Gas Hydrates - Fuel for the Future and Other
Uses
by Bruce Torrie
Department of Physics, University of Waterloo
The picture on the cover of this issue shows a
burning piece of methane hydrate. It is estimated that there
is twice as much fuel in the form of methane hydrate as there
is in all the other carbon based sources put together. That
is twice as much as the combined total for coal, oil and
ordinary natural gas. So what is methane hydrate? If water
at high pressures and low temperatures is in the presence
of a former gas such as methane, then the water molecules
can form cages around the gas molecules and the cages
combine to form a crystalline solid. Proper conditions for
this to occur are found in the coastal trench off the western
side of Vancouver Island , other similar trenches throughout the world, under the permafrost in the MacKenzie River
delta and similar locations elsewhere.
little distortion of the H-O-H angle. A larger cage is formed
by the addition of two hexagons to the smallest cage as
shown in Fig. 1(right). With both types of cages there are
excess hydrogen atoms which can form bonds to other
cages. In the crystalline structure there are 2 small cages
for every 6 large cages in the basic building block or unit
cell to give approximately one methane molecule to every
five water molecules if every cage contains a methane
molecule. Usually there are empty cages so the density of
methane molecules is reduced but the density is still much
greater than it would be in a gas even at high pressures.
The structure described above is known as structure-I but other structures are possible depending on the
size of the former molecule. Structure-II is made up of
dodecahedra and hexakaidecahedra in the ratio 16:8.
Hexakaidecahedra have four hexagons incorporated with
the twelve pentagons to give a 5 1264 cavity. If we look at the
inert gases as examples of former gases, then He and Ne are
too small to stabilise the cavities so no hydrates form with
these gases. 512 and 5 1264 cavities form around Ar and Kr
to give a type-II structure. Xe stabilises 512 and 51262
cavities to give a type-I structure. The two structures
mentioned are the most common ones, but Canadians
working at Chalk River discovered a third type labelled H
and others are possible.
Methane hydrate made the CBC news on April 17,
2001. Researchers from the University of Victoria had their
interest piqued when a fishing trawler hauled up a strange
find last December. Instead of fish, the fishermen snagged
a strange white ice that bubbled. In keeping with their trade,
the fishermen shovelled the ice back into the ocean but
based on photographs the ice appears to be methane
hydrate. The scientists propose to use a mini-submarine
to investigate the ocean bottom in the region looking for
underwater gas vents where the large chunk might have
formed.
Methane hydrate looks like ordinary ice and is
sometimes promoted as the ice that burns. The methane
comes from the decay of organic matter at moderate depths
or from the pressure cooking of organic matter at greater
depths. When the gas comes in contact with cold water at
a depth of 300m or greater, then the cages are formed and
arranged into a crystalline solid with a clathrate structure.
Methane hydrate is made up of two types of cages. The
smaller cage consists entirely of pentagons as shown in
Fig. 1 (left). Water molecules have a nearly tetrahedral
angle H-O-H so that five molecules form a hydrogenbonded pentagon with oxygen atoms at the vertices and
Page 2
Figure 1: Left - dodecahedron (512 - twelve pentagons)
Right - tetrakaidecahedron (51262 - twelve
pentagons, two hexagons)
Gas hydrates, or specifically chlorine hydrate, have
been found to exist since the work of Humphrey Davy as
reported in 1811 and corroborated by Michael Faraday in
1823. Other former gases were discovered in the 1800's
Phys 13 News / Winter 2002
including Br2, SO2, CS2, CHCl3, CO2, etc. with methane, CH4,
being added to the list by Villard in 1888. Practical problems
involving methane hydrates came to the fore in the 1930's
when it was realised that natural gas hydrates, mainly
methane hydrates, were blocking gas transmission lines.
The high pressures in the lines were enough to stabilize the
hydrates even at temperatures above the ice point. This
discovery led to the regulation of the water content in
natural gas pipelines and research into methods of preventing hydrate formation.
An announcement of the discovery of hydrates in
Siberian permafrost came in 1965 followed by production
of gas in 1969. This production was made practical by the
existence of equipment for handling ordinary natural gas
that is also found in the same region. Gas hydrate deposits
were discovered along the north shore of Alaska and in the
MacKenzie delta in the early 1970's but these areas have
remained undeveloped. Recently an exploratory hole was
drilled in the MacKenzie delta by a joint Japanese-American-Canadian team with the object of gaining experience
that can be used in drilling operations in the coastal waters
of Japan. The Japanese have a particular interest in
developing this source of energy since they lack conventional sources. A mapping of the undersea resources
throughout the world is being carried out using sonar
techniques. A particularly strong reflection occurs from
the lower surface of a hydrate body when conventional
natural gas is trapped under the body as is often the case.
Recent soundings of the hydrate body off the west coast
of Vancouver Island were regarded as newsworthy enough
to appear in the Globe and Mail. Estimating the extent of
a hydrate body is difficult because only one surface
reflects. Also the trapped gas may make a significant
contribution to the total supply of methane available at a
given site.
Several practical problems are being vigorously
investigated at the present time. How can the methane be
released from hydrate bodies and brought to the surface in
a controlled fashion? Hydrates are very unstable and can
dissociate if poked in an effort to exploit oil and conventional gas resources that lie near or under the hydrate body.
