 
1
2
 1.2  53.9 3 
  0.06 2
4
 265 W
So the percentage of the power used by the fan is:

265
 100  14%
1875
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 a
small wind turbine capable of providing power for their home.
22
Real Life Examples in Dynamics
KINEMATICS OF RIGID BODIES
6.
Topic: Angular velocity & acceleration
Engage:
Take a yoyo into class and also an empty cable spool with a length
of string wound on to it. You should be able to get cable spools
from your Department of Electrical Engineering, alternatively ask
your campus maintenance staff if their electrical team can collect
them 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 in
the figure below) and ask the students to predict which way the spool will move when
you pull the string and whether the string will be wound or unwound. Collect the options
and take a class vote on it before pulling the string carefully so that the spool does not
slip. This also works with the yoyo but is less easy to see in a large class.
O
A
B
C
Explain:
Velocity is constant all along the unwound string (assume the
string is pulled at 150 mm/s) so vB  0.15 m/s 
0.15
v
(Angular velocity of spool,   B 
 3 rad/s 
rBO 50  10 3
100
mm
150
m
The spool does not slip so, at contact between spool and table,
the velocity is zero, i.e. vC  0
m
Draw the kinematic diagram:
O
B
vB
C
i.e. spool moves in the direction it is being pulled!


Velocity of center of spool, vO  rOC  150  103  3  0.45 m/s 
So since vO  vB the string will be wound up at a rate equal to the difference, i.e. 0.3 m/s.
23
Real Life Examples in Dynamics
Elaborate:
Consider a yoyo falling under gravity when the unwinding
of the string causes it to rotate giving it gyroscopic
stability, 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 the
yoyo then, when it is fully unwound, the rotational energy
in the yoyo will cause it to start winding back up the string
again. The yoyoist will have to pull up on the string
slightly in order to replace the energy loss due to friction.
In some designs of yoyo, the string is not firmly attached to
the axle, in order to allow the yoyo to spin freely or ‘sleep’
when the string is full unwound, with these designs the
yoyoist needs to give a tug to start it winding up again. The
tug has the same effect as pulling the string around the
spool.
A
O
B
aO
vO
mg
The center of mass of a yoyo will fall at a constant acceleration due to gravity: aO  9.81
m/s2 so at the end of its 1.15m string it will have achieved a velocity given by:
v22  v12  2as
and
vO 2 
2as  2  9.81  1.15  4.75 m/s
The diameter of the axle is typically 10mm so the rate of rotation is given by,

4.75
vO

 950 rad/s 
rBO 5  10 3
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 range
available, at the instant when the pedal is at the top of its rotation.
vO
vP
vC
2rTO 2rCO
2rPD
rPF
24
Real Life Examples in Dynamics
Solution:
First each student will need to make some measurements on their bike. Typical
measurements 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:  wheel 
vO
6.7

 19.7 rad/s
rTO 0.340
Velocity of the chain and the outer radius of the gear at the rear wheel,
vC  rCO wheel  0.032  19.7  0.63 m/s
The chain will have the same velocity at the pedal gear, so the rotational speed of the
v
0.63
pedals will be  pedal  C 
 8.4 rad/s
rPD 0.075
And the velocity of the pedal will be v P  rPF  pedal  0.175  8.4  1.5 m/s
So your foot would be moving forward with an instantaneous velocity of 1.5m/s ( 3.3
mph) which is approximately walking speed!
25
Real Life Examples in Dynamics
PLANE MOTION OF RIGID BODIES
7.
Topic: Forces & accelerations
Engage:
Ride your bike into class, apply the front brakes too hard so that the rear wheel just
begins to lift off the ground. Wear a helmet!
Explore:
Dismount from your bike and standing on the floor,
demonstrate how applying the front brakes when
gripping the handle-bars causes the rear wheel of the
bike to lift off the floor. Use a mountaineering
karabiner or other similar clip to suspend a backpack of
books on the bike saddle. Invite members of the class,
standing on the floor, to repeat the demonstration with
different numbers of books in the backpack and with
backpack suspended at different locations. Talk about
the moment of inertia as a measure of the resistance of
a body (the bike and backpack in this case) to angular
acceleration (tendency for the rear wheel to lift in this
case) in the same way that mass is a measure of body’s
resistance to acceleration (i.e. Newton’s Second Law).
Explain:
When the front brake is locked so that there is no motion of the front wheel relative to the
bike, then the bike (including the front wheel) and backpack rotate about the point on the
front wheel that is in contact with the road, i.e. A
G
2m
1.
A
The moment of inertia of a body of mass, m about an axis is:
I   r 2 dm
m
26
Real Life Examples in Dynamics
where r is the perpendicular distance from the axis to an arbitrary element dm. The
moment of inertia for a solid cuboid of height h, width, w depth d and mass m is:
Ih 

