MEI Conference, July 2008
Six more gems in mechanics
I have selected some more examples that have intrinsic interest and are nice demonstrations
of the application of a principle or technique.
1
Two particles are projected at the same time and the only force acting on them is that
of gravity. Particle A has initial position r0 and velocity u; particle B has initial
position R0 and velocity U; all units are SI. The particles meet no obstructions.
What is the condition that the particles collide with each other?
M1
2
In many self-service cafes, plates are stored ready for use in a stack where the top
plate is at counter level however many plates are in the stack. This may be achieved
very simply by suspending the plates in a rack supported by suitable elastic strings, as
shown in the diagram. How does it work?
M3
3
Two objects, A and B, slide on a uniform plane inclined at an angle α to the
horizontal. They are joined together by a light, rigid coupling that is parallel to the
plane. Object A has mass mA and resistance to motion FA; object B has mass mB and
resistance to motion FB. When the objects are being pulled up the plane by a force of
magnitude P, the force in the coupling is T. If FA and FB are independent of α and
the value of g, then so is the equation connecting P and T.
M1
4
A fly of mass m stands on the edge of a circular disc of mass M. The disc is on a
smooth horizontal table. Initially the disc is at rest. Describe 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.
M2
5
A particle moves on a rectangular horizontal table (not necessarily smooth) that has
smooth raised edges. If the coefficient of restitution between the particle and each
of the edges is the same then after contact with two adjacent edges the particle is
travelling parallel to its direction before its first contact.
M2
6
DO is a 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.
In an alternative formulation, O is the highest point of the circle and the particle
starts from rest and travels down a chord OD.
M1
[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)]
David Holland
1 of 1
version 3; MEI Conference 2008
MEI conference 2008
Six more gems in Mechanics
Notes
1 (M1)
Two particles are projected at the same time and
the only force acting on them is that of gravity.
Particle A has initial position r0 and velocity u;
particle B has initial position R0 and velocity U;
all units are SI. The particles meet no obstructions.
What is the condition that the particles collide with
each other?
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’.
David Holland
1 of 12
version 3; MEI Conference 2008
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.
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 towards one another.
u
A (r0)
U
r0 – R0
B (R0)
– u
U
U–u
David Holland
2 of 12
version 3; MEI Conference 2008
2 (M3)
In many self-service cafes, plates are stored ready
for use in a stack where the top plate is at counter
level however many identical plates are in the
stack.
This may be achieved very simply by suspending
the plates in a rack supported by suitable springs,
as shown in the diagram.
All that is necessary is that the springs have a
combined stiffness so that when one plate is added,
the springs extend by an amount equal to the
increase in height of the stack.
Hooke’s Law states that the total extension of a
spring or elastic string and the tension in it are
directly proportional.
Because this is a linear relationship, addition of a
plate to the stack will increase the total tension in
springs by the weight of the plate and cause exactly
the same extension in the springs however many
plates are already in the stack.
David Holland
3 of 12
version 3; MEI Conference 2008
weight of the plates
1 plate added
1 plate added
extension of the springs
Suppose the combined stiffness of the springs is
k N m -1, the weight of each plate is w N and the
addition of a plate increases the height of the stack
by y m. We have, by HL, that the extension of the
springs, e m, caused by adding a plate is given by
w
w = ke so e = and we need to adjust the springs
k
to make e as close as possible to y.
In practice, this seems to be done by changing the
number of springs. Of course, one doesn’t accept
solutions with e > y as this will lead to the top of
the pile getting further below the counter as more
plates are added to the stack.
Many students find this example counter-intuitive.
They ‘feel’ that just one plate could be above the
counter level but that a stack of, say, twenty plates
could be below it.
David Holland
4 of 12
version 3; MEI Conference 2008
3 (M1)
Two objects, A and B, slide on a uniform plane
inclined at an angle α to the horizontal. They are
joined by a light, rigid coupling that is parallel to
the plane. Object A has mass mA and resistance to
motion FA; object B has mass mB and resistance to
motion FB. When the objects are being pulled up
the plane by a force of magnitude P, the force in
the coupling is T.
If FA and FB are independent of α and the value of
g, then so is the equation connecting P and T.