A worst case scenario visualises a huge gas bubble floating up from the ocean floor to sink a drilling ship or platform
by reducing its buoyancy. How can this problem and less
dramatic problems be avoided? The answers to these
questions are evolving as are novel uses to be made of
hydrates. It is possible to drill into or under a hydrate body
in a horizontal direction so that the drill ship is well to the
side away from dangerous bubbles. As mentioned earlier,
Phys 13 News / Winter 2002
hydrates have a high energy density and may be the
preferred form in which to ship methane from offshore
sources. Natural gas could be converted to gas hydrates
and these could be shipped in a metastable state by ship.
The methane hydrate pictured at the start of this article is
in a metastable state with methane gas rising out of the solid
to fuel the flame. A Norwegian group is exploring the details
of the shipping process as it relates to undersea gas
deposits off the coast of Norway and they estimate that
there could be a 20% saving by shipping hydrates rather
than liquid gas. The disposal of CO2 is a serious problem
since it acts as a greenhouse gas when released into the
atmosphere. In hydrate form, the gas could be deposited
at the bottom of the ocean where it would remain stable.
Small scale experiments to test this means of disposal have
successfully been carried out in the Pacific Ocean.
At a more exotic level, it has been suggested that the
asteroid that plummeted into the Yucatan Peninsula and
ended the age of dinosaurs also destabilised the ocean
floor and caused the release of huge quantities of methane
that produced a great inferno that was not dinosaurfriendly. The stability of the ocean floor hydrate deposits
under calmer circumstances depends on the depth of the
water. At the peak of an ice age, much of the ocean water
is locked up in glaciers so the pressure at the ocean floor
drops and methane is released by the destabilised hydrates. Methane is also a greenhouse gas so the released
gas would contribute to global warming and end the ice age.
Is this a realistic senario? You never know until you apply
some numbers to the problem. Models have been constructed to represent various hypotheses and the most
common conclusion is that the effect is small.
The role of physicists and chemists in hydrate
research has been to explore the basic structures of the
hydrates and to study the dynamics of their growth and
dissociation. A group at the National Research Council in
Ottawa has been particularly active in the field for many
years using dielectric, neutron scattering, nuclear magnetic resonance, Raman scattering measurements and
molecular dynamics simulations. A group at Waterloo with
backgrounds in X-ray and neutron scattering, Raman
scattering and microbalance techniques has recently developed an interest in gas hydrates.
For further information about "gas hydrates" do a
web search by entering the two words. There is a vast
amount of information available on the web.
Page 3
Swingers, Sliders and Toppler: The Physics of
Falling Rods
by Phil Eastman, John Vanderkooy and
Anthony Anderson
Department of Physics, University of Waterloo
Abstract
Three cases of a rod falling under gravity and
rotating from a vertical position under different conditions
are analysed. In the first two, in which frictionless surfaces
are assumed, it is found that contact with a wall is lost at an
angle to the horizontal of θ1 = 41.8° = sin -1(2/3). This result
is proved for a uniform rod but is generally true. In the third
case, friction between rod and floor is introduced. It is found
that if slipping has not yet occurred, the direction of the
frictional force reverses at the above angle θ1. It is also
shown that slipping must occur at an angle greater than or
equal to θ2 = 19.5° = sin -1(1/3), no matter how large the
coefficient of static friction may be. Simple equipment for
demonstrating these results is suggested.
is small in comparison to its length, so that initially its centre
of mass is vertically above the axis of rotation.
Up to the point of slippage, the centre of mass will
follow a circular arc about O. It will have a tangential
acceleration component, perpendicular to the rod, given by
aT = αL/2, and a centripetal acceleration component, along
the rod towards O, given by aR = ω2 L/2, where α and ω are
angular acceleration and velocity, respectively. The torque,
measured about O, is given by
τ = Mg(L/2)cosθ = Ioα
where Io is the moment of inertia of the rod about O, equal
to ML2/3. Hence,
aT = Mg(L/2)2cosθ/(ML2/3) = (3gcosθ)/4
The topic of rotational dynamics, with its concepts
of torques, moments of inertia and angular velocities,
accelerations and momenta, is new and challenging for
many students. It is therefore important that the students
are introduced to as many examples as possible, especially
if these relate to simple phenomena that can be easily
demonstrated with readily available equipment. In this
paper, we discuss some simple applications of rotational
dynamics to rods, which fall under gravity from a vertical
position in a variety of ways. Some interesting results,
several of which are not intuitively obvious, are derived,
and, for one example, the unusual behaviour of the frictional
force is discussed.
I-The Swinger
A uniform rod of mass M and length L initially stands
vertically, contacting both wall and floor, as shown in
Figure 1. It then swings in a clockwise direction about the
corner point O, and at a certain value of the angle θ,
measured from the horizontal, it will lose contact with the
wall and slide to the right. In the following analysis, we will
neglect friction and find the position and speed of the centre
of mass when this occurs. Although not strictly necessary,
for simplicity we will also assume that the width of the rod
Page 4
(2)
We now equate the loss of gravitational potential energy to
the gain in rotational kinetic energy:
Mg(L/2)(1−sinθ) = Ioω2/2
Hence,
aR=ω2L/2=2(L/2)2Mg(1−sinθ)/(ML2/3)=3g(1−sinθ)/2
Introduction
(1)
(3)
(4)
The horizontal components of aT and aR are related to
the force from the wall as follows:
H = MaTsinθ − MaRcosθ = M[(3gcosθ)(sinθ)/4 −
3g(1-sinθ)(cosθ)/2]
(5)
When H = 0, the rod loses contact with the wall. Equation
(5) then gives, after some cancellation and with θ = θ1,
sinθ1 = 2(1−sinθ1) or sinθ1 = 2/3, corresponding toθ1= 41.8°.
