Transcript
This booklet is a sequel to a similar col-
lection of problems on kinematics. Sim-
ilarly to that collection the aim here is
to present the most important ideas us-
ing which one can solve most (
>
95%) of
olympiad problems on mechanics. Usu-
ally a problem is stated first, and is fol-
lowed by some relevant ideas and sug-
gestions (letter ‘K’ in front of the number
of an idea refers to the correspondingly
numbered idea in the kinematics book-
let). The answers to the problems are lis-
tedattheendofthebooklet. Theyarepre-
cededbyquitedetailedhints(nofullsolu-
tions), but think carefully before reading
the hints as a last resort!
The guiding principle of this booklet
argues that almost all olympiad problems
are “variations” on a specific set of top-
ics — the solutions follow from corres-
ponding solution ideas. Usually it is not
very hard to recognize the right idea for
a given problem, having studied enough
solution ideas. Discovering all the ne-
cessary ideas during the actual solving
would certainly show much more creativ-
ity and offer a greater joy, but the skill of
conceiving ideas is unfortunately difficult
(or even impracticable) to learn or teach.
Moreover, it may take a long time to reach
a new idea, and those relying on trying it
during an olympiad would be in disad-
vantage in comparison to those who have
mastered the ideas.
In science as a whole, solution ideas
play a similar role as in olympiads:
most scientific papers apply and combine
known ideas for solving new (or worse,
old) problems, at best developing and
generalising the ideas. Genuinely new
good ideas occur extremely rarely and
many of them are later known as master-
pieces of science. However, as the whole
repertoire of scientific ideas encompasses
immensely more than mere mechanics, it
is not so easy to remember and utilise
them in right places. The respective skill
is highly valued; an especial achievement
would be employing a well-known idea
in an unconventional (unexpected, novel)
situation.
In addition to ideas, the booklet also
presents “facts” and “methods”. The dis-
tinction is largely arbitrary, some facts
could have been called methods and vice
versa. In principle, an “idea” should have
wider and/or more creative applications
than a “fact”; a “method” is a universal
and conventionalized “idea”.
Several sources have been used for the
problems: Estonian olympiads’ regional
and national rounds, journal “Kvant”,
Russian and Soviet Union’s olympi-
ads; some problems have been modified
(either easier or tougher), some are “folk-
lore” (origins unknown).
For problems on statics the solution is
usually standard: we have to write down
the condition of force balance for the
x
-,
y
- and (if necessary)
z
-components; often
the condition of torque balance must be
added. Usually the main ingenuity lies in
IDEA
1:
for instance, the reaction force between
two bodies or the tensile force in a string
(or a rod). To zero as many forces as pos-
sible it is worthwhile to note that
a)
the
axes may not be perpendicular;
b)
if the
system consists of several bodies, then a
different set of axes may be chosen for
each body.
IDEA
2:
For example, if we choose the pivot to
be at the contact point of two bodies,
thenthemomentarmsofthefrictionforce
between the bodies and of their reaction
force are both zero.
IDEA
3:
x
y
An equation for the torques can be writ-
ten about any pivot point (“axis” of rota-
tion). In principle, we could write several
equations for several pivots at the same
time, but together with the equations for
the forces
the maximum number of linearly
independent equations equals the number of
degrees of freedom of the body
(three in the
two-dimensional case, as the body can
rotate in a plane and shift along the
x
-
and
y
-axis). Accordingly, all is fine if we
writeoneforcesequationandtwotorques
equations (or just three torques equations
— as long as the pivots do not lie on a
straight line); on the other hand, if we
wrote two equations of both types, then
one of the four equations would always
be a redundant consequence of the three
others and needless to write down.
So, an equation for the force balance
may be replaced by an equation for the
torque balance about an additional pivot.
Such a substitution may turn out to be
useful if the unwanted (uninteresting)
forces are unparallel, because a choice of
a projection axis can zero only one force
in the balance of forces, while a choice of
a pivot for the torques can zero two forces
at once.
PROB
1.
An end of a light wire rod is
bent into a hoop of radius
r
. The straight
part of the rod has length
l
; a ball of mass
M
is attached to the other end of the rod.
The pendulum thus formed is hung by the
hoop onto a revolving shaft. The coef-
ficient of friction between the shaft and
the hoop is
µ
. Find the equilibrium angle
between the rod and the vertical.
r
l
M
ω
µ
Here we mainly need idea 2 with some
simplification offered by
FACT
1:
α
tan
α
=
µ
PROB
2.
On an incline with slope angle
α
thereliesacylinderwithmass
M
, itsaxis
being horizontal. A small block with mass
m
isplacedinsideit. Thecoefficientoffric-
tion between the block and the cylinder is
µ
; the incline is nonslippery. What is the
maximum slope angle
α
for the cylinder
to stay at rest? The block is much smaller
than the radius of the cylinder.
α
m
M
Here we can again use fact 1 and idea 2 if
we add
IDEA
4:
Further, the net force (or torque) is the
sum of forces (torques) acting on the con-
stituents (the effort is eased as the internal
forces are needless — they cancel each
other out). In our case, it is useful to as-
semble such a whole system from the cyl-
inder and the block.
PROB
3.
Three identical rods are connec-
ted by hinges to each other, the outmost
ones are hinged to a ceiling at points
A
and
B
. The distance between these points
is twice the length of a rod. A weight of
mass
m
is hanged onto hinge
C
. At least
how strong a force onto hinge
D
is neces-
sarytokeepthesystemstationarywiththe
rod
CD
horizontal?
B
m
A
C
D
F
Again we can use idea 2. The work is also
aided by
FACT
2:
Indeed, the net external force
⃗
F
onto
either point of application of the forces
must point along the rod, as its torque
with respect to the other point of applic-
ation must be zero. In addition to the ex-
ternal forces, the point is acted on by ten-
sionforce
⃗
T
thatmustcompensatetherest
of the forces, so
⃗
F
=
−
⃗
T
.
Some ideas are very universal, espe-
cially the mathematical ones.
IDEA
K5:
for example, the shortest path from a
point to a plane is perpendicular to it.
PROB
4.
What is the minimum force
needed to dislodge a block of mass
m
rest-
ing on an inclined plane of slope angle
α
,
if the coefficient of friction is
µ
? Investig-
ate the cases when
a)
α
=
0;
b)
0
<
α
<
arctan
µ
.
α
F
m
IDEA
5:
Fact 1, or rather its following generalisa-
tion, turns out to be of use:
FACT
3:
arctan
µ
Thisfactisalsobeneficialinthenextprob-
lem.
PROB
5.
Ablockrestsonaninclinedsur-
facewithslopeangle
α
. Thesurfacemoves
with a horizontal acceleration
a
which lies
in the same vertical plane as a normal vec-
tor to the surface. Determine the values of
the coefficient of friction
µ
that allow the
block to remain still.
α
a
µ
Here we are helped by the very universal
IDEA
6:
To clarify: in a translationally moving ref-
erence frame we can re-establish New-
ton’s laws by imagining that every body
with mass
m
is additionally acted on by
an inertial force
−
m
⃗
a
where
⃗
a
is the ac-
celeration of the frame of reference. Note
that that the fictitious force is totally ana-
logous to the gravitational force and (as
an aside) their equivalence is the corner-
stone of the theory of general relativity
(more specifically, it assumes the inertial
and gravitational forces to be indistin-
guishable in any local measurement).
IDEA
7:
PROB
6.
A cylinder with radius
R
spins
around its axis with an angular speed
ω
. On its inner surface there lies a small
block; the coefficient of friction between
the block and the inner surface of the cyl-
inder is
µ
. Find the values of
ω
for which
the block does not slip (stays still with re-
spect to the cylinder). Consider the cases
where
(a)
the axis of the cylinder is hori-
zontal;
(b)
the axis is inclined by angle
α
with respect to the horizon.
ω
α
IDEA
8:
m
ω
2
⃗
R
ω
R
Warning:
in this idea, the axis of rotation
must be actual, not instantaneous.
For the
last problem, recall idea K5 and fact 3; for
part
(b)
, add
IDEA
9:
PROB
7.
A hollow cylinder with mass
m
and radius
R
stands on a horizontal sur-
face with its smooth flat end in contact the
surface everywhere. A thread has been
wound around it and its free end is pulled
with velocity
v
in parallel to the thread.
Find the speed of the cylinder. Consider
two cases:
(a)
the coefficient of friction
between the surface and the cylinder is
zero everywhere except for a thin straight
band (much thinner than the radius of the
cylinder) with a coefficient of friction of
µ
, the band is parallel to the thread and
its distance to the thread
a
<
2
R
(the fig-
ure shows a top-down view);
(b)
the coef-
ficient of friction is
µ
everywhere.
Hint:
any planar motion of a rigid body can
be viewed as rotation around an instant
centre of rotation, i.e. the velocity vector
of any point of the body is the same as if
the instant centre were the real axis of ro-
tation.
v
a
µ
This is quite a hard problem. It is use-
ful to note
IDEA
10:
Also remember ideas 1 and 2. The latter
can be replaced with its consequence,
FACT
4:
Another useful fact is
FACT
5:
From time to time some mathematical
tricks are also of use; here it is the prop-
ertyofinscribed angles(Thales’theorem),
FACT
6:
The property of inscribed angles is
also useful in the next problem, if we add
(somewhat trivial)
IDEA
11:
PROB
8.
A light wire is bent into a right
angle and a heavy ball is attached to the
bend. The wire is placed onto supports
with height difference
h
and horizontal
distance
a
. Find the position of the wire
in its equilibrium. Express the position as
the angle between the bisector of the right
angle and the vertical. Neglect any fric-
tion between the wire and the supports;
the supports have little grooves keeping
all motion in the plane of the wire and the
figure.
α
a
h
PROB
9.
A rod with length
l
is hinged
to a ceiling with height
h
<
l
. Under-
neath, a board is being dragged on the
floor. The rod is meant to block the move-
ment the board in one direction while al-
lowing it move in the opposite direction.
What condition should be fulfilled for it
to do its job? The coefficient of friction is
µ
1
between the board and the rod, and
µ
2
between the board and the floor.
α
µ
1
µ
2
Let’s remember fact 3: if the relative
sliding between two bodies has a known
direction, then the direction of the sum of
the friction and reaction force vectors is
always uniquely determined by the coef-
ficient of friction. If a force makes one of
the bodies move in such a way that the
reaction force grows, then they jam: the
larger the forces we try to drag the bod-
ies with, the larger friction and reaction
forces restrain them.
IDEA
12:
PROB
10.
Four long and four half as
long rods are hinged to each other form-
ing three identical rhombi. One end of
the contraption is hinged to a ceiling,
the other one is attached to a weight of
mass
m
. The hinge next to the weight is
connected to the hinge above by a string.
Find the tension force in the string.
m
This problem is the easiest to solve us-
ing
the method of virtual displacement
.
METHOD
1:
∆
x
T
∆
x
∆Π
T
=
∆Π
/
∆
x
Generalisation:
if some additional external
forces
⃗
F
i
(
i
=
1,2,...) act on the sys-
temwiththedisplacementsoftheirpoints
of action being
δ
⃗
x
i
, while the interest-
ing string or rod undergoes a virtual
lengthening of
∆
x
, then
T
= (
∆Π
−
∑
i
δ
⃗
x
i
·
⃗
F
i
)
/
∆
x
.
The method can also be used for find-
ing some other forces than tension (for
example, in problems about pulleys): by
imaginarily shifting the point of action of
the unknown force one can find the pro-
jection of this force onto the direction of
the virtual displacement.
PROB
11.