m 2
w d2
12

where Ih is the moment of inertia about the axis through the center of mass parallel to the
height dimension. So, for a pile of textbooks in a backpack of mass 3.8 kg and of
dimensions 0.120.210.26m:
Ih 




m 2
3.8
w  d2 
0.212  0.262  0.035 kg.m2
12
12
When the moment of inertia, IG about the body’s center of mass is known then the
moment of inertia, I about any other parallel axis is given by the parallel axis theorem:
I  I G  mb 2
where b is the perpendicular distance between the axes. So if we assume that the center of
mass of the backpack is at the location, G in the diagram above then, its moment of
inertia about A, IA is given by:
I A  I G  mb 2  0.035  (3.8 1.2 2 )  0.035  5.47  5.51 kg.m2
We neglected the moment of inertia of the backpack but this is inconsequential because
the second term in the expression above dominates and location of the backpack and the
mass of the books has a huge effect on the resistance to rotational acceleration.
Elaborate:
The equation of rotational motion is given by:
M
G
 I G
where the moments MG are considered about an axis through the center of mass and  is
the angular acceleration of the body.
When a child rolls down a steep grassy bank (roly-poly) we can calculate their
acceleration by using the above expression and a few simplifying assumptions. Let us
assume 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


r

C
N
C
F

ma

27
Real Life Examples in Dynamics
Considering motion about C, the equation of rotational motion is:
M
c
 mgr sin   (ma)r  I
Now, the linear acceleration is, a= r and the radius of gyration, k is defined by I = mk2,
so
mgr sin   (mr )r  mk 2
rearranging

r g sin 
r2  k2
and
a
r 2 g sin  0.15 2  9.81  sin 45

 3.67 ms-2
r2  k2
0.15 2  0.8 40 
If the child starts from rest and the bank is 5m high then their velocity at the bottom can
be calculated from:
v 2  u 2  2as
i.e.
v  2as  2  3.67  5  6.1 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 the
driving force is at 45 degrees to the cutting board. The cutting wheel and handle have a
total mass of 500g. The blade is of radius 50mm with an average thickness 0.5mm and
the cylindrical handle is 180mm long with a radius of 15mm. The coefficient of friction
between the pizza and the blade is 0.01 and it can be assumed that the pizza exactly
opposes the magnitude of the driving force with a force acting towards the axis of the
cutting wheel at 10mm above the cutting board. Calculate the constant driving force
required to cut through a pizza 30cm wide exiting at a velocity of 1.5m/s from a standing
start – is this viable for a 12 year old child?
Solution:
Volume of pizza cutter wheel,
V  r 2 t   0.05 2  0.0005  3.93 10 6 m3
So for stainless steel of density, = 8000 kg.m-3,
the mass of the pizza cutter wheel is
m  V  8000  3.93 10 6  0.031 kg
and the moment of inertia of the wheel about its
axis of rotation is given by:
28
Real Life Examples in Dynamics
mr 2 0.031 0.052

 3.93  10 5 kg.m2
2
2
I
The mass of the handle is 0.469kg (= 0.5-0.031) and its moment of inertia about an axis
parallel to the axis of rotation of the wheel and though its center of mass is
I





0.469
m
3r 2  h 2 
3  0.015 2  0.18 2  0.00129 kg.m2
12
12
And using the parallel axis theorem to shift this to be about the center of rotation of the
cutter wheel:
I  I G  mb 2  0.00129  0.469  0.05 2   0.0025 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 an
entry velocity of zero, using:
“ v 2  u 2  2as ” we have
a
v 2  u 2 1.52

 3.75 ms-2
2s
0.6

a  r so
and for the cutter wheel
a 3.75

 75 rad/s2
r 0.05
Drawing a free body diagram for the cutter wheel:
mg
D

0.
5

ma

0.4
FP
0.1

P
FN
C
C
N
For motion about C, consider the equation of rotational motion:
M
c
 0.5 D cos  0.1P cos   0.3P sin   0.1Fp sin   0.3Fp cos   (ma)r  I  mr 2  I
now, D = P = (Fp/) so re-arranging we can obtain:
 mr 2  I 
D
0.5 cos  (0.1cos   0.3 sin  )   (0.1sin   0.3 cos  )
and substituting values for the parameters, to obtain:
D



