Mathematics in Education and Industry
MEI
Mathematics in Education and Industry
MEI
Big ideas
• Mathematical modelling
– use of vectors
– use of algebra and calculus
– use of diagrams
• Conservation of linear momentum
• Conservation of mechanical energy
• Centre of mass
• Invariance and independence
MEI conference 2011
Interesting ideas to enliven A2 Mechanics
Mathematics in Education and Industry
MEI
Stimulating interest
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Stimulating interest
• Standard problems taken a little further to yield interesting and
surprising results
• Stimulation by experiment and simulation
– exploring the situation and forming conjectures
• Some examples are readily accessible to stronger students
given a ‘push’
• Some examples could be worked collaboratively in class with
the teaching more actively guiding
Mathematics in Education and Industry
Mathematics in Education and Industry
MEI
• Attempts to generalise
• Use of diagrams, algebra, trigonometry and graphing
• I hope the following examples will engage and encourage
Mathematics in Education and Industry
MEI
Outline of the examples
Outline of the examples
1. Use of vectors and 3-D graphing representation to investigate
the condition for two projectiles to collide.
2. Some surprising examples of independence for a sliding
object:
• on a rough surface: its mass and the distance slid (and
time taken) to come to rest from a given speed
• from rest smoothly from the top, down a chord of a
vertical circle: the time taken and the angle of the chord
• from rest down a uniformly rough path in a vertical plane:
the speed at the bottom depends on the vertical and
horizontal displacements not the path taken
3. Two approaches to a linear momentum problem
4. The use of algebra and calculus to investigate the most
efficient way of moving a packing case on a rough horizontal
floor – an alternative approach
5. The use of simulation and algebraic techniques to investigate
some properties of projectile trajectories
1
A fly stands on the edge of a circular disc which is
initially at rest on a smooth horizontal table.
What happens to the disc and the fly relative to the
ground as the fly walks across a diameter of the
disc at a constant speed u relative to the disc?
Suppose the disc has mass M and the fly mass m.
fly
m
u
v
V
disc
mass M
+ ve direction
velocity of fly relative to the table
velocity of disc relative to the table
Suppose the velocity of the fly and disc relative to
the table are v and V, respectively, as shown in the
diagram.
As the table is smooth we may conserve linear
momentum horizontally.
PCLM in the + ve direction gives
mv + MV = 0
also v – V = u
so m( u + V) +MV = 0 giving mu + mV + MV = 0
mu
Mu
and v 
and these are the
so V  
mM
mM
velocities of the disc and fly relative to the table.
David Holland
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Another way of arguing this is to use the fact that,
in the absence of an external horizontal force, the
common centre of mass of the fly and the disc does
not move.
Suppose the disc has radius R and that the fly is a
distance x across the diameter AB.
x
m
A
B
M
R
Let the distance of the common centre of mass of
the fly and the disc, G, be a distance xG from A.
(m  M ) xG  mx  MR so xG 
mx  MR
mM
and relative to A the speed of G is xG 
mx
.
mM
mu
.
mM
Now this speed is relative to A. As G is stationary,
mu
it must be that the disc has a velocity 
,
mM
where the +ve direction is that of the fly.
But x  u so xG 
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A proposition by Galileo
If from the highest or lowest point in a vertical
circle there be drawn any inclined planes meeting
the circumference the times of descent along these
chords are each equal to the other.
Galileo Galilei: Dialogues Concerning Two
Sciences; Day 3, Theorem VI, Proposition VI
(1635)
A post-Newton proof
DO is the chord of a circle in a vertical plane with
O at the lowest point. A smooth ring slides on a
thin straight rod joining D and O and is at rest
when released from D.
The time taken for the ring to slide from D to O is
independent of the position of D on the circle.
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D
R
C
mg

N
r

horizontal
O
OD is at to the horizontal; the circle has centre C
and radius r; N is the mid-point of OD; the ring
has mass m.
angle OCN = angle DCN so OD = 2r sin 
Let the acceleration of the particle down DO be a.
N2L in the direction DO gives
mg sin   ma , so a  g sin  .
Using s  ut  12 at 2 on the motion from D to O
gives 2r sin   0  12  g sin  t 2
r
4r
and t  2
(since t > 0).
so t 
g
g
Hence t is independent of the position of D on the
circle.
2
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A packing case on a horizontal surface is being pulled by a
string. Will the case slide or will it tip? If it slides, what is the
optimum angle for the string to make the tension as small as
possible?
T
R

F
W
Assume that the case is in equilibrium and does not tip
T cos  F  0
R  T sin   W  0
(1)
(2)
If the case is on the point of sliding
F  Fmax   R
(3)
Resolve 
Resolve 
Use (1) and (2) to substitute for F and R in (3)
T cos   (W  T sin  )
W
 T (cos   sin  )  W  T 
cos   sin 
For the minimum value of T we need
(4)
dT
0
d
dT ( sin    cos )

2
d
 cos   sin  
dT
 0   sin    cos  0  tan   
d
Hence, the optimum value for  is the angle of friction, .
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(5)
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W
, with tan   
cos   sin 
W
W sec
=
, with tan    .