A rope with mass
m
is hung
from the ceiling by its both ends and a
weight with mass
M
is attached to its
centre. The tangent to the rope at its either
end forms angle
α
with the ceiling. What
is the angle
β
between the tangents to the
rope at the weight?
α
α
β
m
M
FACT
7:
In addition, we can employ
IDEA
13:
In fact, here we do not need the idea as a
whole, but, rather, its consequence,
FACT
8:
In problems about ropes one may some-
times use
IDEA
14:
This allows us to write down the con-
dition of torque balance for the hanging
portion of the rope (as we know the ho-
rizontal coordinate of its centre of mass).
The next problem illustrates that ap-
proach.
PROB
12.
A boy is dragging a rope
with length
L
=
50m along a horizontal
ground with a coefficient of friction of
µ
=
0.6, holding an end of the rope at height
H
=
1m from the ground. What is the
length
l
of the part of the rope not touch-
ing the ground?
PROB
13.
A light rod with length
l
is
hinged in such a way that the hinge folds
in one plane only. The hinge is spun with
angular speed
ω
around a vertical axis. A
small ball is fixed to the other end of the
rod.
(a)
Find the angular speeds for which
the vertical orientation is stable.
(b)
The
ball is now attached to another hinge and,
in turn, to another identical rod; the upper
hinge is spun in the same way. What is
now the condition of stability for the ver-
tical orientation?
ω
ω
a) b)
l
l
l
For answering about the stability of
an equilibrium, usually the following fact
works best.
IDEA
15:
∆
x
∆
ϕ
Incidentally use all formulae of approx-
imate calculation known from mathemat-
ics (sin
ϕ
≈
ϕ
and others);
IDEA
16:
f
(
x
+
∆
x
)
≈
f
(
x
) +
f
′
(
x
)
∆
x
[+
f
′′
(
x
)
∆
x
2
2
]
(
x
+
∆
x
)(
y
+
∆
y
)
≈
xy
+
x
∆
y
+
y
∆
x
The case
(b)
is substantially more diffi-
cult as the system has two degrees of free-
dom (for example, the deviation angles
∆
ϕ
1
and
∆
ϕ
2
of the rods). Although idea
15 is generalisable for more than one de-
grees of freedom, apparently it is easier to
start from idea 11.
IDEA
17:
x
=
y
=
0
Π
(
x
,
y
)
Π
(
x
,
kx
)
k
PROB
14.
If a beam with square cross-
section and very low density is placed in
water, it will turn one pair of its long op-
posite faces horizontal. This orientation,
however, becomes unstable as we increase
its density. Find the critical density when
this transition occurs. The density of wa-
ter is
ρ
v
=
1000kg/m
3
.
IDEA
18:
Indeed, consider a body with density of
the liquid and shape identical to the part
of the given body that is immersed in
the liquid. Of course it must be in equi-
librium when placed in water: whatever
point we choose to measure torques from,
the sum of moments from pressure forces
is always equal to the opposite value of
torquefromgravity. Whencalculatingthe
moments from buoyancy in this question,
it is useful to keep in mind that we can
give negative mass to bits of some body:
if two bits overlap that have the same
density with different signs, they add up
to zero density. The last suggestion can be
formulated in a more general way:
IDEA
19:
This quantity can be mass density (like in
this case), charge or current density, some
force field etc. Often this trick can be com-
bined with
IDEA
20:
This goal can be reached by applying
idea 19, but also by using appropriate
reference frames, dividing the process of
solving into several phases (where some
phases use symmetric geometry), etc.
PROB
15.
A hemispherical container is
placed upside down on a smooth hori-
zontal surface. Through a small hole
at the bottom of the container, water is
then poured in. Exactly when the con-
tainer gets full, water starts leaking from
between the table and the edge of the con-
tainer. Find the mass of the container
if water has density
ρ
and radius of the
hemisphere is
R
.
M
R
IDEA
21:
The latter is given by
pS
, where
p
is pres-
sure of the liquid near the tabletop and
S
is area of the container’s open side.
PROB
16.
A block is situated on a slope
with angle
α
, the coefficient of friction
between them is
µ
>
tan
α
. The slope
is rapidly driven back and forth in a way
that its velocity vector
⃗
u
is parallel to both
the slope and the horizontal and has con-
stant modulus
v
; the direction of
⃗
u
re-
verses abruptly after each time interval
τ
.
What will be the average velocity
w
of the
block’s motion? Assume that
g
τ
≪
v
.
x
y
z
u
t
v
−
v
τ
u
α
IDEA
22:
⟨
X
⟩
˜
X
X
=
⟨
X
⟩
+
˜
X
METHOD
2:
In this particular case, the choice of zeroth
approximation needs some explanation.
The condition
g
τ
≪
v
implies that within
one period, the block’s velocity cannot
change much. Thereforeif the block is ini-
tially slipping downwards at some velo-
city
w
and we investigate a short enough
time interval, then we can take the block’s
velocity to be constant in zeroth approx-
imation, so that it is moving in a straight
line. We can then move on to phase two
and find the average value of frictional
force, based on the motion obtained in
phase one.
PROB
17.
Let us investigate the extent to
which an iron deposit can influence water
level. Consider an iron deposit at the bot-
tom of the ocean at depth
h
=
2km. To
simplify our analysis, let us assume that
it is a spherical volume with radius 1km
with density greater from the surround-
ing rock by
∆
ρ
=
1000kg/m
3
. Presume
that this sphere touches the bottom of the
ocean with its top, i.e. that its centre is
situated at depth
r
+
h
. By how much
is the water level directly above the iron
deposit different from the average water
level?
IDEA
23:
If this was not the case, the potential en-
ergy of the liquid could be decreased by
allowing some particles on the surface to
flow along the surface to where their po-
tential energy is smaller.
IDEA
24:
The principle of superposition still holds
and a sphere’s potential only has a dif-
ferent factor: instead of
Q
/4
πε
0
r
in elec-
trostatics the gravitational potential of a
sphere with respect to infinity is
ϕ
=
−
GM
/
r
; the minus sign comes from the
fact that masses with the same sign [“+”]
attract.
PROB
18.
A horizontal platform rotates
around a vertical axis at angular velo-
city
ω
. A disk with radius
R
can freely
rotate and move up and down along a
slippery vertical axle situated at distance
d
>
R
from the platform’s axis. The disk
is pressed against the rotating platform
due to gravity, the coefficient of friction
between them is
µ
. Find the angular ve-
locity acquired by the disk. Assume that
pressure is distributed evenly over the en-
tire base of the disk.
d
R
ω
r
IDEA
25:
Thus
⃗
ω
3
=
⃗
ω
1
+
⃗
ω
2
, where
⃗
ω
1
is angu-
lar velocity of the reference frame,
⃗
ω
2
an-
gular velocity of the body in the rotating
frame of reference and
⃗
ω
3
that in the sta-
tionary frame. In this question, we can
use fact 5, ideas 2, 8, 10 and also
IDEA
K5:
METHOD
3:
Within an infinitesimal bit (period),
quantities changing in space (time) can be
taken constant (in our case, that quantity
is the direction of frictional force vector).
If necessary (see the next question), these
quantities may be summed over all bits
— this is called integration.
PROB
19.
A waxing machine consists
of a heavy disk with mass
M
densely
covered with short bristles on one side, so
that if it lies on the floor, then its weight
is evenly distributed over a circular area
with radius
R
. An electrical motor makes
the disk rotate at angular velocity
ω
, the
user compensates for the torque from fric-
tional forces by a long handle. The same
handle can be used to push the machine
back and forth along the floor. With what
force does the machine have to be pushed
to make it move at velocity
v
? Assume
that angular velocity of the disk is large,
ω
R
≫
v
, and that the force needed to
compensate for the torque can be neg-
lected. The coefficient of friction between
the bristles and the floor is
µ
.
Here we need fact 5, ideas K5 and 19 and
additionally
IDEA
26:
These pairs of points are often symmetric-
ally located. Idea 20 is relevant as well.
PROB
20.
A hexagonal pencil lies on a
slope with inclination angle
α
; the angle
between the pencil’s axis and the line of
intersectionoftheslopeandthehorizontal
is
ϕ
. Under what condition will the pencil
not roll down?
α
ϕ
IDEA
27:
What (which vector) could be expressed
in terms of its components in our case?
The only promising option is the small
shift vector of centre of mass when its
starts to move; ultimately we are only in-
terested in its vertical component.
PROB
21.
A slippery cylinder with ra-
dius
R
has been tilted to make an angle
α
between its axis and the horizontal. A
string with length
L
has been attached to
the highest point
P
of some cross-section
of the cylinder, the other end of it is tied to
a weight with mass
m
. The string takes
its equilibrium position, how long (
l
) is
the part not touching the cylinder? The
weight is shifted from its equilibrium pos-
ition in such a way that the shift vector is
parallel to the vertical plane including the
cylinder’s axis; what is the period of small
oscillations?
l
P
IDEA
28:
PROB
22.
A uniform bar with mass
m
and length
l
hangs on four identical light
wires. The wires have been attached to the
bar at distances
l
3
from one another and
are vertical, whereas the bar is horizontal.
Initially, tensions are the same in all wires,
T
0
=
mg
/4. Find tensions after one of the
outermost wires has been cut.
l
l
/3
IDEA
29:
Let us note that this statement is in ac-
cordance with idea 3 that gives the num-
ber of available equations (there can be no
more unknowns than equations). In this
particular case, we are dealing with ef-
fectively one-dimensional geometry with
no horizontal forces, but the body could
rotate (in absence of the wires). Thus
we have two degrees of freedom, corres-
pondingtoverticalandrotationalmotion.
Since the wires are identical, they must
have the same stiffness as well; the word
“wire” hints at large stiffness, i.e. deform-
ations (and the inclination angle of the
bar) are small.
A large proportion of dynamics problems
consist of finding the acceleration of some
system or forces acting between some
bodies. There are several possible ap-
proaches for solving these questions, here
we consider three of them.
METHOD
4:
x
y
z
We need the same number of equations as
we have unknowns; following idea 1 can
help to reduce that number.
PROB
23.
A block with mass
M
lies on
a slippery horizontal surface. On top of it
there is another block with mass
m
which
in turn is attached to an identical block
by a string. The string has been pulled
across a pulley situated at the corner of
the big block and the second small block
is hanging vertically. Initially, the system
is held at rest. Find the acceleration of the
big block immediately after the system is
released. You may neglect friction, as well
as masses of the string and the pulley.
M
m
m
This question can be successfully
solved using method 4, but we need two
more ideas.
IDEA
30:
IDEA
31:
Ifbodiesstartatrestorifmotionisalonga
straight line, then the same relation holds
between accelerations, since the relation
for shifts can be differentiated w.r.t. time.
This relation is usually relatively simple,
but in some problems it is easy to make a
mistake.
METHOD
5:
Method5isusefulinmanyquestionscon-
cerning wedges, where it can be difficult
to write out the condition for an object to
stayonthewedgeinthelaboratoryframe.
Applying idea 31 is also often easier in
the wedge’s frame of reference than in the
laboratory frame. Since the body defining
the reference frame is at rest, we can write
out the condition(s) of equilibrium for it.
FACT
9:
PROB
24.
A wedge has been made out
of a very light and slippery material.
Its upper surface consists of two slopes
making an angle
α
with the horizontal
and inclined towards one another. The
block is situated on a horizontal plane;
a ball with mass
m
lies at the bottom
of the hole on its upper surface. An-
other ball with mass
M
is placed higher
than the first ball and the system is re-
leased. On what condition will the small
ball with mass
m
start slipping upwards
along the slope? Friction can be neglected.