75  0.5  0.05 2  0.0025
0.281

 3.1 N
1 1
4
9
3 12
0
.
09

 (  )  0.01(  )
2 2 50 50
50 50
So, the driving force required is 3.1N which is very small.
29
Real Life Examples in Dynamics
PLANE MOTION OF RIGID BODIES
8.
Topic: Work and energy
Engage:
Take a bag of marbles into class. Empty it onto
the bench so that marbles scatter everywhere
and fall off the bench. Invite students to collect
them 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 talk
through to the class their understanding of the energy conversions. Remind them about
translational kinetic energy ( KE  mv 2 2 ) and tell them about the concept of rotational
kinetic energy ( KE  I o  2 2 ) about center of mass of a marble.
Explain:
Consider the moment of inertia about the center of mass for a marble of mass 5g & radius
15mm:
I
2 2 2
mr  0.005  0.0152  4.5  10 7 kgm2
5
5
The kinetic energy as it rolls along the bench at 0.5 m/s:
KE 
1 2 1 2 1 2 2 2 2 7
mv  I   mr  mr   mr 2 2
2
2
2
5
10

where v  r so  
v
0.5

 33.3 rad/s
r 15  10 3
thus, the kinetic energy of a rolling marble is given by:
KE 
7
7
mr 2 2  0.005  0.0152  33.32  8.75  10 4 J
10
10
Elaborate:
Now consider a marble starting from a stationary position on the edge of the bench and
rolling over the edge.
Applying conservation of energy between two positions: KE1  PE1  KE 2  PE 2
30
Real Life Examples in Dynamics
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. KE21  PE12 and substituting:
7
mr
10
2

2
2

  12  mgr (1  cos  )
and the marble is initially stationary so  1  0 and  22 
10 g (1  cos  )
7r
Free body diagram at point of losing contact, i.e. position 2:
r(1-cos)
FN
n


12
0.

man
mg
mat
N
r sin
So resolving forces normal to the contact force at the corner, noting that when contact is
lost, N  FN  0
F
n
since
 0 so  mg cos   man  0 and  mg cos   man
an  
v 22
g cos 
  r  22 then g cos   r  22 and  22 
r
r
so equating with expression above from conservation of energy:
g cos  10 g (1  cos  )

r
7r
and cos  
10
i.e.   54 
17
The marble loses contact with the bench when it has rolled over by 54 degrees and at this
instant it has a velocity:
v2  r 2 
gr cos  
9 .81  0 .015 
10
 0 .294 m/s
17
at 54 to the horizontal so that it will land at some distance from the bench, depending on
the height of the bench.
31
Real Life Examples in Dynamics
Evaluate:
Ask students to attempt the following examples:
Example 8.1
A yoyo of diameter 6cm and mass 50g has an axle of diameter 10mm around which a
string of length 1.15m is wound. Use the principle of conservation of energy to calculate
its angular velocity of the yoyo when it is allowed to fall and unwind under gravity.
Solution
The moment of inertia of the yoyo is: I 
mR 2 0.05  0.03 2

 2.25  10 5 kg.m2
2
2
Conservation of energy, i.e. from position 1 (start of fall) to position 2 (fully unwound):
KE1  PE1  KE2  PE2
1
1
1
1
mv 12  I  12  mgh 1  mv 22  I  22  mgh
2
2
2
2
or
mg h1  h 2  
so
 
2
1
1
mr 2  22  I  22
2
2
2 mg h1  h 2 

mr 2  I
2  0 .05  9 .81  1 .15
 105 rad/s
0 .05  0 .05 2  2 .25  10  5
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 who
can exert a constant force of 600N. After pushing for 5m, the student is joined by a
second student who exerts a constant force or 900N and after another 2m a third student
joins them and exerts a constant force of 700N but the second student slips and stops
pushing only a 1m after the third joins and 2m later the car crests the incline and
everyone 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:
KE12  PE12  Work Done
i.e.
1 2
Fd
or v2 
mv2  mgh12  
2
2
v2 
 Fd  mgh
m
600 10  900  3  700  30  1500  9.81 0.7   2.46 m/s
1500
The car will crest the hill at 2.5m/s or 5.5 mph).
32
12
Real Life Examples in Dynamics
PLANE MOTION OF RIGID BODIES
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 standing
on each end of it. On your bike freewheel very gently
towards the block of wood. Repeat the experiment until
your front wheel just rolls over the wood block.
Explore:
Discuss the impulse received by your front wheel when it hits the wooden block; note
that the impulse force is of unknown magnitude and direction. Highlight that angular
momentum just before and just after impact must be conserved. Discuss the conservation
of energy during the event and that kinetic energy at the impact must at least equal the
potential energy when the wheel is on top of the wooden block in order for the wheel to
roll over the block.
Actually this example is easier if you consider a basketball rolling over the block of
wood, 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 practical
interest, 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 after
impact with a wooden block.
Explain:
Free body diagrams:
m(vA )2
A
A
+
m(vA )1
B
A
=
mgt
B
b
r
d