From equation (3), ω2 = MgL(1−sinθ1)/(ML2/3) = 3g(1−2/3)/
L or ω = −(g/L) 1/2. The speed of the centre of mass is given
by v c = ωL/2 = −(gL) 1/2/2.
This analysis can be generalised for non-uniform
rods and applied to the practical case of a tapered ladder, for
example. If the moment of inertia is left as I o (and not replaced
by ML2/3) and the position of the centre of mass is given by
Rc (and not L/2) in the above equations, then it can be shown
that both terms will cancel to give the same value of θ1 = 41.8°
for loss of contact with the wall [1].
It is also of interest to see if there is an angle at which
the vertical force V exerted by the floor on the rod would go
to zero, since this would correspond to loss of contact with
the floor. The appropriate equation of motion is as follows:
Mg −V = MaRsinθ + MaTcosθ
(6)
Phys 13 News / Winter 2002
Using equations (2) and (4), we obtain after some
algebraic manipulation
V = Mg(1− 6sinθ + 9sin 2θ)/4 = Mg(1 – 3sinθ)2/4
(7)
It can be seen that starting atθ = 90°, V = Mg. As the
rod swings, V steadily decreases with θ and has a value of
Mg/4 at θ = θ1. This result shows that contact with the floor
is not lost before θ = θ1. At this point, the rod leaves the wall
and the equations of motion change. However, if the rod
were pivoted at the corner, equation (7) shows that V would
go to zero at θ = θ2 = sin−1(1/3) = 19.5°. V would not become
negative but would increase to a value of Mg/4 atθ = 0°, as
the rod hits the floor.
Figure 1: The Swinger; wall and floor shown in grey, initial
rod position in white, and intermediate position in black.
II - The Slider
In this case, the rod slides down the wall and along
the floor to the right with the angle θ decreasing, as shown
in Figure 2. We again assume frictionless surfaces and find
the value of θ for which contact at the wall is lost, using a
somewhat different approach which avoids the use of the
τ = Iα equation (since the axis of rotation is not fixed). If we
again choose the corner as origin, then the co-ordinates of
the centre of mass are given by
xc = (Lcosθ)/2 and y c = (Lsinθ)/2
(8)
We differentiate to obtain the velocity components:
v cx = −Lsinθ(ω/2) and v cy = Lcosθ(ω/2)
Figure 2: The Slider: notation as for Figure 1.
(9)
with ω = dθ/dt, which is negative since θ is decreasing with
time. We note that
Rc =(xc 2 + y c 2)1/2 = L/2
(10)
which shows that the path of the centre of mass is again a
circle centred at O, until contact with the wall is lost, just as
in the case of the swinger. Moreover, the speed of the
centre of mass is given by
v c = (v cx2 + v cy 2)1/2 = Lω/2
(11)
Differentiating equation (9) gives the following equations
for the acceleration components:
Figure 3: The Toppler: notation as for Figure 1, but with no
wall. The direction of F, the frictional force, is that for large
θ (see text).
Phys 13 News / Winter 2002
acx = −Lcosθ(ω2/2) − Lsinθ(dω/dt)/2
(12a)
acy = −Lsinθ(ω2/2) + Lcosθ(dω/dt)/2
(12b)
Page 5
The only forces acting on the rod are gravity and the
normal reactions at the floor and wall, since we are assuming
zero friction. If the rod loses contact with the wall, then the
horizontal force H will vanish at a certain value of θ = θ3:
H = Macx = −ML(ω2cosθ3 + sinθ3dω/dt)/2 = 0 which gives
ω2cosθ3 = −sinθ3(dω/dt)
(13)
We now apply the conservation of energy principle
motion: MgL(1−sinθ)/2 = Mvc 2/2 + Ic ω2/2 = M(ωL/2)2/2+
(ML2/12)ω2/2 =ML2ω2/6 which can be re-written as follows:
3g(1−sinθ) = Lω2
In this third case, a uniform rod is placed vertically
on its end on the floor, well away from any walls, in a
position of unstable equilibrium. If the floor is frictionless,
no horizontal forces act, and the rod will topple in such a
way as to keep its centre of mass falling vertically. The
motion is much more interesting if friction is present. We
let the coefficient of static friction between the rod and floor
be µ, the normal reaction force be N, the frictional force be
F and, as before, the angle between the rod and floor be θ,
as shown in Figure 3.