M
m
α α
The final method is based on using
gener-
alised coordinates
and originates from the-
oretical mechanics. There its description
requires relatively complicated mathem-
atical apparatus, but in most problems it
can be used in a much simpler form.
METHOD
6:
ξ
¨
ξ
ξ
Π
Π
(
ξ
)
ξ
K
=
M
˙
ξ
2
/2
M
¨
ξ
=
−
Π
′
(
ξ
)
/
M
.
Here, a dot denotes differentiation w.r.t.
time and dash w.r.t. coordinate
ξ
. Indeed,
due to conservation of energy
Π
(
ξ
) +
M
˙
ξ
2
/2
=
Const. Differentiating that
w.r.t. time and using the chain rule, we
obtain
Π
′
(
ξ
)
˙
ξ
+
M
˙
ξ
¨
ξ
=
0. We reach
the aforementioned formula after divid-
ing through by
˙
ξ
.
PROB
25.
A small block with mass
m
lies
on a wedge with angle
α
and mass
M
. The
block is attached to a rope pulled over a
pulleyattachedtothetipofthewedgeand
fixed to a horizontal wall (see the figure).
Find the acceleration of the wedge. All
surfaces are slippery (there is no friction).
m
α
M
a
= ?
Full solution of this problem is given
in the hints’ section to illustrate method 6
PROB
26.
A wedge with mass
M
and
acute angles
α
1
and
α
2
lies on a horizontal
surface. A string has been drawn across
a pulley situated at the top of the wedge,
its ends are tied to blocks with masses
m
1
and
m
2
. What will be the acceleration of
the wedge? There is no friction anywhere.
m
2
m
1
α
1
α
2
M
It may seem that there is more than one
degree of freedom in this question: the
wedge can move and the string can shift
w.r.t. the wedge. However, we are saved
by
IDEA
32:
x
x
We can use this circumstance to reduce
the effective number of degrees of free-
dom. In our particular case, the system
consists of two components and thus the
shift of component can be expressed by
that of the other.
IDEA
33:
x
X
C
=
∑
x
i
m
i
/
∑
m
i
,
m
i
i
x
i
X
C
=
∫
xdm
/
∫
dm
dm
=
ρ
(
x
,
y
,
z
)
dV
PROB
27.
Two slippery horizontal sur-
faces form a step. A block with the same
height as the step is pushed near the step,
and a cylinder with radius
r
is placed on
the gap. Both the cylinder and the block
have mass
m
. Find the normal force
N
between the cylinder and the step at the
moment when distance between the block
and the step is
√
2
r
. Initially, the block
and the step were very close together and
all bodies were at rest. Friction is zero
everywhere. Will the cylinder first separ-
ate from the block or the step?
√
−
2
r
m
m
r
It is easy to end up with very complic-
ated expressions when solving this prob-
lem, this may lead to mistakes. Therefore
it is wise to plan the solution carefully be-
fore writing down any equations.
IDEA
34:
But how to find acceleration(s) in that
case? It is entirely possible if we use
method 6, but this path leads to long
expressions. A tactical suggestion: if
you see that the solution is getting very
complicated technically, take a break and
think if there is an easier way. There is a
“coincidence” in this particular problem:
straight lines drawn from the sphere’s
centre to points of touching are perpen-
dicular; can this perhaps help? It turns
out that it does.
IDEA
35:
Let us remind what we learned in kin-
ematics:
IDEA
K29:
v
2
/
R
v
R
ε
R
ε
The centre of mass of the cylinder under-
goes rotational motion, method 6 is ne-
cessary to find angular acceleration — but
we hoped to refrain from using it. An im-
provement on idea 1 helps us out:
IDEA
36:
We can easily find the cylinder’s velocity
(and thus the radial component of accel-
eration) if we use
IDEA
37:
forces changing in time (force acting on
a moving point, moving inclined plane)
change energy as well. Idea 31 helps to
write out conservation of energy (relation
between bodies’ velocities!). To answer
the second question, we need
IDEA
38:
Also, review idea 31 for horizontal com-
ponents of accelerations.
PROB
28.
Lightwheelswithradius
R
are
attached to a heavy axle. The system rolls
alongahorizontalsurfacewhichsuddenly
turns into a slope with angle
α
. For which
angles
α
willthewheelsmovewithoutlift-
ing off, i.e. touch the surface at all times?
Mass of the wheels can be neglected. The
axle is parallel to the boundary between
horizontalandslopedsurfacesandhasve-
locity
v
.
m
α
v
IDEA
39:
If normal force has to be negative at that
point, then the body lifts off; the critical
value is zero — compare with idea 38).
Also, review ideas 1, 37 and K29.
PROB
29.
A block with mass
M
lies
on a horizontal slippery surface and also
touches a vertical wall. In the upper sur-
face of the block, there is a cavity with the
shape of a half-cylinder with radius
r
. A
small pellet with mass
m
is released at the
upper edge of the cavity, on the side closer
to the wall. What is the maximum velocity
oftheblockduringitssubsequentmotion?
Friction can be neglected.
r
m
M
IDEA
40:
IDEA
41:
You will also need idea 37.
IDEA
42:
0
=
dv
dt
=
a
PROB
30.
A light rod with length 3
l
is
attached to the ceiling by two strings with
equal lengths. Two balls with masses
m
and
M
are fixed to the rod, the distance
betweenthemandtheirdistancesfromthe
ends of the rod are all equal to
l
. Find the
tension in the second string right after the
first has been cut.
m
l
M
l
l
There are several good solutions for this
problem, all of which share applying
idea 34 and the need to find the angu-
lar acceleration of the rod. Firstly, angu-
lar acceleration of the rod can be found
from method 6 by choosing angle of ro-
tation
ϕ
to be the generalised coordinate.
Secondly, we may use Newton’s 2nd law
for rotational motion: we find the torque
on the rod about the point of attachment
of the second string and equate it to
I
ε
with angular acceleration
ε
and moment
of inertia
I
=
ml
2
+
4
Ml
2
. More gener-
ally,
IDEA
43:
s
M
=
I
ε
I
s
I
=
∑
m
i
r
2
i
=
∫
r
2
·
dm
=
∫
r
2
ρ
·
dV
r
i
i
s
K
=
1
2
I
ω
2
Once the angular acceleration is found, in
order to apply the idea 34 it may be help-
ful to use
IDEA
44:
⃗
F
=
d
⃗
P
dt
⃗
P
⃗
F
⃗
M
=
d
⃗
L
dt
⃗
L
⃗
M
In our case this last method is fruitful
when applied both to forces and torques.
Another solution method is to con-
sider the rod and the balls as three dif-
ferent (interacting) bodies. Then the balls’
accelerations can be found as per idea 31;
one can also employ
IDEA
45:
Clearly if this were not true, a non-zero
force would generate an infinite accelera-
tion for a massless body.
PROB
31.
An inextensible rough thread
with mass per unit length
ρ
and length
L
is thrown over a pulley such that the
length of one hanging end is
l
. The pul-
ley is comprised of a hoop of mass
m
and
radius
R
attached to a horizontal axle by
light spokes. The initially motionless sys-
tem is let go. Find the force on the axle
immediately after the motion begins. The
friction between the pulley and the axle is
negligible.
R
Why not proceed as follows: to find the
force, we will use idea 34; the acceleration
of the system will be found using Method
6. To apply idea 34 most handily, let us
employ
IDEA
46:
⃗
F
=
M
⃗
a
C
⃗
a
C
This idea is best utilised when a part of
the system’s mass is motionless and only
a relatively small mass is moved about
(just like in this case: the only difference
after a small period of time is that a short
length of thread is “lost” at one end and
“gained” at the other end). Obviously
idea 32 will be useful here, and idea 19
will save you some effort. Bear in mind
that in this case we are not interested in
the centre of mass coordinate per se, but
only in its change as a function of time;
therefore in the expression for this co-
ordinate we can omit the terms that are
independent of time: their time derivat-
iveswillvanish. Thetime-dependentpart
of the centre of mass coordinate should be
expressed using the same coordinate that
we will use with Method 6 (since Method
6 will produce its second derivative with
respect to time). A technical bit of ad-
vice may help: a vector is specified by
(a)
its magnitude and direction;
(b)
its projec-
tions onto coordinate axes in a given co-
ordinate system;
IDEA
47:
Above all, this applies when the direction
of the vector is neither known nor appar-
ent. In this instance, we should find
F
x
and
F
y
in a suitable coordinate system.
PROB
32.
A thread is thrown over a pul-
ley. At its both ends there are two blocks
with equal masses. Initially the two blocks
are at the same height. One of them is
instantaneously given a small horizontal
velocity
v
. Which of the two blocks will
reach higher during the subsequent mo-
tion? The pulley’s mass is negligible.
v
This problem is really tough, because the
key to the solution is a very specific and
rarely used
IDEA
48:
Here the centre of mass can move about
a little bit, but in the longer term (aver-
aged over one period of the pendulum-
like motion of the kicked block — cf. idea
22) it is motionless: the blocks have the
same mass and if one of them rises, then
in the expression for the centre of mass
this will be compensated by the descent
of the other block. This is also true for
the horizontal coordinate of the centre of
mass, but it is enough to consider the ver-
tical coordinate only to solve the problem.
Let us also bring up the rather obvious
FACT
10:
The solution algorithm is then as follows:
we write down Newton’s 2nd law for
(a)
the system made out of two blocks and
(b)
one block; we average both equations
and use the equality apparent from
(a)
to find the average tension in the thread,
which we then substitute into equation
(b)
. Based on idea 22, we partition the ten-
sion in the threadinto the average and the
high-frequency component and use idea
16.
PROB
33.
A system of blocks sits on
a smooth surface, as shown in the fig-
ure. The coefficient of friction between
the blocks is
µ
, while that between the
blocks and the surface is
µ
=
0.
m m
x
M M
F
The bottom right block is being pulled by
a force
F
. Find the accelerations of all
blocks.
IDEA
49:
For example, if we are to assume that
there is no slipping between two touch-
ing bodies, then they could be treated as
a whole. Then one should find the fric-
tional force
F
h
between the bodies and de-
termine when the assumption holds, or
when is
F
h
less that the maximum static
friction force
µ
N
.
PROB
34.
A billiard ball hits another sta-
tionary billiard ball. At which collection
of points could the stationary ball be po-
sitioned such that it would be possible
to achieve the situation where both balls
will fall into two (different) pockets on the
table? The collisions are perfectly elastic,
the balls are perfectly slippery (hence the
rotation of the balls is negligible).
IDEA
50:
To prove this, note that the three velocity
vectors (velocity before and the two ve-
locities after the impact) form a triangle
because of the momentum conservation
law. The conservation of energy means
that the sides of the triangle satisfy Py-
thagore’s theorem. A special case of this
result is (see the problem after next)
FACT
11:
PROB
35.
An absolutely elastic and slip-
pery billiard ball is moving with velocity
v
towardtwomotionlessidenticalballs. The
motionless balls are touching and their
centres lie on a straight line that is per-
pendicular to the incoming ball’s velocity
vector. The moving ball is directed ex-
actly toward the touching point of the two
balls. Which velocity will the incoming
ball have after the collisions? Consider
two scenarios:
(a)
the incoming ball hits
exactly in the middle between the balls;
(b)
its trajectory is a little bit off and it hits one
of the stationary balls marginally earlier.
v
To answer the first question, it is neces-
sary to use
IDEA
51:
Also, donotforgetidea37! Forthesecond
question, let us use
IDEA
52:
PROB
36.
n
absolutely elastic beads are
sliding along the frictionless wire. What
is the maximum possible number of col-
lisions? The sizes of the beads are negli-
gible, and so is the probability that more
than two beads will collide at the same
time.