Momentum just before impact
A
A
B
∫Fdt
Momentum just after impact
Impulses during impact
33
Real Life Examples in Dynamics
Let’s assume that the basketball does not slip on the wooden block or rebound from it but
essentially rolls over it by rotating about the corner B. Then there are two impulses or
forces acting on the basketball: (a) its weight and (b) the impulse from the wooden block
at 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 not
know 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]12 = [System Momentum]2
m v A 1 r  b   I A  1  m v A 2 r  I A  2
For simplicity, consider the basketball as a hollow sphere with mass, m = 0.6kg and
radius, r =0.012m:
IA 
so
2mr 2
3
v A 1 r  b  
v
r
 ,
and
2r va 1
2r va 2
 v A 2 r 
3
3
 3b 
Re-arranging*: v A 2  v A 1 1  
 5r 
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. v A 3  0 :
KE 2  PE 2  KE 3  PE3
 mv A 22 I 22 
`

  0  0  mgb
2 
 2
Substituting for I and   v r  : v A 2 
2
6mgb
5
Thus combining with conservation of angular momentum*:
6mgb
 2b 
 v A 1 1  
5
 3r 
hence
v A 1
6mgb
5 

 2b 
1  
 3r 
6  0.6  9.81  0.02
 2  0.02 
1 

 3  0.12 
5  0.422 m/s
Note that as the radius of the rolling object gets larger the required velocity becomes
smaller, so it easier to get a larger diameter wheel up a curb than a small one.
34
Real Life Examples in Dynamics
Evaluate:
Ask students to attempt the following examples:
Example 9.1
A girl of mass 50 kg is setting off along a level sidewalk on her skateboard. If she can
push off with one foot with a constant force of 150N for 0.5 seconds, calculate her speed
after this push. If, almost immediately she can push again for another 0.5 seconds, then
calculate her speed after this second push assuming the force applied is sinusoidal with a
maximum of 150N.
Solution:
Change in Momentum = Linear Impulse
mv
or
2
 mv 1 
t2
  Fdt
t1
so, in the first push with a constant force:
v2 
Ft 150  0.5

 1.5 m/s ( 3.36 mph)
m
50
and in the second push:
1
so
 150sin t  dt  50  1.5
v2 
0.5
v2 
150   cos 1  cos 0.5  75
 2.5 m/s (5.6mph)
50
50
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 25 to the horizontal, bounce at an angle of
15, if it has a diameter of 66mm and mass of 57g.
Solution
m(vA)1
m(vA)2


+
=


Momentum just before impact
∫Fdt

∫Ndt




Momentum just after impact
External impulse during impact
35
Real Life Examples in Dynamics
Conservation of angular momentum about B:
m v A 1 cos 25  I  A 1  0  m v A 2 cos 15  I  A 2
assume a perfectly elastic impact, so
v A 1
 v A 2 and  A 1   A 2
and that the tennis ball is a hollow sphere, i.e. I 
then
and
4mr 2
 A 1  mr v A 1 cos 15  cos 25
3
 
 A 1  3 v A 1 cos15  cos 25  3  35.8cos15  cos 25  48.5
or about 463 rpm!
36
2mr 2
note that 80mph  35.8m/s
3
4r
4  0.033
rads/s
Real Life Examples in Dynamics
THREE DIMENSIONAL RIGID BODY MOTION
10.
Topic: Kinematics of rigid bodies in three-dimensions
Engage:
If you play the violin, then bring it to class
and play an excerpt from your favorite piece;
or if you don’t play, maybe you could ask a
student in your class or a colleague in the
music school.
And, or search in
www.YouTube.com for “Janine Jansen
Mozart” and select Mozart Violin Concerto 5
(2of 5) for a good close-up of the violin being
played6. Play it a second time and ask the
students play along on the “air violin”.
6
http://www.youtube.com/watch?v=5QlyMNMLlfI
Explore:
Again in www.YouTube.com search under ‘Robot violinist’
and 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. It
would be relevant to introduce examples of engineering
applications of robots, e.g. welding robots8; again search in
www.YouTube.com for suitable videos (also see “Robot of
the Year 2007”9).
Ask students to explore in pairs the degrees of freedom and
axes of rotation in their own arms. Ask them to identify the
degrees of freedom used in playing the “air violin”.
7
http://www.youtube.com/watch?v=EzjkBwZtxp4
http://www.youtube.com/watch?v=362vMN7Ra4w
9
http://www.youtube.com/watch?v=Q8WnAN9jmEc
8
Explain
You have seven degrees of freedom in your arm: at the shoulder, elbow and wrist, as
shown in the picture. Use your own arm, or ask a student to help you. If you have
brought a violin to class then talk about the motions involved in moving the bow to the
instrument.
37
Real Life Examples in Dynamics
Elaborate
A simple robotic arm for a violin playing robot is shown in the figure. To engage the
bow with the instrument, the arm will rotate about the vertical axis shown with the elbow
at E fixed. To avoid damage to the bow and strings we need to calculate the angular
velocity and acceleration of the bow.
y
b
P
S
u
O
f