(14)
We differentiate this expression to obtain dω/δτ
-3gωcosθ = 2Lω(dω/dt) or (dω/dt) = −3gcosθ/2L
III - The Toppler
(15)
If we now substitute equations (14) and (15) withθ = θ3 into
(13), we obtain 3g(1−sinθ3)(cosθ3)/L = (3gsinθ3cosθ3)/2L
which simplifies to sinθ3 = 2/3 or θ3 = θ1 = 41.8°. From
equation (14), we get ω = −(g/L) 1/2 and hence from equation
(11), v c = −(gL) 1/2/2
Perhaps not too surprisingly, these results are
identical to those obtained in part I, although the motion is
quite different. For both cases, we could in principle now
find the time required for the rod to rotate from θ = 41.8° to
0° and then use this to find how far the centre of mass
travels (at constant speed because no horizontal forces are
now acting) before it hits the ground. Although the motion
is now more complicated, it is still valid to use the τ = Iα
equation, provided that dynamical quantities are evaluated
about the centre of mass [2]. However, we have been
unable to obtain an analytical solution for this case. A
numerical simulation for a one metre rod gives the value of
t = 0.179 seconds, which leads to ∆x = v cxt = (gL) 1/2sinθ1
(t/2) = (gL) 1/2(t/3) = 0.187 metres for this distance. If we add
this to the centre of mass co-ordinate xc at the instant the
rod loses contact with the wall, given by equation (8), we
obtain the following for the distance measured from the
corner: x = xc + ∆x = (Lcosθ1)/2 + ∆x = 0.373 + 0.187 = 0.560m.
In other words, the left end of the rod will be 6.0 cm from the
corner when the rod first reaches the horizontal position.
If we assume there is sufficient friction to prevent
slippage, the conservation of energy expression is as
before - equation (14). The radial (centripetal) acceleration
of the centre of mass is given by
aR = ω2L/2 = 3g(1−sinθ)/2
We use the torque equation (1) to obtain the corresponding
tangential acceleration:
aT =αL/2 = (3gcosθ)/4
Page 6
(17)
In the vertical direction, Newton’s second law gives
Mg−N = MaTcosθ + MaRsinθ =
(3Mgcos2θ)/4 + 3Mg(1-sinθ)(sinθθ)/2
(18)
In the horizontal direction, we have
F = MaTsinθ − MaRcosθ = (3Mgcosθsinθ)/4 −
3Mg(1-sinθ)(cosθ)/2 which simplifies to
F = (3Mgcosθ)(3sinθ−2)/4
(19)
This last equation is interesting because it shows
that for large θ (rod nearly vertical) the frictional force is
positive (i.e. in the direction shown in figure 3). But when
sinθ =2/3 (our old friend, θ1=41.8°), it becomes zero, and for
smaller angles it reverses its direction. The normal reaction
force, N, cannot reverse sign, of course, but equation (18)
shows that it can become zero. This occurs when
Mg[1−(3cos2 θ)/4 – 3(1−sinθ)(sinθ)/2] = 0
Seasoned physics problem solvers will recognise
the value of θ1 (or its equivalent measured from the vertical)
as the angle at which an object sliding from rest at the top
of a frictionless hemisphere will lose contact – for example
in Canada, we often use a hockey puck sliding down an
igloo. Readers might like to check that this classic problem
[3] uses many of the same equations discussed above.
(16)
(20)
which simplifies to sinθ =1/3 or θ =19.5° (our other old friend,
θ2). The physical significance of this is that no matter how
great µ is, there will always be a point of slippage at some
angle not less than 19.5°.
Phys 13 News / Winter 2002
and then acts to the left, decelerating the centre of mass. It
would, after all, be decidedly non-physical for friction to
continue to accelerate the rod to the right! Since friction
always opposes any motion, real or potential, of the rod at
the contact point with the floor, there is an initial tendency
to slide back (i.e. to the left) followed by one to slide forward.
In any case, sliding is inevitable at an angle greater than or
equal to 19.5°, no matter how large the value of µ. Although
we have not proved this analytically, for the case of a small
value for µ, there will be an initial motion of the point of
contact to the left, followed by motion of it to the right.
Figure 4: Plots of N/Mg and F/Mg versusθ for the Toppler.
10
9
Coefficient of Friction
8
7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
80
90
Angle (degrees)
Figure 5: Plot of µ = F/N versus θ for the Toppler.
In Figure 4, we have plotted F/Mg and N/Mg versus
the angle θ. This shows that as θ decreases, F initially
increases, reaches a maximum at about 63.3°, as can be
readily shown by differentiating equation (19), then decreases to zero at 41.8° and is negative for smaller angles.
N, on the other hand, decreases to zero at 19.5° (with dN/
dθ = 0) and then begins to increase. When the rod is about
to slip on the floor, the frictional force has its maximum
value, given by F = µN. We have shown in Figure 5 the value
of µ = F/N, for which slipping occurs, as a function of θ.
Starting at zero for θ = 90°, it increases to a maximum value
of 0.371 at θ = 54.9° = sin -1(9/11) and then decreases to zero
at θ1 = 41.8°. The frictional force F then reverses direction
and µ increases in magnitude becoming infinite atθ2 = 19.5°.
We see that if µ is less than 0.371, slipping will occur
with the frictional force acting to the right as shown in
Figure 3. However, if µ is greater than 0.371, F acts initially
to the right (as it must do since it is the only horizontal force
Phys 13 News / Winter 2002
Demonstrations
Many of the interesting results discussed above
may be readily demonstrated, at least qualitatively, with
simple equipment. A metre rule, with a small quasi-frictionless
roller attached to one end may be used to illustrate the
“swinger” and determine the angle at which contact with the
wall is lost and the position of the centre of mass when the
rule hits the floor. This device is also suitable for demonstrating the frictionless case of the “toppler” with the centre
of mass falling vertically. Similarly, a rule with rollers at both
ends will simulate the frictionless contacts with wall and
floor for the “slider”. A teflon-tipped rule on a teflon sheet
serves to illustrate the low friction case of the “toppler” with
the reversal of the direction of motion at the contact point
nicely demonstrated. Finally, a rule tipped with rough sand
paper resting on a similar strip will show that even with a
very high coefficient of friction, motion of the contact point
will occur at an angle close to 20° to the horizontal.