IDEA
53:
Here is an auxiliary question: what
would the elastic collision of two balls on
an
x
−
t
diagram look like?
PROB
37.
A plank of length
L
and mass
M
is lying on a smooth horisontal surface;
on its one end lies a small block of mass
m
. The coefficient of friction between the
block and the plank is
µ
. What is the min-
imal velocity
v
that needs to be imparted
to the plank with a quick shove such that
during the subsequent motion the block
would slide the whole length of the board
andthenwouldfallofftheplank? Thesize
of the block is negligible.
L
µ
µ
=0
m
M
v
This problem has two more or less equi-
valent solutions. First, we could solve it
using idea 6. Second, we could use ideas
37 and 51, further employing
IDEA
54:
Indeed, the friction force has a constant
magnitude and, as seen in the reference
frame of the support, it is always parallel
to displacement.
PROB
38.
The given figure has been pro-
duced off a stroboscopic photograph and
it depicts the collision of two balls of equal
diameters but different masses. The arrow
notes the direction of motion of one of the
balls before the impact. Find the ratio of
the masses of the two balls and show what
the direction of motion for the second ball
was before the impact.
IDEA
55:
To be more specific: when two bodies in-
teract, the vector of the impulse is equal
tothevectorialdifferenceoftheirtwomo-
menta. Cf. idea 5.
FACT
12:
FACT
13:
PROB
39.
There are two barrels (
A
and
B
) whose taps have different design, see
figure. The tap is opened, the height of the
water surface from the tap is
H
. What ve-
locity does the water stream leave the bar-
rels with?
H
H
A B
IDEA
56:
Itcouldnotbeotherwise: theanswersare,
after all, different. It pays to be attentive
here. While designing the tap
A
, there
was a clear attempt to preserve the lam-
inarity of the flow: energy is conserved.
However, if, motivated by method 3, we
were to write down the momentum given
to the stream by the air pressure during
an infinitesimal time
dt
—
pSdt
(where
S
is the tap’s area of cross-section), we
would see that, owing to the flow of wa-
ter,
p
̸
=
ρ
g
(cf. dynamical pressure,
Bernoulli’s law!). On the other hand, for
tap
B
the laminar flow is not preserved;
therewillbeeddiesandlossofenergy. We
could nonetheless work with momentum:
we write the expression for the pressure
exerted on the liquid by the walls of the
barrel (generally the pressures exerted by
the left and the right hand side walls of
the barrel cancel each other out, but there
remains an uncompensated pressure
p
=
ρ
gH
exerted to the left of the cross-section
of the tap
S
).
PROB
40.
Sand is transported to the con-
struction site using a conveyor belt. The
length of the belt is
l
, the angle with re-
spect to the horizontal is
α
; the belt is
driven by the lower pulley with radius
R
,
powered externally. The sand is put onto
the belt at a constant rate
µ
(kg/s). What
is the minimal required torque needed
to transport the sand? What is the ve-
locity of the belt at that torque? The
coefficient of friction is large enough for
the sand grains to stop moving immedi-
ately after hitting the belt; take the ini-
tial velocity of the sand grains to be zero.
α
µ
l
R
FACT
14:
For this problem, idea 56 and methode 3
will come in handy in addition to
IDEA
57:
σ
v
=
σ
(
x
)
v
(
x
)
For a flow of incompressible (constant
density) liquid in a pipe, such a density
is
σ
=
ρ
S
and therefore
vS
=
Const. For
a region of space where the flow is dis-
charged — a sink — the mass increases:
dm
dt
=
σ
v
— this equation, too, could be
called the condition for continuity.
PROB
41.
A ductile blob of clay falls
against the floor from the height
h
and
starts sliding. What is the velocity of the
blob at the very beginning of sliding if
the coefficient of friction between the floor
and the blob is
µ
? The initial horizontal
velocity of the blob was
u
.
IDEA
58:
µ
Indeed,
∆
p
⊥
=
∫
N
(
t
)
dt
(integrated over
the duration of the impact) and
∆
p
∥
=
∫
µ
N
(
t
)
dt
=
µ
∫
N
(
t
)
dt
.
PROB
42.
A boy is dragging a sled by
the rope behind him as he slowly ascends
a hill. What is the work that the boy does
to transport the sled to the tip of the hill
if its height is
h
and the horizontal dis-
tance from the foot of the hill to its tip is
a
? Assume that the rope is always parallel
to the tangent of the hill’s slope, and that
the coefficient of friction between the sled
and the snow is
µ
.
h
a
FACT
15:
Clearly, to apply the fact 15, one will need
idea 3.
PROB
43.
An empty cylinder with mass
M
is rolling without slipping along a
slantedsurface, whose angleof inclination
is
α
=
45
◦
. On its inner surface can slide
freely a small block of mass
m
=
M
/2.
What is the angle
β
between the normal
to the slanted surface and the straight line
segment connecting the centre of the cyl-
inder and the block?
β
α
m
M
Clearly the simplest solution is based on
idea 6, but one needs to calculate the kin-
etic energy of a rolling cylinder.
IDEA
59:
K
=
K
c
+
M
Σ
v
2
c
/2
K
c
M
Σ
⃗
P
=
M
Σ
⃗
v
c
⃗
P
c
≡
0
⃗
L
=
L
c
+
⃗
r
c
×
⃗
P
I
=
I
0
+
M
Σ
a
2
I
s
I
0
s
a
We will have to compute angular mo-
mentum already in the next problem, so
let us clarify things a little.
IDEA
60:
⃗
L
=
∑
⃗
L
i
i
⃗
L
i
=
⃗
r
i
×
⃗
p
i
L
i
=
h
i
p
i
=
r
i
p
ti
h
i
=
r
i
sin
α
i
p
ti
=
p
i
sin
α
If in a three-dimensional space the an-
gular momentum is a vector, for a mo-
tion in a plane this vector is perpendicu-
lar to the plane and is therefore effectively
a scalar (and thus one can abandon cross
products). It is often handy to combine
ideas 59 and 60: we do not divide the sys-
tem into particles but, instead, into rigid
bodies (
L
=
∑
L
i
), we compute the mo-
ment of inertia
L
i
of each body according
to idea 59: the moment of inertia of the
centre of mass plus the moment of inertia
as measured in the centre of mass frame.
IDEA
61:
l
1
12
Ml
2
2
5
MR
2
2
3
MR
2
1
2
MR
2
a
1
6
Ma
2
If the the rotation axis does not go
through the centre of mass, then one can
(a)
findthemomentofinertiawithrespect
to the axis of interest using the parallel-
axis (Steiner) theorem;
(b)
apply idea 59
to calculate kinetic energy or angular mo-
mentum (in which case it is only enough
to know the moment of inertia with re-
spect to the centre of mass).
PROB
44.
A rod of mass
M
and length 2
l
is sliding on ice. The velocity of the centre
of mass of the rod is
v
, the rod’s angu-
lar velocity is
ω
. At the instant when the
centre of mass velocity is perpendicular to
the rod itself, it hits a motionless post with
an end. What is the velocity of the centre
ofmassof therodafter theimpact if
(a)
the
impact is perfectly inelastic (the end that
hits the post stops moving);
(b)
the impact
is perfectly elastic.
v
ω
M
2
l
In case of an absolutely elastic collision
one equation follows from energy conser-
vation; if the collision is inelastic, then
another condition arises: that of a mo-
tionless end of the rod. Still, we have
two variables. The second equation arises
from
IDEA
62:
Indeed, during the impact the body’s
motion is affected by the normal and
frictional forces, but both are applied
through the point of impact: their lever
arm is zero. If a body is moving in a
gravitational or similar field, then in the
longer term the angular momentum with
respect to the point of impact may be-
gin to change, but immediately before
and after the collision it is nonetheless the
same(gravityisnottoostrongasopposed
to the normal forces that are strong yet
short-lived; even though gravity’s lever
arm is non-zero, it cannot change the an-
gular momentum in an instant).
PROB
45.
If one hits something rigid —
e.g. a lamppost — with a bat, the hand
holding the bat may get stung (hurt) as
long as the impact misses the so-called
centre of percussion of the bat (and hits
either below or above such a centre). De-
termine the position of the centre of per-
cussion for a bat of uniform density. You
may assume that during an impact the bat
is rotating around its holding hand.
METHOD
7:
Phrased like that, it may seem that the
method is rather pointless. However,
converting and interpreting real-life scen-
arios —
modelling
the problem — is one of
the most challenging and interesting as-
pects of physics. It is interesting because
it supplies more creative freedom than
solving an existing model using well-
established ideas. Still, this freedom has
limits: the model has to describe the real-
ity as best as possible, the approxima-
tions have to make sense and it is desir-
able that the model were solvable either
mentally or with aid of a computer. For
a given problem, there is not much free-
dom left and the business is simplified:
there clear hints as to sensible assump-
tions. Let us begin translating: “A rigid
rod of length
l
and uniform density is ro-
tating around one end with the angular
velocity
ω
, the rotation axis is perpendic-
ular to the rod. At a distance
x
from the
axis there is a motionless post that is par-
allel to the axis of rotation. The rod hits
the post.” Now we encounter the first
obstacle: is the impact elastic or inelastic?
This is not brought up in the text of the
problem. Let us leave it for now: maybe
we can get somewhere even without the
corresponding assumption (it turns out
that this is the case). Now we encounter
the central question: what does it mean
for the hand “not to get stung”? We know
it hurts when something hits our hand —
ifthissomethinggetsanimpulsefromthe
hand during a short period of time (the
impact), as this implies a large force. The
hand is stationary, so the hand-held end
of the bat should come to halt without re-
ceiving any impulse from the hand. Thus
our interpretation of the problem is com-
plete: “Following the impact, the rotation
is reversed, 0
≥
ω
′
≥ −
ω
; during the
impact the axis of rotation imparts no im-
pulse on the rod. Find
x
.” The penultim-
ate sentence hints at the usage of idea 62.
PROB
46.
A massive cylinder of radius
R
and mass
M
is lying on the floor. A
narrow groove of depth
a
has been chis-
elled along the circumference of the cylin-
der. A thread has been wrapped around
the groove and is now being pulled by its
free end, held horizontally, with a force
F
.
The cylinder is positioned such that the
thread is being freed from below the cyl-
inder. With what acceleration will the cyl-
inder start moving? The friction between
the floor and the cylinder is large enough
for there to be no slipping.
F
M
a
R
F
⊙
There are multiple ways to tackle this
problem, butletususethefollowingidea.
IDEA
63:
I
ε
=
M
To prove this idea, recall idea 6: kinetic
energy appears when work is done,
K
=
1
2
I
ω
2
=
M
ϕ
(
ϕ
is the angle of rotation of
the body,
ω
=
d
ϕ
/
dt
). If the moment
of inertia with respect to the instantan-
eous axis of rotation
I
does not depend on
time, then
dK
/
dt
=
1
2
Id
ω
2
/
dt
=
I
ωε
=
dM
ϕ
/
dt
=
M
ω
, which gives
I
ε
=
M
.
PROB
47.
A ball is rolling along a hori-
zontal floor in the region
x
<
0 with ve-
locity
⃗
v
0
= (
v
x
0
,
v
y
0
)
. In the region
x
>
0
there is a conveyor belt that moves with
velocity
⃗
u
= (
0,
u
)
(parallel to its edge
x
=
0). Findthevelocityoftheball
⃗
v
= (
v
x
,
v
y
)
with respect to the belt after it has rolled
onto the belt. The surface of the conveyor
belt is rough (the ball does not slip) and is
level with the floor.