arm
W

E
x
We can establish a coordinate system which is fixed relative to the arm with the x-axis in
the plane SEW and the y-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 and
 arm     0 j . Thus, we can consider the position vector of W relative to O:
rW / O   f cos   u cos   i  ( f sin   u sin  ) j
now,
“ v A  v B  ω  rA / B ”
so
v W  v O  ω arm  rW / O  0 
and
vW  0  f cos   u cos   k
i
j
k
0
f cos   u cos 
0
 f sin   u sin 
0
0

P
S
O
arm
38
rel

E

rel
W
Real Life Examples in Dynamics
The coordinate system is fixed with respect to the arm so the angular velocity of the bow
with respect to the coordinate system is:
ω rel   rel cos   i   rel sin   j
and the position of P, halfway along the bow, relative to W is:
rP / W  b sin   i  (b cos  ) j
so, the velocity of the point, P is given by:
i
v P  v W  ω bow  rP / w   0  f cos   u cos   k   rel cos 
b
 sin 
2
j
k
 0   rel sin 
0
b
 cos 
2
0

b
b

 

v P   0  f cos   u cos     rel cos   cos     0  rel sin   sin   k
2
2

 


For a bow of length 75cm held by an arm with upper and forearm lengths of 25cm and
30cm respectively at angles of  = 15 and  = 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:
v P   0.75  0.3 cos 20  0.25 cos15  0.5 cos 20  0.375 cos 20   0.75  0.5 sin 20  0.375 sin 15
v P   0.393  0.117  0.482  0.99 m/s
The acceleration of the bow,  is given by:
α
d y
d x
d z
i
j
k  Ωω
dt
dt
dt
though the components of  are constant, the acceleration is not zero but
α bow  Ω  ω bow 
i
0
rel cos 
j
arm
arm  rel sin 
k
0
0
 armrel cos  k
 0.75  0.5  cos 20  0.35 rad/s about the z-axis.
39
Real Life Examples in Dynamics
Evaluate
P
Example 10.1
c=0.18m
An industrial robot uses its arm in the position
shown to roll out adhesive tape on the underside of
a structure as shown in the figure. If the arm CB
rotates about the vertical axis through C at a
constant angular velocity of 3 rad/s find the velocity
at which the tape is being laid out when = 35.
B
y
Solution
d=0.3m

C

A
m
0. 2
b=
P

b
c
rel
A
rel
Establish a coordinate system that is fixed
relative to the arm with the x-axis coincident
with CB and the y-axis co-incident with the axis
of rotation, i.e with an origin at C. Then
d

C
B
arm    0 j
and we can consider the position vector of A
relative to C:
rA / C  c  b cos  i  (c sin  ) j
now
and
i
0
0
k
0
c  b cos
 c sin 
0
v A  v C  ω arm  rA / C  0 
j
v A  0 c  b cos  k
The coordinate system is fixed with respect to the arm so the angular velocity of the disc
with respect to the coordinate system is:
ω rel  rel cos  i  rel sin   j
For the point of contact, P there is no motion due to the adhesive on the tape so v P  0 .
Now, the position of P relative to A is:
d
d

rP / A   sin  i  ( cos ) j
2
2

So, the velocity of the point, P is given by:
40
x
Real Life Examples in Dynamics
i
j
v P  v A  ω disk  rP / A  0 c  b cos   k   rel cos 0   rel sin 
d
d
 sin 
 cos 
2
2
k
0 0
0
 2c 2b