cos (1   tan  ) (1   tan  )
Tmin =
W 1   2
W

Hence Tmin =
1 2
1 2
(6)
To show this is a minimum.
T is continuous for 0   
Tmin 
Now
and
W
1 
2

1 2

1 
David Holland
2

2
, T (0)  W , T  2   W and
.
 1 so Tmin < W = T  2 
  so Tmin  W  T  0  so the result follows.
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An alternative method that does not require calculus
T
R
Fig. 1
F
W
S
R
S is the total
surface contact
force
Fig. 2
F
Replace R and F with S
T
S
Fig. 3
W
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Define the angles of S and T as shown
T
S


Fig. 4
W
In equilibrium, the
triangle of forces is
T
Fig. 5

W
S

T
There are many possible
values of T and 
W
S
Fig. 6

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When friction is limiting, tan () =This value of
is usually written as 
R
S
Fig. 7

R
T is least when the string is perpendicular to
the direction of the total surface contact force.
The triangle of forces is now
T

W
Fig. 8
S

the solution with minimum T is when = 
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Now let us consider the condition for tipping
T
R

F
A
b
x
B
a
a
W
With the tension inclined above the horizontal, as shown above,
tipping could take place about the line through A perpendicular
to the plane of the diagram; this will occur when x = 0. If the
tension is inclined below the horizontal then tipping could take
place about the line through B perpendicular to the plane of the
diagram this will occur when x = 2a.
With the system in equilibrium, the general equation for c.w.
moments about A is
Wa  Tb cos  2aT sin   Rx  0
With the system on the point of tipping about the line through A
perpendicular to the plane of the diagram, x = 0 and
Wa  Tb cos  2aT sin   0
Wa
(7)
so T 
2a sin   b cos
To decide whether the case slips or slides with a given tension at
a given angle, the values of T for sliding (equation (4)) and
tipping (equation (5)) are compared.
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For slipping before tipping
Wa
W

and as the tensions are
cos   sin  2a sin   b cos
positive quantities this gives
a  tan   b  a  0
a  b 1 b
 
so tan  
 a
a
etc
It is also interesting to note that using the total surface contact
force, S, as above, there are only three forces acting on the case.
If the case is in equilibrium, these three forces S, W and T must
be concurrent.
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For an object sliding until it comes to rest on a uniformly rough
plane surface, the distance slid and the time taken depend on the
initial speed and the coefficient of friction but not on the mass of
the object.
initial speed, u
R
acceleration, a
mass
m

F
mg
Using Newton’s second law down the plane
mg sin   F  ma
Resolving perpendicular to the plane
R  mg cos   0
Given that the object is sliding, F  Fmax   R
Thus mg sin    mg cos   ma
and a  g sin    g cos  . Hence a is constant and does not
depend on m.
In the situation we are considering, the object comes to rest and
so a < 0. This means that mg sin    mg cos  so   tan  or
the object will never stop.
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Using the constant acceleration results gives answers not
involving m.
Using v = u +at with v = 0, t  
u
a
u2
Using v  u  2as with v = 0, s  
2a
2
2
1
Using s  vt  12 at 2 with v = 0 gives s   at 2 which shows
2
that the connection between s and t does not require knowledge
of the initial speed (which can be hard to measure).
Note that the result is valid for horizontal surfaces (take = 0).
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Many problems easily solved by energy methods
are tackled by students who falsely (or without
justification) modify the problem so that may apply
constant acceleration formulae. This is a case
where it works!
Suppose that a particle of mass m moving in a
vertical plane slides down a uniformly rough slope.
Suppose that the vertical displacement downwards
is H, the horizontal distance travelled is L and the
coefficient of friction is . If the path of the
particle is any smooth curve and the particle is at
all times in contact with the slope, the KE gained
by the particle is mg(H – L).
To start with, suppose a particle of mass m slides
down a small straight section of plane AB, as
shown in the diagram. We shall neglect air
resistance and assume that the coefficient of
friction between the particle and the plane has a
constant value of  and the particle never loses
contact with the plane. The vertical and horizontal
displacements of the particle in travelling from A
to B are h and l respectively. The normal reaction
is R and the frictional force F.
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A
R
F
h
mg