One configuration is shown in the diagram. All
units are SI.
RB
FB
RA
FA
B
T
mBg
A
α
P
mAg
Let the acceleration up the plane be a
N2L up the plane on B gives
P − T − FB − mB g sin α = mBa
(1)
N2L up the plane on A gives
T − FA − mA g sin α = mA a
(2)
David Holland
5 of 12
version 3; MEI Conference 2008
Eliminating a gives
mB ( T − FA − mA g sin α ) = mA ( P − T − FB − mB g sin α )
so
mB (T − FA ) − mBmA g sin α
= mA ( P − T − FB ) − mA mB g sin α
giving mB (T − FA ) = mA ( P − T − FB ) .
If FA and FB are independent of α and g then so is
this result.
⎛ m ⎞
m
Solving for P gives P = ⎜ 1 + B ⎟ T + FB − B FA (3)
mA
⎝ mA ⎠
The cases when FA = 0 and FB = 0 with α = 0 and
α = 90° are easily worked out from 1st principles.
A further result
Suppose FA and FB are ordinary frictional forces
and the coefficient of friction between each object
and the plane is μ.
In this case FA = μ mA g cos α and FB = μ mB g cos α
⎛ m ⎞
and substituting in (3) gives P = ⎜ 1 + B ⎟ T which
⎝ mA ⎠
is independent of α and g.
David Holland
6 of 12
version 3; MEI Conference 2008
4 (M2)
A fly of mass m stands on the edge of a circular
disc of mass M. The disc is on a smooth horizontal
table. Initially the disc is at rest. Describe 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.
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
7 of 12
version 3; MEI Conference 2008
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
M
B
R
Let the distance of the common centre of mass of
the fly and the disc, G, be a distance xG from A.
xG =
mx + MR
m+M
and so the speed of G is xG =
mx
relative to A.
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 xG =
David Holland
8 of 12
version 3; MEI Conference 2008
5 (M2)
A particle moves on a rectangular horizontal table
(not necessarily smooth) that has smooth raised
edges. If the coefficient of restitution between the
particle and each of the edges is the same then
after contact with two adjacent edges the particle is
travelling parallel to its direction before its first
contact.
This is a well-known result in the case where the
collisions with the barrier are perfectly elastic and
the table is smooth but this more general case is
also true.
Consider a collision with a smooth barrier where
the coefficient of restitution is e.
⎛u⎞
⎛u⎞
before
⎜ ev ⎟ after
⎜ −v ⎟
⎝ ⎠
⎝ ⎠
j
α
β
i
It is a well-known result that tan β = e tan α .
Suppose that the velocity of the particle just before
the collision is ui – vj and the velocity just after the
collision is xi + yj.
By PCLM in the i direction x = u.
By NEL in the j direction y = ev.
Hence the velocity after the collision is ui + evj.
v
ev
so tan β = e tan α .
Now tan α = and tan β =
u
u
David Holland
9 of 12
version 3; MEI Conference 2008
At this stage we note that the result is independent
of the speed of the particle.
C
δ
Applying the result gives
γ
β
D
tan β = e tan α
and tan δ = e tan γ
α
B
A
Now γ = 90 − β so
tan δ = e tan γ = e tan(90° − β ) = e cot β
e
e
Also e cot β =
=
= cot α
tan β e tan α
Thus tan δ = cot α and so δ + α = 90° and AB is
parallel to DC.
It doesn’t matter if the particle slows up between
the collisions.
On a snooker table the situation is more
complicated but players use the result as a guide.
David Holland
10 of 12
version 3; MEI Conference 2008
6 (M1)
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.
[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)]
David Holland
11 of 12
version 3; MEI Conference 2008
D
R
C
mg
α
N
r
α
horizontal
O
OD is at α 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 α
Let the acceleration of the particle down DO be a.
N2L in the direction DO gives
mg sin α = ma , so a = g sin α .
Using s = ut + 12 at 2 on the motion from D to O
gives 2r sin α = 0 + 12 g sin α t 2
r
4r
so t =
and t = 2
(since t > 0).
g
g
Hence t is independent of the position of D on the
circle.
2
David Holland
12 of 12
version 3; MEI Conference 2008