hence, rel  0   cos  sin  
d
d


and
ω disk  Ω  ω rel   rel cos  i  0   rel sin  j
finally, the tape is used at
i
j
k
vT  ω disc  rP / A  rel cos 0  rel sin 
d
d
 sin 
 cos
2
2
0
0
vT  0 c  b cos  k  3  0.18  0.2 cos 35  1.03 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:
a b
The determinant of a 3x3 matrix, A  d e
 g h
c
f 
i 
is given by detA   aei  afh  bfg  bdi  cdh  ceg
and can remember using the diagonals as follows:
+
+
+
-
-
-
a
d
g
b
e
h
c a
f d
i g
b
e
c
f
h
i
41
Real Life Examples in Dynamics
THREE DIMENSIONAL RIGID BODY MOTION
11.
Topic: Kinetics of rigid bodies in three-dimensions
Engage:
Take a child’s spinning top into class. In fact wood ones are
cheap enough for you to take a handful into class so that you can
pass them around in case the students have forgotten what they
are like. Encourage the students to spin the tops. Also take the
front wheel of your bicycle.
Explore:
Spin your top sufficiently fast that its spin axis remains vertical – this is known as
sleeping. 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 about
the vertical axis which is known as precession.
In practice, the spin rate of the top will continue to decrease due to friction but in steady
precession it is assumed that the rate of spin is constant.
Z
z
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 the
top.
Re-iterate that three angles are required to define
the three-dimensional motion of a rigid body
relative to an axis: one angle to define the rotation
of the body about the axis and two angles to define
the 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 as
Euler’s angles and the equations of motions can be
expressed in terms of them, i.e.
M
M
M
42

.

y
Y
mg
X
x
 I xx  I zz  I xx   2 sin  cos   I zz sin 
y
 I xx  sin   2 cos   I zz    cos 
z
 I zz

 
  cos  sin  
h


x
Real Life Examples in Dynamics
Generally these equations have to be solved numerically; however when the precession
rate is constant these equations reduce to:
M
x
 I zz  I xx   2 sin  cos   I zz sin  ,
M
Now, the mass of the top exerts a moment about the origin of:
y
 0 and
M
x
M
z
 0.
 mgh sin 
and hence,
mgh sin   I zz  I xx   2 sin  cos   I zz sin 
or
 
mgh sin   I zz  I xx   2 sin  cos 
I zz sin 
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:
m  V  
and
R 3h
3
 700 
  0.064
3
 0.0095 kg
 3  0.06 2 3  0.06 2 
 3h 2 3R 2 
  2.57  10 5 kg.m2
  0.0095  

I xx  I yy  m

5
20 
20 

 5
I zz 
3mR 2 3  0.0095  0.06 2

 1.03  10 5 kg.m2
10
10
Now, estimating the nutation to be about 15 and the precession rate to be about ten
revolution per minute (= 1.05 rads/s):
 
0.0095  9.81  43  0.06 sin 15  1.03  2.56  10 5  1.052 sin 15 cos15
 712 rad/s
1 10 5  1.05 sin 15
Or about 113 revolutions/second which 6800 r.p.m.
43
Real Life Examples in Dynamics
Elaborate:
Z

z
Engage the students by spinning your bicycle
wheel and holding by the axle stubs in front of
you so the axle is horizontal. Tilt the axle by
raising one end of the axle and lower the other
end so that you experience gyroscopic moments
restoring the axle to its original position. Explain
to the students about the forces you are
experiencing. Perhaps have a student come to the
front of the class and hold the wheel while you
spin it for them.

zk
X
y
x
Y
 sin  i
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 
ω   K   j   k
where i, j, and k are unit vectors along the local axes and K is the unit vector along the
fixed Z axis, such that K   sin  i  cos  k
hence, the angular velocity is given by:


ω   sin  i   j     cos  k
The local axes of rotation, xyz are also the axes of inertia, so the angular momentum, Ho
will be the sum of the moments of inertia multiplied by the appropriate component of the
angular velocity, i.e.


H 0   I  sin  i  I  j  I    cos  k
where I and I  are the moments of inertia about the spin axis and fixed axes respectively.
Newton’s second law of motion in terms of moments is:
M
O

H
O
and it can be shown that:
M
O
   ΩH
 H
O xyz
O
  is the rate of change of HO with respect to rotating axes xyz and  is the
where H
O xyz
angular velocity of the rotating axes xyz. Now, these axes nutate and precess but do not
spin thus,
Ω   sin  i   cos  k
44
Real Life Examples in Dynamics
Now consider the case where, the nutation, , rate of
precession,  and rate of spin,  are all constant then
the angular momentum, HO is also constant, i.e.
H O xyz  0 then
 MO  Ω  HO