IDEA
64:
Indeed, the points where the normal force
and the gravity are applied are on the
same straight line with the forces them-
selves and their sum is zero, meaning that
their net torque is also zero; the force of
friction is lying in the plane of the surface,
andsoitsleverarmwithrespecttoanaxis
in the same plane is zero.
PROB
48.
A “spring-dumbbell” com-
prises two balls of mass
m
that are con-
nect with a spring of stiffness
k
. Two such
dumbbells are sliding toward one another,
the velocity of either is
v
0
. At some point
the distance between them is
L
(see fig.).
After which time is the distance between
them equal to
L
again? The collisions are
perfectly elastic.
L
v
0
v
0
IDEA
65:
Note:
this is a rather general idea, division
into simpler steps can be useful if rapid
(almost instantaneous) processes can oc-
cur in a dynamical system; see next prob-
lem for an example (also recall idea 51)
PROB
49.
Small grains of sand are slid-
ing without friction along a cylindrical
trough of radius
R
(see fig.). The inclin-
ation angle of the trough is
α
. All grains
have initial velocity zero and start near
point
A
(but not necessarily at the point
A
itself). What should be the length of the
trough such that all grains would exit it at
the point
B
?
L
α
A
B
IDEA
66:
2
π
PROB
50.
A coat hanger made of wire
with a non-uniform density distribution
is oscillating with a small amplitude in
the plane of the figure. In the first two
cases the longer side of the triangle is ho-
rizontal. In all three cases the periods of
oscillation are equal. Find the position of
the centre of mass and the period of oscil-
lation.
42cm
10cm
Background info:
A finite-size rigid
body that oscillates around a fixed axis
is known as the physical pendulum. Its
frequency of small oscillations is easy to
derive from the relation
I
¨
ϕ
=
−
mgl
ϕ
,
where
I
is the moment of inertia with re-
spect to the axis of oscillation and
l
is the
distance of the centre of mass from that
axis:
ω
−
2
=
I
/
mgl
=
I
0
/
mgl
+
l
/
g
(here
we employ the parallel-axis/Steiner the-
orem, see idea 59). The
reduced length
of the physical pendulum is the distance
˜
l
=
l
+
I
0
/
ml
such that the frequency of
oscillation of a mathematical pendulum
of that length is the same as for the given
physical pendulum.
IDEA
67:
˜
l
Proof: the formula above could be rewrit-
ten as a quadratic equation to find the
length
l
corresponding to the given fre-
quency
ω
(i.e. to the given reduced length
˜
l
=
g
/
ω
2
):
l
2
−
l
˜
l
+
I
0
/
m
=
0. Accord-
ing to Vieta’s formulae, the solutions
l
1
and
l
2
satisfy
l
1
+
l
2
=
l
, so that
l
1
and
l
2
=
˜
l
−
l
1
result in the same frequency of
oscillations.
PROB
51.
A metallic sphere of radius
2mmanddensity
ρ
=
3000kg/m
3
ismov-
ing in water, falling freely with the accel-
eration
a
0
=
0,57g. The water density is
ρ
0
=
1000kg/m
3
. With what acceleration
would a spherical bubble of radius 1mm
rise in the water? Consider the flow to be
laminar in both cases; neglect friction.
IDEA
68:
Using method 6 we find that in the case
(A) the kinetic energy of the system
K
=
1
2
v
2
(
m
+
αρ
0
V
)
, where the constant
α
is
a number that characterizes the geometry
of the body that correspond to the extent
of the region of the liquid that will move
(compared to the volume of the body it-
self). If a body is acted on by a force
F
,
then the power produced by this force is
P
=
Fv
=
dK
dt
=
va
(
m
+
αρ
0
V
)
. Thus
F
=
a
(
m
+
αρ
0
V
)
: the effective mass of
the body increases by
αρ
0
V
. In the prob-
lemabove, theconstant
α
forthespherical
body can be found using the conditions
given in the first half of the problem.
In case (B), if we assume that the ve-
locity of the body is constant, we find
K
=
1
2
v
2
ρ
0
(
α
Svt
)
, where
S
is the cross-
sectional area of the body and
α
S
is the
cross-sectional area of the turbulent ‘tail’.
This
α
, again, characterizes the body.
From here, it is easy to find
Fv
=
dK
dt
=
α
2
v
3
ρ
0
S
, which gives
F
=
α
2
v
2
ρ
0
S
.
PROB
52.
A stream of water falls against
a trough’s bottom with velocity
v
and
splits into smaller streams going to the left
and to the right. Find the velocities of
both streams if the incoming stream was
inclined at an angle
α
to the trough (and
the resultant streams). What is the ratio of
amounts of water carried per unit time in
the two outgoing streams?
α
This is a rather hard problem. Let us first
state a few ideas and facts.
IDEA
69:
p
+
ρ
gh
+
1
2
ρ
v
2
=
p
h
v
FACT
16:
To solve the second half of the problem,
the following is needed:
IDEA
70:
⃗
F
=
d
⃗
P
dt
+
⃗
Φ
P
−
⃗
Φ
P
⃗
Φ
P
⃗
Φ
P
The momentum flux of the flowing liquid
could be calculated as the product of mo-
mentum volume density
ρ
⃗
v
with the flow
rate (volume of liquid entering/leaving
the system per unit time).
What is the open system we should be
considering in this case? Clearly, a system
that would allow relating the incoming
flow rate
µ
(kg/s) to the outgoing fluxes
(
µ
l
ja
µ
r
) using the formula above: a small
imaginary region of space that would in-
clude the region where the stream splits
into two.
FACT
17:
PROB
53.
Find the velocity of propaga-
tion of small waves in shallow water.
The water is considered shallow if the
wavelength is considerably larger than
the depth of the water
H
. Thanks to
this we can assume that along a vertical
cross-section the horizontal velocity of all
particles
v
h
is the same and that the hori-
zontal velocity of water particles is signi-
ficantly smaller than the vertical velocity.
The smallness of the waves means that
their height is significantly smaller than
the depth of the water. This allows us
to assume that the horizontal velocity of
the water particles is significantly smaller
than the wave velocity,
u
.
IDEA
71:
(An alternative approach is to linearise
and solve a system of coupled partial dif-
ferential equations.)
PROB
54.
A small sphere with mass
m
=
1g is moving along a smooth surface, slid-
ing back and forth and colliding elastic-
ally with a wall and a block. The mass of
the rectangular block is
M
=
1kg, the ini-
tial velocity of the sphere is
v
0
=
10m/s.
What is the velocity of the sphere at the
instant when the distance between the
sphere and the wall has doubled as com-
pared with the initial distance? By how
many times will the average force (aver-
aged over time) exerted by the sphere on
the wall have changed?
IDEA
72:
I
x
p
x
Let us be more precise here. The closed
contourisproducedasaparametriccurve
(the so-called phase trajectory)
x
(
t
)
,
p
x
(
t
)
if we trace the motion of the system dur-
ing one full period
T
. The phase traject-
ory is normally drawn with an arrow that
indicated the direction of motion. The
adiabatic invariant is not exactly and per-
fectly conserved, but the precision with
which it is conserved grows if the ratio
τ
/
T
grows, where
τ
is the characteristic
time of change of the system’s paramet-
ers.
Adiabatic invariant plays an instru-
mental role in physics: from the adiabatic
law in gases (compare the result of the
previous problem with the adiabatic ex-
pansion law for an ideal gas with one de-
gree of freedom!) and is applicable even
in quantum mechanics (the number of
quanta in the system — e.g. photons —
is conserved if the parameters of the sys-
tem are varied slowly).
PROB
55.
Astraighthomogeneousrodis
being externally supported against a ver-
tical wall such that the angle between the
wall and the rod is
α
<
90
◦
. For which
values of
α
can the rod remain stationary
whenthussupported? Considertwoscen-
arios: a) the wall is slippery and the floor
is rough with the friction coefficient
µ
; b)
the floor is slippery and the wall is rough
with the friction coefficient
µ
.
PROB
56.
A light stick rests with one
end against a vertical wall and another on
a horizontal floor. A bug wants to crawl
down the stick, from top to bottom. How
should the bug’s acceleration depend on
its distance from the top endpoint of the
stick? The bug’s mass is
m
, the length of
the stick is
l
, the angle between the floor
and the stick is
α
and the stick’s mass is
negligible; both the floor and the wall are
slippery (
µ
=
0). How long will it take the
bugtoreachthebottomofthestickhaving
started at the top (from rest)?
α
l
x
a
PROB
57.
A wedge with the angle
α
at
the tip is lying on the horizontal floor.
There is a hole with smooth walls in the
ceiling. A rod has been inserted snugly
into that hole, and it can move up and
down without friction, while its axis is
fixed to be vertical. The rod is suppor-
ted against the wedge; the only point with
friction is the contact point of the wedge
and the rod: the friction coefficient there is
µ
. For which values of
µ
is it possible to
push the wedge through, behind the rod,
by only applying a sufficiently large hori-
zontal force?
α
µ
F
PROB
58.
Sometimes a contraption is
used to hang pictures etc. on the wall,
whose model will be presented below.
Against a fixed vertical surface is an im-
movable tilted plane, where the angle
between the surface and the plane is
α
.
There is a gap between the surface and
the plane, where a thin plate could be fit.
The plate is positioned tightly against the
vertical surface; the coefficient of friction
between them can be considered equal to
zero. In the space between the plate and
the plane a cylinder of mass
m
can move
freely, its axis being horizontal and paral-
lel to all considered surfaces. The cylin-
der rests on the plate and the plane and
the coefficients of friction on those two
surfaces are, respectively,
µ
1
and
µ
2
. For
whichvaluesofthefrictioncoefficientsthe
plate will assuredly not fall down regard-
less of its weight?
α
F
µ
= 0
µ
2
µ
1
m
PROB
59.
On top of a cylinder with a
horisontal axis a plank is placed, whose
length is
l
and thickness is
h
. For which
radius
R
of the cylinder the horizontal po-
sition of the plank is stable?
R
l
PROB
60.
A vessel in the shape of a
cylinder, whose height equals its radius
R
and whose cavity is half-spherical, is
filled to the brim with water, turned up-
side down and positioned on a horizontal
surface. The radius of the half-spherical
cavity is also
R
and there is a little hole in
the vessel’s bottom. From below the edges
of the freely lying vessel some water leaks
out. How high will the remaining layer of
water be, if the mass of the vessel is
m
and
the water density is
ρ
? If necessary, use
the formula for the volume of a slice of a
sphere (see Fig.):
V
=
π
H
2
(
R
−
H
/3
)
.
h
H
V
PROB
61.
A vertical cylindrical vessel
with radius
R
is rotating around its axis
withthe angular velocity
ω
. Byhow much
does the water surface height at the axis
differ from the height next to the vessel’s
edges?
PROB
62.
A block with mass
M
is on a
slippery horizontal surface. A thread ex-
tends over one of its corners. The thread is
attached to the wall at its one end and to a
little block of mass
m
, which is inclined by
an angle
α
with respect to the vertical, at
the other. Initially the thread is stretched
and the blocks are held in place. Then
the blocks are released. For which ratio of
masses will the angle
α
remain unchanged
throughout the subsequent motion?
α
m
M
PROB
63.
Two slippery (
µ
=
0) wedge-
shaped inclined surfaces with equal tilt
angles
α
are positioned such that their
sides are parallel, the inclines are facing
each other and there is a little gap in
between (see fig.). On top of the surfaces
are positioned a cylinder and a wedge-
shaped block, whereas they are resting
one against the other and one of the
block’ssidesishorizontal. Themassesare,
respectively,
m
and
M
. What accelerations
willthecylinderandtheblockmovewith?
Find the reaction force between them.
PROB
64.