Acknowledgements
We express our thanks to Jim Brandon, Don De Smet and
Reg Moore for helpful discussions and to Lauren Lettress
for technical assistance.
References
[1] D. De Smet (Personal communication).
[2] See, for example, J.B.Marion and W.F.Horniak,
Physics for Science and Engineering (1st Ed.) pages
324-5. Saunders College Publishing, Philadelphia,
1982.
[3] See, for example, D.Halliday and R.Resnick, Funda
mentals of Physics (3rd Ed.) Chapter 8, problem 42.
John Wiley and Sons, New York, 1988.
Page 7
The Optimal Arc Angles for Shooting a
Basketball
The minimum speed, v min, that can send the ball to the
hoop center is equal to
gx , which is obtained by setting
θ = 45°. When v is greater than v min , twoθ values are obtained
for each v value.
by K. Frank Lin
[email protected]
Since Dr. James Naismith invented the basketball
game in 1891, many players have asked the question,
“What is the optimal arc angle for shooting a basketball?”
Many intelligent people would have guessed that the
optimal angle is 45°. Based on the fact that the ball would
have a larger hole to fall into if the ball approached the hoop
more vertically, some people would say it is better to shoot
with a high arc. Most good shooters do shoot the ball with
a high arc angle. But how high is high enough? As a
recreational player for over 35 years, I will try to find the
answer to this question using physics knowledge,
mathematics, and computer simulation. For purposes of my
analysis, air resistance to the movement of the ball will be
considered
For the purpose of finding the optimal arc angle, we
can consider the ball is shot in the vertical plane that passes
through the player and the hoop center.
Using a computer program (nicknamed Isaac), the
combinations of θ and v that would allow the bottom of the
ball to land at a designated position can be plotted out as
a curve on a graph with the shot speed v as the horizontal
axis and the angle θ as the vertical axis. Figure 2 was
produced with the values of x set equal to 13.25 feet, 14 feet,
and 14.75 feet.
Figure 2: The "shot height" is a constant 10 ft.
Figure 1: A simplified case, the ball is released at the
same height as the hoop.
Consider the case when the ball is released at the
same height as the hoop (Figure 1 with y = 0). When the
ball is released, the trajectory of the basketball is a parabola.
If the hoop center is x ft. away when the ball is released, then
the shot speed v and the upward arc angle θ must be related
by equation 1.
x = horizontalspeed ⋅ 2t1 =
υ cosθ ⋅ 2υ sinθ
= υ sin2θ / g
2
g
...1
Where g is 9.8 m/sec2, and t 1 is the time for the ball
to reach the maximum height.
Page 8
In figure 2, the right, center, and left curves are for
the ball to land at the front rim, the hoop center, and the back
rim, respectively. The crescent area between the left and the
right curves indicates the acceptable combinations of θ and
v that would allow the ball to land between the front and
back of the rim, which would be considered a good shot.
Players never have perfect control over the speed
and angle that they shoot the ball. To have a high percentage
of good shots, certainly it is wise to shoot the ball in the zone
where the acceptance area is the largest. From figure 2, the
largest acceptance area is located at the front end of the
crescent, with θ equal to 45°. If x is set to different values (but
not too small), similar curves are produced by Isaac.
Phys 13 News / Winter 2002
Hence, we have reached a conclusion. When the
ball is released at the same height as the hoop, the optimal
arc angles for shooting a basketball is 45°. At this angle, the
speed that sends the ball to the center is the minimum speed
required by the ball to reach the hoop center at any angle.
General Case: The ball is released y feet lower than the hoop.
You will note that as in the special case, equation 6
indicates that there are generally two θ values for each v
value.
When υ 4 − 2 gyυ 2 − g 2 x2 < 0 in equation 6, there
is no possible angle for the ball to reach B.
In most situations the hoop is y ft. higher than the
height of the ball (Figure 1). Taking a clue from the simplified
case, let us find the minimum speed and the associated arc
angle to send the ball from A to B.
The trajectory of the basketball can be defined by
the following set of equations:
4
2
2 2
υ − 2 gyυv − g x = 0 then υ = υo p and θ = θ op . Since
2
2
(
υ op = g y +
2
tanθ op =
(TheballisatHmax .)
1
1
2
K3
2
K4
2
1
K2
2
υop
y =
(t
1
2
+ t2 )( t 1 − t 2 ) =
g
(t
1
2
+ t 2 ) [ 2t 1 − ( t 1 + t 2 ) ]
g
x
⋅
2 υ cosθ
= xtanθ −
= xtanθ −
g
 2υ sinθ
 g
⋅
x
⋅
−
x
υ cosθ
y+
2
=
2
)
x +y
2
gx
...7
2
...8
x
(
tanθ op = tanβ + sec β
tanβ + sec β =
1 + sin β
cos β
=
1 − cos (90° + β )
sin (90° + β )
= tan
tan β =
(
y
x
90° + β
2
)
)
β
2
Using Isaac, the combinations of θ and v can again
be plotted and the curves generated as in the special case.
In figure 3, x has again been set to 13.25 feet, 14 feet, and
14.75 feet and the player shot height to 7 feet.