Z

z
k
 K
X

 MO  I   cos  I  cos  sin  j
MO
x
y
This is the couple that must be applied to maintain the
steady precession, i.e.       0 assuming the that
center of gravity of the wheel is fixed in space (F=0).
Y
Z
Now, for the situation in the demonstration the precision
and spin axes are at right angles, i.e. = 90 so
 K
M
 k
MO
y
x
z
O
I j
When this is applied about an axis perpendicular to the
axis of spin the wheel will precess about an axis
perpendicular to both the axis of spin and the couple, i.e.
the global Z-axis.
This behavior makes it easier to balance on your bicycle
at higher speeds due to the stabilizing effects of the
spinning wheels. It is also known as gyroscope motion.
Evaluate:
Ask students to attempt the following examples:

Example 11.1
The Earth can be thought of as a spinning top with a spin
axis through the North and South Poles which slowly
rotates, or precesses so that the North Pole draws out a
circle in space. This precession is very slow (one degree
every 71.6 years) and is known as the Precession of the
Equinoxes. It was probably observed first by Hipparchus
in around 137BC and results in a slow change in the
position of the sun with respect to the stars at an equinox.
This is important for calendars and their leap year rules.
N

MO
O
S
The Precession of the Equinoxes is caused by a (force) couple acting on the earth due to
the gravitational attraction of the sun and moon. Assuming that the earth is an oblate
spheroid of average radius, 3960 miles and mass 5.97421024kg, and that the earth’s
nutation is constant at 23.45, calculate the couple.
45
Real Life Examples in Dynamics
Solution:
Rate of precession,  
Rate of spin,  
1
2
1


 7.73  10 12 rad/s
71.6 360 365  24  3600
2
 7.27  10 5 rad/s
24  3600
Assume a moment of inertia for a sphere,
2
8000 

2  5.9742  10   3960 

s
2mR
5 

I

 9.59  10 37 kg.m2
5
5
24
And
M
M



O
 Ω  H O  I    cos   I  cos   sin  j
O
 I  I  I   cos   sin  j


assuming I  I  then:
M O  I sin   9.59  10 37  7.27  10 5  7.73  10 12 sin 23.45
and finally:
M O  2.14  10 22 Nm
Example 11.2
Ask the students to design a gyroscope stabilizer for an experienced rider on a
motorcycle.
46
Real Life Examples in Dynamics
MECHANICAL VIBRATIONS
12.
Topic: Free and forced vibrations
Engage:
Take a hula hoop and a wooden ruler
into class. Place the ruler overhanging
the bench, lean on the end on the bench
and flick the free end so that the ruler
vibrates. Repeat this a few times and
slide the ruler onto the bench, as it
vibrates, so that the pitch of the noise
changes – 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 quite
hard so tell them this and put it to one side. Pick up your hula hoop and, if you can, do
some hooping (see www.hooping.org for instructions on hula hooping and links on how
to make your own).
Explore:
Allow the hoop to oscillate on your finger, in simple harmonic
motion, and discuss the characteristics of the motion.
;
The moment of inertia of a thin ring or hoop about an axis
through its center is I C  mr 2
and using the parallel axis system the moment of inertia about a
point, E on the hoop is I E  I C  mr 2  2mr 2
The kinetic energy of a hoop oscillating on your finger is the sum of its translational
(=zero) and rotational kinetic energy:
KE 
mvE2 I En2

 mr 2 2
2
2
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 r 1  cos   and for
small angles we can use the series expansion of cos   1   2 2 so:
PE  mgh  mgr
2
2
Now, the total energy of the system is constant:
47
Real Life Examples in Dynamics
KE  PE  mr 2 2  mgr
2
2
= constant
And its derivative yields the equation of motion of the hoop, i.e.
mr 2 2  mgr  0
or


mr 2r  g  0
and  is not always zero, hence:
 
g
 0
2r
This is the equation of motion for the undamped free vibration of a simple block and
spring system which could be used to represent the oscillating hoop. It is known as
simple harmonic motion in which the angular acceleration is proportional to the angular
displacement of the hoop. The natural frequency of oscillation is
fn 
1
2
g
2r
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 ½ Hz  9.81 2 .