Three little cylinders are con-
nected with weightless rods, where there
is a hinge near the middle cylinder, so that
the angle between the rods can change
freely. Initially this angle is a right angle.
Two of the cylinders have mass
m
, another
one at the side has the mass 4
m
. Find
theaccelerationoftheheaviercylinderim-
mediately after the motion begins. Ignore
friction.
90
o
m
m
4
m
a
=?
PROB
65.
A slippery rod is positioned
at an angle
α
with respect to the horizon.
A little ring of mass
m
can slide along the
rod, to which a long thread is attached. A
small sphere of size
M
is attached to the
thread. Initially the ring is held motion-
less, and the thread hangs vertically. Then
the ring is released. What is the accelera-
tion of the sphere immediately after that?
m
M
PROB
66.
A block begins sliding at the
uppermost point of a spherical surface.
Find the height at which it will lose con-
tact with the surface. The sphere is held in
place and its radius is
R
; there is no fric-
tion.
PROB
67.
The length of a weightless rod
is 2
l
. A small sphere of mass
m
is fixed at
a distance
x
=
l
from its upper end. The
rod rests with its one end against the wall
and the other against the floor. The end
that rests on the floor is being moved with
a constant velocity
v
away from the wall.
a)
Find the force with which the sphere
affects the rod at the moment, when the
angle between the wall and the rod is
α
=
45
◦
;
(b)
what is the answer if
x
̸
=
l
?
α
v
m
x
2
l
PROB
68.
A light rod with length
l
is
connected to the horizontal surface with
a hinge; a small sphere of mass
m
is con-
nected to the end of the rod. Initially the
rod is vertical and the sphere rests against
the block of mass
M
. The system is left
to freely move and after a certain time
the block loses contact with the surface of
the block — at the moment when the rod
forms an angle
α
=
π
/6 with the hori-
zontal. Find the ratio of masses
M
/
m
and
the velocity
u
of the block at the moment
of separation.
m
l
M
PROB
69.
At a distance
l
from the edge
of the table lies a block that is connected
with a thread to another exact same block.
The length of the thread is 2
l
and it is
extended around the pulley sitting at the
edge of the table. The other block is held
above the table such that the string is un-
der tension. Then the second block is re-
leased. What happens first: does the first
block reach the pulley or does the second
one hit the table?
l l
PROB
70.
A cylindrical ice hockey puck
with a uniform thickness and density is
given an angular velocity
ω
and a transla-
tional velocity
u
. What trajectory will the
puck follow if the ice is equally slippery
everywhere? In which case will it slide
farther: when
ω
=
0 or when
ω
̸
=
0, as-
suming that in both cases
u
is the same?
PROB
71.
A little sphere of mass
M
hangs at the end of a very long thread;
to that sphere is, with a weightless rod,
attached another little sphere of mass
m
.
The length of the rod is
l
. Initially the
system is in equilibrium. What horizontal
velocity needs to be given to the bottom
sphere for it to ascend the same height
with the upper sphere? The sizes of the
spheres are negligible compared to the
length of the rod.
M
m
v
l
PROB
72.
A block of mass
m
lies on a
slippery horizontal surface. On top of it
lies another block of mass
m
, and on top of
that — another block of mass
m
. A thread
that connects the first and the third block
has been extended around a weightless
pulley. The threads are horizontal and the
pulleyisbeingpulledbyaforce
F
. Whatis
the acceleration of the second block? The
coefficient of friction between the blocks is
µ
.
F
a
=?
m
m
m
PROB
73.
A boy with mass
m
wants
to push another boy standing on the ice,
whose mass
M
is bigger that his own. To
that end, he speeds up, runs toward the
other boy and pushed him for as long as
they can stand up. What is the maximal
distance by which it is possible to push in
this fashion? The maximal velocity of a
run is
v
, the coefficient of friction between
both boys and the ice is
µ
.
PROB
74.
A uniform rod with length
l
is
attached with a weightless thread (whose
length is also
l
) to the ceiling at point
A
.
The bottom end of the rod rests on the
slippery floor at point
B
, which is exactly
below point
A
. The length of
AB
is
H
,
l
<
H
<
2
l
. The rod begins to slide from
rest; find the maximal acceleration of its
centre during subsequent motion.
l
l
H
A
B
PROB
75.
A stick with uniform density
rests with one end against the ground and
with the other against the wall. Initially it
was vertical and began sliding from rest
such that all of the subsequent motion
takes place in a plane that is perpendicu-
lar to the intersection line of the floor and
the wall. What was the angle between the
stick and the wall at the moment when the
stick lost contact with the wall? Ignore
friction.
PROB
76.
A log with mass
M
is sliding
along the ice while rotating. The velocity
of the log’s centre of mass is
v
, its angu-
lar velocity is
ω
. At the moment when
the log is perpendicular to the velocity of
its centre of mass, the log hits a station-
ery puck with mass
m
. For which ratio
of the masses
M
/
m
is the situation, where
the log stays in place while the puck slides
away, possible? The collisions are per-
fectly elastic. The log is straight and its
linear density is constant.
v
ω
M
m
PROB
77.
A ball falls down from height
h
, initially the ball’s horizontal velocity
was
v
0
and it wasn’t rotating. a) Find the
velocity and the angular velocity of the
ball after the following collision against
the floor: the ball’s deformation against
the floor was absolutely elastic, yet there
was friction at the contact surface such
that the part of the ball that was in con-
tact with the floor stopped. b) Answer
the same question with the assumption
thatthevelocitiesofthesurfacesincontact
never homogenized and that throughout
the collision there was friction with coeffi-
cient
µ
.
PROB
78.
A ball is rolling down an in-
clined plane. Find the ball’s acceleration.
The plane is inclined at an angle
α
, the
coefficient of friction between the ball and
the plane is
µ
.
PROB
79.
A hoop of mass
M
and radius
r
stands on a slippery horizontal surface.
There is a thin slippery tunnel inside the
hoop, along which a tiny block of mass
m
can slide. Initially all the bodies are at
rest and the block is at the hoop’s upper-
most point. Find the velocity and the ac-
celerationofthehoop’scentralpointatthe
moment when the angle between the ima-
ginary line connecting the hoop’s central
point and the block’s position and the ver-
tical is
ϕ
.
ϕ
O
A
r
PROB
80.
A block with mass
m
=
10g is
put on a board that has been made such
that, when sliding to the left, the coef-
ficient of friction
µ
1
=
0,3, while when
sliding to the right it is
µ
2
=
0,5. The
board is repeatedly moved left-right ac-
cording to the graph
v
(
t
)
(see fig.). The
graph is periodic with period
T
=
0,01s;
the velocity
v
of the board is considered
t
T
T
/2
v
1
m
/
s
positive when directed to the right. Using
the graph, find the average velocity that
the block will move with.
PROB
81.
A water turbine consists of
a large number of paddles that could be
considered as light flat boards with length
l
, that are at one end attached to a rotat-
ing axis. The paddles’ free ends are po-
sitioned on the surface of an imaginary
cylinder that is coaxial with the turbine’s
axis. A stream of water with velocity
v
and flow rate
µ
(kg/s) is directed on the
turbine such that it only hits the edges of
the paddles. Find the maximum possible
usable power that could be extracted with
such a turbine.
v
l
µ
PROB
82.
A flat board is inclined at an
angle
α
to the vertical. One of its ends is in
the water, the other one is outside the wa-
ter. The board is moving with velocity
v
with respect to its normal. What is the ve-
locity of the water stream directed up the
board?
v
u
PROB
83.
Amotor-drivenwagonisused
to transport a load horizontally by a dis-
tance
L
. The load is attached to the side of
the wagon by a cable of length
l
. Half of
the time the wagon is uniformly acceler-
ated, the other half — uniformly deceler-
ated. Find the values of the acceleration
a
such that, upon reaching the destination,
the load will be hanging down motion-
lessly. You can assume that
a
≪
g
.
PROB
84.
A shockwave could be con-
sidered as a discontinuous jump of the air
pressure from value
p
0
to
p
1
, propagat-
ing with speed
c
s
. Find the speed which
will be obtained, when influenced by the
shockwave,
(a)
a wedge-shaped block: a
prism whose height is
c
, whose base is a
right triangle with legs
a
and
b
and which
is made out of material with density
ρ
;
b)
anbodyofanarbitraryshapewithvolume
V
and density
ρ
.
c
s
c
b
x
p
p
1
p
0
a
PROB
85.
A dumbbell consisting of two
elastic spheres connected with a thin steel
rod is moving parallel to its axis with
a velocity
v
toward another exact same
spheres. Find the velocity of the dumbbell
after a central collision. Is the kinetic en-
ergy of the system conserved?
v
0
1.
Write out the balance of torques for the contact
point
O
of the hoop and the shaft. What is the angle
that the tangent to the shaft at point
O
forms with
the horizon (given that the wire slips on the shaft)?
2.
Write down the equation for the torques for the
cylinder & block system with respect to the contact
point of the cylinder and the inclined plane. What
angle with respect to the horizon is formed by the
tangent to the cylinder constructed at the position of
the little block?
3.
According to the idea 4, consider the system “rod
CD
+ the mass
m
” as a whole; there are four forces
acting on it:
m
⃗
g
,
⃗
F
, and the tension forces of the rods,
⃗
T
AC
and
⃗
T
BD
. The tension forces are the ones which
we don’t know and don’t want to know. Accord-
ing to the idea 2, these will drop out from the bal-
ance of torques acting on the rod
CD
with respect to
the intersection point of
AC
and
BD
. Indeed, due to
the fact 2, the tension force in the rod
AC
is parallel
to
AC
; the same applies to the rod
BD
. Now, what
must be the torque of force
F
? For what direction of
the force will this torque be achieved with the min-
imum magnitude?
4.
Thevectorsumoftheforces
⃗
F
and
m
⃗
g
hastocom-
pensate the sum of the friction and the normal force
⃗
f
=
⃗
N
+
⃗
F
h
, i.e. has to be at an angle arctan
µ
with
respect to the normal to the plane. Let us draw the
force triangle
m
⃗
g
+
⃗
f
+
⃗
F
=
0: the vector
m
⃗
g
can be
drawnimmediately(itsdirectionandmagnitudeare
known), the direction of
⃗
f
can be noted by a straight
line passing through the terminal point of
m
⃗
g
.
⃗
F
has
to connect that straight line to the initial point of
m
⃗
g
.
For which direction is its magnitude minimal?
5.
Go to the reference frame of the inclined surface
(invoke Ideas 6 and 7) and use the same method as
forproblem4(
⃗
a
+
⃗
g
functionsastheeffectivegravity
⃗
g
e
).
6.
Usearotatingreferenceframeassociatedwiththe
cylinder (where the block is at rest, and the centri-
fugal force
⃗
f
t
is constant and pointing downwards).
(a)
The terminal point of the net force of gravity and
centrifugal force is moving on a circle and has to
be equal to the net force
⃗
f
of the normal and fric-
tional forces. What is the maximum allowed angle
between the vectors
⃗
f
t
and
⃗
f
so that there be no
slipping? For which direction of
m
⃗
g
is the angle
between the vectors
⃗
f
t
and
⃗
f
maximal?
(b)
There are
still only three forces; as long as there is an equilib-
rium, these three vectors must form a triangle and
hence, must lay on the same plane. According to the
idea 9, we’ll depict the force balance in this plane,
i.e. in the plane defined by the vectors
⃗
g
and
⃗
f
t
. The
approach used in part (a) can still be used, but the
terminal point of
⃗
f
t
+
m
⃗
g
draws only an arc of a full
circle. Determine the central angle of that arc. De-
pending on the arc length, it may happen that the
maximal angle between the surface normal (= the
direction of
⃗
f
t
) and
⃗
f
is achieved at one of the en-
dpoints of the arc.