2
2 υ cos θ
gx
2υ
2
2
2
( tan θ + 1 )
2
2
(
gx tan θ − 2υ xtanθ + 2 yυ + gx
2
2
v op is the minimum speed required to send the ball
from A to B, and there is one θop value for each v op value.
Substituting in equations 1 and 2 we get:
y=
y +x
⇒ θ op = 45° +
From equation 3 we can obtain the following:
g
2
υop > 0 we obtain the following:
2

 vsinθ = gt

 x = vcosθ ⋅ ( t + t )

g
 y = (t − t )

2
When
2
2
2
)=0
...5
If we take the quadratic root of equation 5 we obtain:
υ ±
2
tanθ =
υ − 2gyυ − g x
4
2
gx
Phys 13 News / Winter 2002
2
2
...6
Figure 3: The "shot height" is at a constant 7 ft.
Page 9
In figure 3, the largest acceptance area is again
located at the front end of the crescent. Notice that the
crescent has shifted up but retains the same general shape
as the special case. Also, the largest acceptance area is
located about the point where θ = θop and v = v op.
A calibrated CCD camera system (nicknamed
ArcMaster) has been used to record the trajectory of the
basketball as it flies through the air. The set-up is shown in
figure 4. The camera head was located 14 feet away from the
player-hoop plane. The camera head was on the same level
as the hoop, which typically is 10 feet high.
Hence we can conclude that, for a vertical offset
between the player shot height and the hoop of y, the
optimal arc angle for shooting the basketball is θop. At this
angle, the speed that sends the ball to the center of the hoop
is v op.
A table of θop and v op values for different shot release
heights versus a standard hoop height of 10 feet can be
calculated and is presented below. You will note that the
optimal angles are all greater than 45°.
Figure 4
Table 1
Optimal Arc Angles for Shooting the Basketball from Typical
Heights
Shot
Height
9'
11'
14'
17'
20'
24'
5’ 5” Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
58.28
21.64
.79
56.12
22.95
.86
53.91
24.85
.96
52.41
26.65
1.05
51.34
28.35
1.13
50.31
30.49
1.23
6’
Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
56.98
21.10
.78
54.99
22.47
.85
52.97
24.43
.95
51.62
26.27
1.04
50.65
28.01
1.13
49.73
30.18
1.23
6’ 6” Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
55.63
20.57
.78
53.83
21.99
.85
52.02
24.01
.95
50.82
25.90
1.04
49.96
27.66
1.12
49.15
29.87
1.23
7’
Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
54.22
20.04
.77
52.63
21.52
.84
51.05
23.60
.94
50.00
25.52
1.04
49.27
27.33
1.12
48.56
29.57
1.23
7’ 6” Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
52.76
19.51
.76
51.40
21.05
.84
50.06
23.19
.94
49.18
25.16
1.03
48.56
26.99
1.12
47.97
29.26
1.23
8’
51.26
18.99
.76
50.15
20.59
.83
49.07
22.78
.94
48.35
24.79
1.03
47.86
26.66
1.12
47.38
28.96
1.22
Shot Angle (Deg.)
Shot Speed (ft/s)
Flight Time (sec.)
The player would get ready for a shot when he heard
a two-click audio signal. He would then have to make a shot
within one second after he heard a third click. The camera
would then record for two and a half seconds after this click.
To increase contrast, a black curtain was placed in a plane
six feet behind the player-hoop plane. A recorded image is
shown in figure 5. Also shown are the best-fitted trajectory
and the calculated shot speed and shot angle.
ArcMaster can keep records of shot angles and shot
speeds along with the shooter’s name. This makes possible
future study of the statistics and trends of shooters.
CCD Camera Measuring System:
Using recorded video images of a basketball shot,
the shot angles and shot speeds can be calculated. If
modern CCD cameras and machine vision technologies are
used, the shot speeds and the shot angles can be calculated
in less than 10 seconds. Comparing them to the optimal shot
angles and shot speeds, the player can attempt necessary
adjustments for his or her next shot. The shooting drill will
be more efficient. The player can improve faster or even
Page 10
Phys 13 News / Winter 2002
TEAMS ENHANCING LEARNING
WITH TECHNOLOGY: THE CLARICA
SCHOLARS PROGRAM
by Tracy Penny Light
LT3, University of Waterloo
The goal of the Clarica Scholars Program is to assist
in the development of teaching with technology to improve
student learning. That goal started to become realized last
summer when teams of high school teachers and students
(2 each per team) from across Canada attended a one-week
program at the University of Waterloo. While on campus
they began to design, develop and evaluate learning
support software to address instructional challenges in
their history and geography classrooms. In the summer of
2002, the program will work with teams to address instructional challenges in Science (chemistry, physics, biology,
and earth sciences) and Canadian Studies (history, geography, english and economics). In addition to the training
week, teams receive continuing assistance to further develop, program, implement and then evaluate their projects
as well as share the resulting learning support systems with
other Clarica Scholars. The Director, Tracy Penny Light,
notes, "this is really a unique program because it gets
teachers and students working together to come up with
ways to help students learn better." The process of
learnware design is not something that can be completed
designing a learning tool as the teams do at Waterloo is a
long one and typically, more successful if extensive time
and thinking are put into the process." Furthermore, it is
not just about technology although learning how to use
different software programs like Flash and Dreamweaver is
an important component of the weeklong program. "Pedagogy should lead the technology," says Penny, "not the
other way around." Because of this approach, projects
prototyped at Waterloo are well on their way to being
cutting-edge learning tools: the teams figure out how to
teach students better, BEFORE the project is developed.