B
l
Since the particle is sliding, we have
F  Fmax   R   mg cos  .
The KE gained by the particle as it travels from A
to B, KE , is given by
KE  AB  mg sin   AB   mg cos   mgh   mgl
since ABsin   h and ABcos   l .
This gives KE  mg  h   l  .
Note that this expression is also valid for the
particle sliding up the slope if is taken to be
negative; the displacement h is now negative while
the displacement l remains positive. This means
that we can apply this result to any sequence of
such sections as shown below. If the vertical
displacements are h1, h2, h3, … and the horizontal
displacements l1, l2, l3, … then if
H   hi and L   li we have
KE  mg  H   L  .
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Of course, we are assuming that the particle has
enough KE initially for it to travel from A to B (i.e.
get over any bumps) and that it never loses contact
with the slopes.
h1
H
h4
h3
A
h2
B
l1
l3
l2
l4
L
We may make the straight line segments as small
as we like and so move to a smooth curve.
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It is, perhaps, not obvious that a particle would gain
exactly the same speed sliding from A to B down
any of the curves (a), (b) and (c). All that is
necessary is that the slope is uniformly rough – the
length and shape of the path from A to B do not
matter.
A
B
As long as the slope from A to B is uniformly
rough, the speed gained is independent of the
length and shape of the path
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Some interesting things about trajectories
Suppose that a number of fireworks are fired from
ground level and travel over horizontal ground.
Suppose also that air resistance is negligible.
No matter what their initial speeds and angles
of projection, all the fireworks that reach their
highest points at the same time do so at the
same height. [If the fireworks turned into
flares when they reached their highest points
you would see a rising horizontal plane of
light.]
If the greatest height reached is H m reached in T s,
then taking + ve as upwards and using
s  vt  12 at 2 gives H  0  12 ( g )T 2 so H  12 gT 2 ,
which does not require separate knowledge of the
speed, u, and angle of projection, . This is
because both H and T depend only on the vertical
component of the initial velocity, y 0 , and
y 0  u sin  : i.e. it is the value of u sin  that
determines both H and T. [Considering the vertical
direction, we have, using v  u  at , that
y 0
and, using v 2  u 2  2as , that
0  y 0  gT so T 
g
y 02
2
0  y 0  2 gH so H 
.]
2g
David Holland
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Suppose that particles are projected with the same
speed from the same point in the same vertical
plane but have different angles of projection. It is
well known that the plane containing the
trajectories may be partitioned into three regions:
points on a parabola (called the parabola of safety)
where there is a single trajectory that passes
through that point: points inside that parabola for
each of which there are two distinct trajectories that
pass through them: points outside the parabola for
each of which there are no trajectories that pass
through them. This is easy to derive and even
easier to show using Autograph.
The parabola of safety may be found from the
equation of the trajectory as follows. Using the
usual symbols, we make tan the subject of the
gx 2
trajectory equation y  x tan   2 1  tan 2  
2u
gx 2
gx 2
2
to give 2 tan   x tan   y  2  0 .
2u
2u
We require equal roots for a single solution so
gx 2 
gx 2 
g 2 u2
2
x  4  2  y  2  giving y   2 x 
.
2u 
2u 
2u
2g
This parabola intersects the
 u2 
 u2 
y – axis at  0,
 and the +ve x – axis at  ,0  .
 g 
 2g 
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No two of the trajectories can intersect if they are
both rising so either one is rising and one is falling
or they are both falling. If one is rising and the
other falling, the point of intersection must be
before the highest point of one of the trajectories.
The locus of the highest points may be found as
follows:
with the usual conventions of the origin at the point
of projection and initial speed and angle of u and
, respectively, the coordinates of the highest
 u 2 sin 2 u 2 sin 2  
point are easily shown to be 
,
.
2
g
2
g


u 2 sin 2
u 2 sin 2  u 2
Put x 
,y 

1  cos 2 
2g
2g
4g
2 gx
4 gy
giving sin 2  2 and cos 2  1  2 .
u
u
2
4 g 2 x 2  4 gy 
Eliminating  gives
 1  2   1 so
u4
u 

2
2 2
2
2


4 g x 16 g u
 4 
 y   1.This is an ellipse with
4
u
u  4g

u2
u2
, semi – minor axis
and
semi – major axis
2g
4g
 u2 
with centre  0,
.
4
g


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Also, no two trajectories can intersect if they both
have angles of projection less than 45° so either
one is less than 45° and one greater or they are both
greater and the trajectory with angle of projection
of 45° is the boundary between these cases.
The diagram below shows the regions inside a
parabola of safety partitioned by the ellipse found
above and the trajectory with angle of projection of
45°.
50
y
40
30
20
10
x
10
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40
50
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What is the condition that two particles projected
at the same time collide with each other?
Assume that the only force acting on the particles is
that of gravity and that they meet no obstructions.
Particle A has initial position r0 and velocity u;
particle B has initial position R0 and velocity U;
all units are SI.
The position vector, r, of A is r = r0 + ut + ½gt²,
the position vector, R, of B is R = R0 + Ut + ½gt².
The particles collide when r = R and t > 0,
thus r0 + ut + ½gt² = R0 + Ut + ½gt²
and so r0 – R0 = t(U – u), t > 0.
This may be stated as ‘the initial position of A
relative to B must be parallel and with the same
sense as the initial velocity of B relative to A’.
Note that, as in many problems involving two
particles moving only under gravity, the value of g
is not relevant. This gives many good ways of
visualising what is going on.
One simple way the condition can be met is by the
particles being projected directly towards one
another. They must always collide.
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u
A (r0)
U–u
r0 – R0
U
B (R0)
Another is when B is projected directly towards A
and A is projected directly away from B. In this
case they will collide only if U > u.
Much more interesting, and less obvious, is when
they are not projected in the same line.
u
A (r0)
U
r0 – R0
B (R0)
– u
U
U–u
David Holland
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