7.
Based on the Fact no. 4, on which line does the
intersection point of the frictional forces have to lie?
What can be said about the two angles formed by
the frictional force vectors and the thread’s direc-
tion. Given the Idea no. 1 (the axis is perpendic-
ular with the tension in the thread)? Now combine
the two conclusions above. Where is the intersection
point of the friction force vectors? What is the direc-
tion of the cylinder’s velocity vectors at the points
where the cylinder rests on the rough band? Where
isthecylinder’sinstantaneousrotationaxis(seehow
to find it in the kinematics brochure)? What is the
velocity vector of the cylinder’s centre point?
(b)
Will the equilibrium condition found above be vi-
olated if the surface is uniformly rough?
8.
Draw a circle whose diameter is the straight line
connecting the points of support. Use Fact no. 6:
which curve can the ball move along? Where is the
bottom-most point of this curve?
9.
Consider the torques acting on the rod with re-
spect to the hinge. For which angle
α
will the net
forceof the normaland frictionalforcespush therod
harder against the board?
10.
By how much will the block descend if the
thread is extended by
δ
?
11.
Let’s assume that the horizontal component of
the tension in the rope is
T
x
. What is the vertical
component of the tension next to the ceiling? Next
to the weight? Write down the condition for the bal-
ance of the forces acting on a) the weight and b) the
system of weight & rope (cf. Idea no. 4).
12.
Seeing as
H
≪
L
, clearly the curvature of the
rope is small, and the angle between the tangent
to the rope and horizon remains everywhere small.
From the horizontal force balance for the rope, ex-
press the horizontal component of the tension force
T
x
as a function of the length
l
(note that while
T
x
remains constant over the entire hanging segment
of the rope, we’ll need its value at the point
P
separ-
ating the hanging and lying segments). Write down
the balance of torques acting on the hanging piece of
the rope with respect to the holding hand (according
to what has been mentioned above, the arm of the
gravity force can be approximated as
l
/2). As a res-
ult, you should obtain a quadratic equation for the
length
l
.
13.
Use Idea 8: change into the reference frame of
the rotating hinge.
a)
Following the idea 15, write
down the condition of torque balance with respect
to the hinge (Idea no. 2) for a small deviation angle
ϕ
. Whichgeneratesabiggertorque,
m
⃗
g
orthecentri-
fugal force? (Note that alternatively, the idea 17 can
be also used to approach this problem).
b)
Follow-
ing the idea 17, express the net potential energy for
the small deviation angles
ϕ
1
and
ϕ
2
using the en-
ergy of the centrifugal force (which resembles elastic
force!) and the gravitational force; according to the
idea 16, keep only the quadratic terms. You should
obtain a quadratic polynomial of two variables,
ϕ
1
and
ϕ
2
. The equilibrium
ϕ
1
=
ϕ
2
=
0 is stable if it
corresponds to the potential energy minimum, i,e, if
the polynomial yields positive values for any depar-
ture from the equilibrium point; this condition leads
to two inequalities. First, upon considering
ϕ
2
=
0
(with
ϕ
2
̸
=
0) we conclude that the multiplier of
ϕ
2
1
has to be positive. Second, for any
ϕ
2
̸
=
0, the poly-
nomial should be strictly positive, i.e. if we equate
this expression to zero and consider it as a quad-
ratic equation for
ϕ
1
, there should be no real-valued
roots, which means that the discriminant should be
negative.
14.
Apply the ideas 15 ja 18 for such a angular po-
sition of the beam, for which the magnitude of the
buoyant force doesn’t change (i.e. by assuming a
balance of vertical forces). From Idea no. 2, draw
the axis through the centre of mass. While comput-
ing the torque of the buoyant force, use Ideas 19,
20; the cross-section of the underwater part of the
beam could be represented as a superposition of a
rectangle and two narrow triangles (one of them of
negative mass).
15.
The container & water system is affected by
the gravity and the normal reaction force of the ho-
rizontal surface on the liquid. Since we know the
pressure of the liquid at the base of the container,
we can express the mass of the container from the
vertical condition for equilibrium.
16.
To compute the first correction using the per-
turbation method we use the Fact 49 and the ref-
erence system of the block sliding down uniformly
and rectilinearly: knowing the magnitude and the
direction of the frictional force we can find its com-
ponent in
⃗
w
and
⃗
u
direction. The sign of the latter
flips after half a period, and so it cancels out upon
averaging.
17.
Let us choose the origin of the vertical
x
-axis to
be a point on the surface of the ocean very far from
the iron deposit. For the zero reference point of the
Earth’s gravitational potential we shall choose
x
=
0
(i.e.
ϕ
earth
=
gx
), forthatoftheirondepositweshall
take a point at infinity. Then, for the points on the
ocean’s surface very far from the iron deposit, the
gravitational potential is zero. It remains to find an
expressionforthepotentialabovetheirondepositas
a function of
x
(using the principle of superposition)
and equate it to zero.
18.
Let us employ the reference frame of the plat-
form. Let us the consider the balance of torques
with respect to the axis of the small disk (then the
lever arm of the force exerted by that axis is zero).
Let us divide the disk into little pieces of equal size.
The frictional forces acting on the pieces are equal
by magnitude and are directed along the linear ve-
locities of the points of the disk (in the chosen refer-
ence frame). Since the motion of the disk can be rep-
resented as a rotation around an instantaneous axis,
then concentric circles of frictional force vectors are
formed (centred at the instantaneous rotation axis).
Clearly, the net torque of these vectors with respect
to the disk’s axis is the smaller, the smaller is the
circles’ curvature (i.e. the farther the instantaneous
rotation axis is): the torque is zero when the instant-
aneous rotation axis is at infinity and the concent-
ric circles become parallel straight lines. An instant-
aneous rotation axis at infinity means that the mo-
tion is translational,
ω
3
=
0 (since the linear velocity
v
=
ω
3
r
of a given point is finite, but
r
=
∞
).
19.
The instantaneous axis of rotation is at a dis-
tance
r
=
v
/
ω
from the disk’s axis. Let’s use the
same imaginary slicing as in the previous problem.
Now compute the component of the net force in the
direction of motion. Notice that the frictional forces
on the points that are symmetrical with respect to
the instantaneous rotation axis balance each other
across a whole circular region of radius
R
−
r
. The
non-balancedregionisunfortunatelyshapedforcal-
culation. Let us imagine extending the "balanced"
region up to
R
(the dashed circle in the figure). The
part of this extended balanced region, where there
is no actual rotating disk underneath (the dark gray
crescent in the figure), could be represented as a su-
perposition of the two disks, one rotating clockwise
and the other – anticlockwise. In that case the clock-
wise component partakes in the balancing, whereas
the anticlockwise component remains unbalanced.
To sum up, two thin crescent-shaped regions remain
unbalanced: one corresponds to the the real disk
(light gray in the figure), the other — to a disk ro-
tating anticlockwise (dark gray); normal to
⃗
v
, the
width of these regions is everywhere equal to
r
. The
net force is the easiest to find by integrating across
the crescent-shaped regions using the polar coordin-
ate
ϕ
:
|
d
⃗
F
|
=
A
·
dS
, where
dS
is the area of the
surface element;
dF
x
=
A
cos
ϕ
dS
=
B
cos
2
ϕ
d
ϕ
,
F
x
=
∫
dF
x
=
B
∫
2
π
0
cos
2
ϕ
d
ϕ
. What are the values
of the constants
A
and
B
?
O
r
20.
Consider the unit vector
⃗
τ
directed along the in-
finitesimal displacement vector of the centre of the
mass at the instant when the pencil begins moving.
Let’s express its coordinates in the Cartesian axes
(
x
,
y
,
z
)
, where
x
is parallel to the pencil and the
(
x
,
y
)
-plane is parallel to the inclined slope. Using
the spatial rotations formulae we represent it in the
newcoordinates
(
x
′
,
y
′
,
z
)
,whicharerotatedwithre-
spect to
(
x
,
y
,
z
)
around the
z
-axis by an angle
ϕ
(so
that the axis
x
′
is horizontal). Using the spatial rota-
tions formulae we express the vector’s
⃗
τ
vertical co-
ordinate
z
′
in the
(
x
′
,
y
′
,
z
′
)
coordinate axes, which
is obtained from the axes
(
x
′
,
y
′
,
z
)
by rotating about
the
x
′
by the angle
α
.
21.
The string connects the two points with the
shortest distance along the cylinder’s side; when
unfolded, the cylinder is a rectangle. Consider the
vertical plane touching the surface of the cylinder
that includes the hanging portion of the string. This
plane and the cylinder touch along a straight line
s
. If you imagine unfolding the cylinder, the angle
between the string and the straight line
s
is equal
to the cylinder’s inclination angle
α
. Given this,
l
is
easy to find. When the weight oscillates, the trace of
the string still stays straight on the unfolded cylin-
der. Therefore the length of the hanging string (and
thus the weight’s potential energy) do not depend
in any oscillatory state on whether the surface of
the cylinder is truly cylindrical or is unfolded into
a planar vertical surface (as long as the spatial ori-
entation of the axis
s
is preserved).
22.
Write down the two equations describing the
balance of force and torques, and then another one
that describes the linear relation between the elong-
ations of the string:
T
1
−
T
2
=
T
2
−
T
3
.
23.
Initially only the vertical forces affect the
hanging block, therefore the initial displacement
vector is also vertical. If the acceleration of the large
block is
a
1
, that of the block on top of it —
a
2
and
that of the hanging block —
a
3
, then
a
1
+
a
2
=
a
3
holds. Now we can write down Newton’s 2nd law
for each body. The fourth and the final unknown is
the tension in the string.
24.
Go to the reference frame of the wedge-block. In
the borderline case, the force of inertia’s and grav-
ity’s net force on the ball
m
is normal to the left
slope (so that the ball stay at rest there). Consider
the net forces acting on the balls. Their compon-
ents normal to the surface they rest on are
⃗
F
⊥
1
and
⃗
F
⊥
2
. These are equal to the normal forces
⃗
N
1
and
⃗
N
2
acting on the balls and therefore have to have equal
magnitudes (
F
⊥
1
=
F
⊥
2
) to ensure that the force bal-
ance is achieved horizontally for the wedge-block.
25.
Let’s take the displacement
ξ
of the wedge as
coordinate describing the system’s position. If the
wedge moves by
ξ
, then the block moves the same
amount with respect to the wedge, because the rope
is unstretchable, and the kinetic energy changes by
Π
=
mg
ξ
sin
α
. The velocity of the wedge is
˙
ξ
and
that of the block is 2
˙
ξ
sin
α
2
(found by adding velocit-
ies, where the two vectors
˙
ξ
are at an angle
α
), there-
fore the net kinetic energy
K
=
1
2
˙
ξ
2
(
M
+
4
m
sin
2
α
2
)
.
Then we find
Π
′
(
ξ
) =
mg
sin
α
and
M
=
M
+
4
m
sin
2
α
2
; their sum gives the answer.
26.
Again, let’s take the wedge’s displacement as
the coordinate
ξ
; if the displacement of the block
alongthesurfaceofthewedgeis
η
, thenthecentreof
mass being at rest gives
η
(
m
1
cos
α
1
+
m
2
cos
α
2
) =
(
M
+
m
1
+
m
2
)
ξ
. From here one can extract
η
as
a function of
ξ
, but to keep the formulae brief it’s
better not to substitute this expression everywhere.
The kinetic energies of the block can be found as
sums of horizontal [
1
2
m
i
(
˙
ξ
−
˙
η
cos
α
i
)
2
] and vertical
[
1
2
m
i
(
˙
η
sin
α
i
)
2
] energies.