The projects, then, are not just examples of how to incorporate technology into the classroom but instead use the
technology to help teach students something that was
either not possible or problematic before the introduction
of the learning tool. Teams in 2001 tackled a wide variety
of instructional challenges, which included learning about
water quality, how to use GIS systems, sovereignty and
writing essays. The work is paying off and Penny notes:
"teams from last summer more than surpassed our expectations of what could be created, both while they were oncampus as well as what they are accomplishing back in their
own schools. That level of success promises to be repeated
this upcoming summer with the teams from Science and
Canadian Studies.
Phys 13 News / Winter 2002
For more information about the Clarica Scholars
Program at the University of Waterloo, you can visit the
website http://LT3.uwaterloo.ca/claricascholars/community, or you can contact Jill Porter, Manager, Clarica Scholars Program by email: [email protected].
QUESTION AND ANSWER CORNER
by Jeff Chen
Department of Physics, University of Waterloo
Q: How do digital thermometers work?
A: The first thing to which a doctor or a nurse subjects a
patient is to measure his/her body temperature. Thermometers are of two types: mercury based or the newer digital
thermometer. The physics behind these different types of
body thermometers are completely different.
A mercury-based thermometer contains a small
amount of mercury enclosed within a small volume, usually
located near the tip of the thermometer (the shiny part). Since
mercury is a liquid that can expand in volume rapidly when
the temperature increases (the phenomena of thermal expansion), one can use the volume of the mercury inside a
thermometer as a indirect measure of temperature. Contained within a very thin capillary (the handle part of the
thermometer) the mercury's length, which varies with temperature, will consequently reflect the object's temperature
with which it is in contact. This type of classical thermometer
has been used not only to find a person's body temperature
but also for temperature measurements in general. Historically, temperature scales used in scientific research are also
based on this technique.
Digital thermometers depend on temperature sensitive resistors and are the product of today’s microchip
technology. The sensing resistor has a resistance which
varies dramatically with small changes in its temperature.
The electric circuit coupled to the sensing resistor detects
the change in its resistance and converts it to an electric
signal, a voltage or current change. A small “computer”
(microchip) inside the thermometer then converts the
changing signal to a numeric readout.
Page 11
NOBEL LAUREATES
DERANGED SCIENTISTS
#5
FRANCE
The solution to the fifth puzzle in this series, with details of the laureates' citations, is as follows:
TREKSAL 6 Alfred KASTLER
Physics
(1966)
For the discovery and development of optical methods
for studying Hertzian resonances in atoms.
LOEBRIDGE 6 Prince Louis-Victor
Raymond DE BROGLIE
Physics
(1929)
For his discovery of the wave nature of electrons.
HAPRACK6 Georges CHARPAK
Physics
(1992)
For his invention and development of particle detectors,
in particular the multiwire proportional chamber.
DRAGRING6 Victor GRIGNARD
Chemistry
(1912)
For the discovery of the so-called Grignard reagent,
which in recent years has greatly advanced the progress
of organic chemistry.
Physics
(1926)
For his work on the discontinuous structure of matter,
and especially for his discovery of sedimentation
equilibrium.
SOLUTION (from underlined letters)
Jean Baptiste PERRIN
Correct entries were received from Susan Hewitt, Russ Flint, C. Lannoy, Christine Wilkolaski, Chris Curran, Brenda Gerein and
Tomas Liko. Our book prize winner, is Susan Hewitt, who describes herself as a Deranged Physics Teacher from Lisgar Collegiate
Institute in Ottawa. A copy of "Seven Ideas that Shoot the Universe" (2nd Edition) by Nathan Spielberg and Bryon D. Anderson,
published by Wiley, has been mailed to you. Congratulations, Susan!
DERANGED SCIENTISTS
NOBEL LAUREATES
# 6 U.S.A.
by Tony Anderson
Unscramble the letters below to form the names of four Nobel Laureates from the country indicated above. Then use the
letters with the asterisks to find the name of a fifth Laureate from the same country. Send your entries to reach us before April
21, 2002. (Please include your full name, affiliation and address).
MAIL:
FAX:
E-MAIL:
A. Anderson, Department of Physics, University of Waterloo, Waterloo, Ontario N2L 3G1, Canada
(519) 746-8115 - attention of A. Anderson
[email protected]
A draw for a book prize will be made from all correct entries. This contest is open to all readers of Phys 13 News. The solution
and the winner's name will be given in a future issue of the magazine.
CLOMSHINE
*
*
*
NOMPOCT
*
*
CHOYSKEL
VANSIDOS
*
*
*
S O L U T I O N:
Page 12
Phys 13 News / Winter 2002
THE SIN BIN
A problem corner intended to stimulate some
reader participation. The best solution to the problem will
receive a book prize. Send your favourite problems and
solutions to our BINkeeper, John Vanderkooy,
[email protected].
Problem 101
The problem below was originally posed by Neil
Isenor, a retiree from our department. Initially I was not sure
that it could be solved in a reasonable way. But with
persistence (and some insight) it does succumb to analysis.
I haven’t yet solved the final equation analytically, and you
may resort to a numerical approach.