The solution to the equation of motion is:
xt   A sin
g
g
t  B cos
t
2r
2r
where A and B must determined from known boundary conditions. The amplitude of the
oscillation is
A2  B 2 .
Explain:
Now, return to the ruler and repeat the
analysis. The kinetic energy stored in
the ruler is
" KE 
y
2
mv
"
2
M
M+dM
dm
dx
d
and for an element this is
d KE  
y 2 dm
2
So to obtain the total strain energy in the
ruler we need to integrate over its length,
i.e.
48
R
d
Real Life Examples in Dynamics
l
KE 
1 2
y dm
2 0
And in one period,  any point on the ruler moves through ŷ so
y 
yˆ

  yˆ
and
KE 
m 2
2
l
 yˆ dx
2
0
where m is the mass per unit length. Now considering the strain energy stored in terms of
the bending moment, M and the curvature of the beam:
U
1
Md
2
and for small deflections:

dy
dx
and
1 d d 2 y


R dx dx 2
also from the theory of beams:
1 M

R EI
l
thus
2
d2y
1
1
U   Md   EI  2  dx
2
2 0  dx 
Neglecting the gravitational potential energy, the total energy of the ruler is:
2
l
l
 d2y 
1
m 2 2
ˆ

 dx =constant

KE  U 
y
dx
EI
2 0
2 0  dx 2 
Elaborate:
from Mechanics of Solids, the deflection of a cantilever can be described as:
y
Pl 3
6 EI
  x 2  x 3 
3     
  l   l  
so the total energy can be demonstrated to be:
 Pl 3 


 3EI 
2
  2 33ml 3EI 

 3  = constant
2l 
 2 140
49
Real Life Examples in Dynamics
At the natural frequency we can equate the energies, so:
 2 33ml
2 140

3EI
140 EI
and  
3
2l
11ml 4
Thus as the ruler gets shorter the frequency goes up. For a wooden mahogany ruler
(E=9,200 MPa) of cross-section 28mm3mm and mass per unit length of 0.0459 kg/m
(=1410-3/0.305):
I
bh 3 0.028  0.0033

 6.3  10 11 m4
12
12
At l=22cm  
140 EI
140  9.2  10 9  6.3  10 11

 261.63 Hz i.e. middle C.
11ml 4
11  0.0459  0.22 4
Whereas a plastic ruler (polycarbonate, E = 2.2 GPa) of cross-section 452mm
(I =310-11 m4) and mass per unit length 0.1076 kg/m requires a length of 10.3cm to
produce 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 the
end of a whip antenna of length 0.7m.
s
Solution
For the ball, apply Newton’s Second Law in the direction of
motion, i.e. tangential to the support, F from the aerial:
F
t
 mat so
using kinematics,
 mg sin   mat
d 2x
at  2  x
dt
l

and from the geometry, for small angles,
x  l
so we can rewrite the equation of motion as
g
 mg sin   ml or     0
l
an
assuming for small angles sin    and so the natural
frequency is
n 
50
g
9.81

 3.74 Hz
l
0.7
at

mg
F
Real Life Examples in Dynamics
Example 12.2
The natural period of a building or structure can be lengthened to alleviate the highest
earthquake forces using isolating bearings that consist of a ball carrying the structural
weight and rolling on a concave surface (e.g. www.earthquakeprotection.com). For a
bridge, sixteen such bearings are to be used in which the high strength steel ball has a
radius of 10cm and the concave bearing surface a radius of curvature of 1.5m, find the
natural frequency of a bearing.
If the bearing is shaken by an earthquake with a seismic frequency of 40Hz that produces
sinusoidal motion of the ground of 150mm what will be the form of the equation of
motion of the bearing.
Solution:
and
 

mv 2 I 2

2
2
R
r
mr 2
v

also I 
Rr Rr
2
so substituting for v,  and I:
3mR  r  
KE 
4
2
R
Rr
KE 
r

2
(R-r)cos

Kinetic energy at bottom of bearing:
h
The potential energy gained when the ball is displaced from bottom of the bearing is:
PE  mgh  mg R  r  1  cos  
now 1  cos    2 sin
2

2

2
2
for small angles, so PE  mg R  r 
2
2
2
3mR  r   2
 gmR  r 
4
2
2
Thus the energy equation is:
KE  PE 
And differentiating with respect to time:
2
3R  r  
 g R  r   0 or R  r  3R  r    2 g   0
2


2g
 0
3R  r 
Since  is not always zero then:
 
And the natural frequency is:
n 
2g
2  9.81

 2.16 Hz
3R  r 
31.5  0.1
The earthquake will generate a forcing term of the form F0 sin  t so that the equation of
motion will be  
2g
  F0 sin 80 t .
3R  r 
51
Junior Dynamics Course: Suggested exemplars within lesson plans
NOTES FOR INSTRUCTORS ON EXAMPLE APPLICATIONS
Prepared as part of the NSF-supported project ( #0431756) entitled:
“Enhancing Diversity in the Undergraduate Mechanical Engineering
Population through Curriculum Change”
Edited by Eann A Patterson, University of Liverpool