Figure 1
Consider three identical frictionless barrels of
diameter 1 metre packed into a triangular formation on a flat
surface, as shown in the diagram. When they are released
from rest, what is the final speed of the two lower barrels?
Figure 2
When sliding starts, what fraction of the belt length is
hanging on the long side?
Another way to pose the problem is to ask: what is
the height of the upper barrel above the surface when the
two lower barrels have just lost contact with the upper one?
Problem 100 from the last issue:
The problem this issue is a variant of one proposed
by Neil Isenor, who retired from our Physics Department
some years ago. It requires some careful thought and
analysis!
A thin flexible belt of uniform mass density hangs
over a “rough” horizontal cylinder, hanging equally over
both sides. The cylinder radius R is small compared to the
length L of the belt. The coefficient of friction between the
belt and cylinder is µ=0.5. The cylinder is slowly turned
until the belt begins to slide.
Phys 13 News / Winter 2002
You will need to calculate the varying tension in the belt as
it goes around the cylinder. As a hint, you should find that
this tension varies exponentially with the wrap angle!
There is an implicit assumption in this problem that the belt
must be allowed to “creep” over the cylinder such that the
friction force is always the maximum allowed by the normal
force. That makes sense, since the final state of the belt is
incipient slippage. Happy calculating!
The Solution!
There were essentially correct solutions sent in by Chris
Curran, Leigh Palmer and Norman Cowan. Thanks to you
stalwarts! Chris gets the book prize this issue.
Figure 1 shows the belt draped over a pulley, with labels for
the variables involved. We need to work out the details of
the friction for an element of the belt around the cylinder,
as shown in the free-body diagram in Figure 2. The small
angle dθ encompasses a length dl=R dθ of the belt, acted
Page 13
on by tangential tensions T on the left and T+dT on the
right. The force dN is the normal force of the pulley on the
element of belt. To first order we can ignore the change in
tension dT to work out the force equilibrium. Each tension
is at an angle d θ/2 with respect to the tangent to the pulley
in the middle of the segment. Thus to first order
2 T sin(dθ/2) = dN
and since sin(α)≈α for small angles we get T d θ = dN. If we
assume that the force of friction is the maximum allowed,
then the increase in the tangential tension dT is
dT = µ dN = µ T dθ.
This results in
dT/T = µ dθ,
which solves as an exponential relationship
µθ
T = T1 e ,
where T1 is the tension as the belt begins its wrap around
the cylinder, as shown in the figure. Incidentally, the
foregoing theory explains why a sailor can wrap a rope
around a harbour bollard several times and stop the motion
of an ocean liner. The exponential really builds up in a few
turns!
The belt leaves the cylinder on the other side π
radians away, where the tension is T2, so
µπ
T2/T1 = e .
These tensions must support the dangling ends of the belt,
and since we have assumed that the friction force is the
maximum static value allowed, the lengths so calculated are
the solution to the incipient slippage problem. We set
l1+l2=L, the total length of the belt, because we can ignore
the radius of the cylinder with respect to L.
From the Editor:
Many thanks to the kind readers who responded
with comments to last summer's survey in Physics 13
News. It appears that by and large we have the right
editorial balance. Nevertheless many of you did express
an interest in more articles involving research projects and
lab activities for students, with a view on the new curriculum. We couldn't agree more and in this issue we'll start
with an article on "Swingers, Sliders and Topplers" in the
spirit of such favourite demos. We welcome articles by
anyone who would like to share their favourite physics
demo.
In case you are not aware, Physics 13 News is
present in a limited form on the web at:
http://www.science.uwaterloo.ca/physics/p13news .
On a trial basis we plan to make complete future editions
available on this site and look forward to hearing which
format, electronic or printed (or possibly both) you think
is the preferred medium.
Finally, a number of subscriptions have been
cancelled (at least two notices have gone out but some
may have been missed) because a renewal has not been
received for some time. If you find yourself among these,
please don't hesitate to contact us to renew your subscription. You can find subscription information in each issue
and at the above web site.
Phys 13 News is published four times a year by the Physics
Department of the University of Waterloo. Our policy is
to publish anything relevant to high school and first-year
university physics, or of interest to high school physics
teachers and their senior students. Letters, ideas, and
articles of general interest with respect to physics are
welcome by the editor. You can reach the editor by email
at: [email protected]. Alternatively you can send all
correspondence to:
If λ is the mass per unit length of the belt, then the tensions
must satisfy
T1 = λ l1 g = λ (L−l2) g,
T2 = λ l2 g.
Substituting these values for the tensions into the earlier
exponential relationship gives the solution for l 2 in terms of
L (Notice that λ and g don’t matter since we need the ratio
of the tensions):
−µπ
l2 = L / (1 + e
Editor:
Ghnter Scholz
Editorial Board:
Tony Anderson, Mike Fich, Jan
Kycia, Jim Martin, Donna Strickland
and John Vanderkooy
Publisher:
Judy McDonnell, Debbie Guenther
).
Putting in µ=0.5, we find the ratio l2/L is 0.828, so as the
cylinder is turned, 82.8% of the belt is hanging over the side
before it slips.
Page 14
Phys 13 News, Physics Department
University of Waterloo
Waterloo, Ontario
N2L 3G1
Printing by Susan Schaffer and the University of
Waterloo Graphic Services Department.
Phys 13 News / Winter 2002
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