27.
When writing down energy conservation, note
that the block’s velocity is twice the cylinder’s ve-
locity horizontal component and that the latter is
equal to the vertical component, too (why?). Project
Newton’s 2nd law onto the axis that passes through
the top corner of the step and the cylinder’s centre:
this axis is perpendicular both to the normal force
between the block and the cylinder and to the cylin-
der’s tangential acceleration. Second question: the
ratio of two normal forces is constant (why? what
is it equal to? Hint: compare the horizontal acceler-
ations of the cylinder and the block and remember
Newton’s 2nd law), therefore they will be equal to
zero at the same instant.
28.
By projecting Newton’s 2nd law on the axis in
the direction of the normal force we see that the nor-
mal force is the smallest at the bottommost point of
the trajectory’s arch-shaped part. (There, the centri-
petal acceleration is the largest, gravitational force’s
component along the axis is the smallest).
29.
The energy of the "pellet & block" system is al-
ways conserved; momentum will only start to be
conserved once the pellet passes the bottommost
point. When it arrives there for the second time, the
block’s velocity is maximal (why?).
30.
Let’s apply Idea no. 44 for
⃗
P
: the system’s
net momentum is
P
=
ω
lm
+
2
ω
lM
, net force
F
=
(
m
+
M
)
g
−
T
. The same using rotational consid-
erations: with respect to the leftmost ball’s initial
position, the angular momentum is
l
(
2
ω
l
)
M
(velo-
city is 2
ω
l
, the velocity’s lever arm —
l
); net torque
is
(
T
+
Mg
)
l
. Now, for the formula given in Idea
no. 44 we need the angular acceleration
ε
=
˙
ω
.
Let’s find it using Method no. 6:
Π
=
l
ϕ
(
m
+
2
M
)
,
K
=
1
2
˙
ϕ
2
l
2
(
m
+
4
M
)
. Another solution route: the
ratio of accelerations is 1:2; there are four unknowns
(two normal forces, acceleration and string tension);
equations: three force balances (for either ball and
the rod) and one torque balance (wrt the left end-
point of the rod).
31.
Method no. 6: for the generalized coordinate
ξ
we can use the displacement of the thread’s end-
point. Ideas no. 32,20: the change of the system’s
CM
y
-coordinate is
ξρ
h
/
M
(
h
— the difference in
the heights of the thread’s endpoints,
M
— the net
mass of the system; assume that
ξ
≪
h
). For the
x
-coordinate it’s 2
ξρ
R
/
M
.
32.
⟨
T
(
1
+
cos
α
)
⟩
=
2
mg
,
T
=
⟨
T
⟩
+
˜
T
, where
|
˜
T
| ≪
T
. Based on the Idea no. 16 we ignore the
tiniest term
⟨
˜
T
α
2
⟩
and note that
⟨
α
2
⟩
>
0.
33.
We have to consider two options: either all the
bodies move together, or the rightmost large block
moves separately. Why cannot the situations occur
where
(a)
all three components move separately, or
(b)
the left large block moves separately?
34.
After the collision the ball’s trajectories are or-
thogonal crossing straight lines; the angle with re-
spect to the initial trajectory is determined by how
much the collision was off-centre.
35.
For slightly non-central motion: what will be
the direction of momentum of the ball that was first
to be hit? Now apply the Idea no. 50 again. Cent-
ral motion: express the velocities after the collision
via the horizontal component of the momentum
p
x
that has been transferred to one of the balls. What is
the transferred vertical component
p
y
? Energy con-
servation provides us an equation to find
p
y
(it is
convenient to express the energy as
p
2
/2
m
).
36.
Thegraphlookslike
n
intersectingstraightlines;
the intersection point of a pair of straight lines cor-
responds to a collision of two balls (the graph of
either ball’s motion is a jagged line; at a collision
point the angles of the two jagged lines touch one
another so that it looks as if the two straight lines
intersect).
37.
Initial velocities in the centre of mass:
mv
m
+
M
,
Mv
m
+
M
, final velocities are zero; friction does work:
µ
mgL
.
38.
Based on the figure we immediately obtain
(to within a multiplicative constant) the magnitudes
and directions of the momenta, but not which mo-
mentum is which ball’s. It is necessary to find out
where the ball marked with an arrow will proceed
after the collision. Fact no. 13 will help choose from
the three options.
39.
Energy: in time
dt
the distribution of the liquid
will change: there is still some water at the centre,
butacertainmass
dm
hasbeendisplacedfromabove
to the level of the tap (and then through the tap),
so the change in the system’s potential energy is
gH
·
dm
. Momentum: the water in the barrel obtains
the total momentum
ρ
gHS
·
dt
from the walls. This
momentum is passed on to the stream of water with
the mass
ρ
Sv
·
dt
.
40.
Energy is not conserved: the grains of sand
slip and experience friction. In time
dt
the sand
landing on the conveyor belt receives momentum
dp
=
v
·
dm
=
v
µ
·
dt
fromthebelt: theforcebetween
the freshly fallen sand and the belt is
F
1
=
dp
/
dt
.
The sand already lying on the belt experiences the
gravitational force
mg
which is compensated by the
component of the friction parallel to the belt,
F
2
=
mg
cos
α
, where
m
=
σ
L
is the mass of the sand on
the belt and
σ
v
=
µ
. The minimization has to be
done over
v
.
41.
During the collision
∆
p
⊥
=
√
2
gh
.
42.
Consider a short section of the path along the
hill with length
dl
. In addition to the change in the
potential energy work is done to overcome friction,
dA
h
=
µ
mg
tan
α
·
dl
. WE find
dA
h
=
C
·
dx
, where
C
is a constant. Summing over all such little path
increments
dl
we find
A
h
=
C
∆
x
.
43.
The kinetic energy
K
=
m
2
˙
x
2
+
M
˙
x
2
, where
x
is the displacement along the slanted surface;
Π
=
(
M
+
m
)
sin
α
. Having found the acceleration
a
we
change into a reference frame (of the cylinder) mov-
ing with acceleration
a
(Ideas no. 6 and 7), where the
block is being displaced along the effective accelera-
tion due to gravity — as low as possible.
44.
According to the Ideas no. 59 and 60, the
angular momentum of the rod before the collision
is
L
0
=
Mlv
−
1
3
Ml
2
ω
; after the collision
L
1
=
Mlv
′
−
1
3
Ml
2
ω
′
;
L
1
=
L
2
. The expression for energy
is
K
=
1
2
Mv
2
+
1
6
Ml
2
ω
2
. The condition for being at
the end:
v
′
+
l
ω
′
=
0 (we consider
ω
to be positive if
the rotation is in the direction marked in the figure).
45.
The angular momentum with respect to the
impact point before the collision:
mv
(
x
−
l
2
)
−
I
0
ω
,
where
v
=
ω
l
2
and
I
0
=
1
12
ml
2
.
46.
The instantaneous rotation axis passes the con-
tact point of the cylinder and the floor; its distance
from the centre of mass does not change, so we can
use Idea no. 63;
I
=
3
2
mR
2
.
47.
Let us direct the
z
axis upward (this will fix the
signs of the angular momenta). The final moment of
inertia with respect to the
x
-axis is
−
7
5
mv
y
R
−
muR
and with respect to the
y
-axis is
7
5
mv
x
R
.
48.
Immediately after the first collision the centres
of masses of both dumbbells are at rest, the velocit-
ies of the colliding balls reverse direction, the non-
colliding balls’ velocities don’t change. Both dumb-
bells act like pendula and complete half an oscilla-
tion period, after which the second collision occurs
– analogous to the first one.
49.
The grains of sand perform harmonic oscilla-
tions in the plane perpendicular to the cylinder’s
axis — like a mathematical pendulum of length
l
=
R
in the gravitational field
g
cos
α
; along the axis
thereisuniformacceleration(
a
=
g
sin
α
). Focussing
occurs if the time to cross the trough along its axis is
an integer multiple of the oscillation’s half-period.
50.
Observing the equilibrium position we con-
clude that the centre of mass lies on the symmetry
axis of the hanger. The three suspension points must
be located on the two concentric circles mentioned
by Idea no. 67. Therefore one of the circles must
accommodate at least two points out of the three,
whilethecircles’centre(thehanger’scentreofmass)
must lie inside the region bounded by the hanger’s
wires on its symmetry axis. There is only one pair of
circles that satisfies all these conditions. Computing
the radii
l
1
and
l
2
of the circles using trigonometry
we determine the reduced length of the pendulum
l
1
+
l
2
and, using that, the oscillation period.
51.
The effective mass of the moving water can be
found using the acceleration of the falling ball. For
the rising bubble the effective mass is exactly the
same, the mass of the gas, compared to that, is neg-
ligibly small.
52.
The water stream could be mentally divided
into two parts: the leftmost stream will turn to the
left upon touching the trough, the rightmost — to
the right. Thus, two imaginary ’water tubes’ form.
In either tube the static pressure is equal to the ex-
ternal pressure (since there is the liquid’s outer sur-
face in the vicinity): according to Bernoulli’s law, the
velocity of the liquid cannot change. Based on the
conservation of momentum horizontally, the mo-
mentum flows of the left- and right-flowing streams
have to add up to the original stream’s momentum
flow’s horizontal component. Note that due to con-
tinuity,
µ
=
µ
v
+
µ
p
.
53.
Due to continuity
(
u
+
v
)(
H
+
h
) =
Hu
Const,
where
h
=
h
(
x
)
is the height of the water at point
x
and
v
=
v
(
x
)
is the velocity. We can write down
Bernoulli’s law for an imaginary ’tube’ near the sur-
face (the region between the free surface and the
stream lines not far from the surface):
1
2
ρ
(
u
+
v
)
2
+
ρ
g
(
H
+
h
) =
1
2
ρ
u
2
+
ρ
gH
=
Const. We can ig-
nore that small second order terms (which include
the factors
v
2
or
vh
)
54.
The phase trajectory is a horizontal rectangle
with sides
L
and 2
mv
, where
L
is the distance from
the block to the wall; the adiabatic invariant is thus
4
Lmv
.
55.
Consider the balance of torques. For the net
force vectors of the normal and frictional forces,
when you extend them, their crossing point must be
above the centre of mass.
56.
Let’s write down Newton’s 2nd law for rota-
tional motion with respect to the crossing point of
the normal forces: the angular momentum of the
bugis
L
=
mvl
sin
α
cos
α
, thespeedofchangeofthis
angular momentum will be equal to the torque due
to gravity acting on the bug (the other forces’ lever
arms are zero). When computing the period, note
that the acceleration is negative and proportional to
the distance from the bottom endpoint, i.e. we are
dealing with harmonic oscillations.
57.
The blocking occurs if the net force of normal
and frictional forces pulls the rod downwards.
58.
Once the blocking occurs we can ignore all
the forces apart from normal and frictional ones.
Suppose it has occurred. Then the net frictional
and normal forces acting from the left and from
the right have to balance each other both as forces
and torques, i.e. lie on the same straight line and
have equal magnitudes. Thus we obtain the angle
between the surface normal and the net force of fric-
tion and normal force.
59.
Consider the direction of the torque acting on
the plank with respect to the point of contact, when
theplankhasturnedbyanangle
ϕ
: thecontactpoint
shifts by
R
ϕ
, the horizontal coordinate of the centre
of mass shifts by the distance
h
2
ϕ
from the original
position of the contact point.
60.
The only force from the surface on the sys-
tem vessel & water is equal to the hydrostatic pres-
sure
ρ
gh
π
R
2
; it balances the gravitational force
(
m
+
ρ
V
)
g
. Note that
H
=
R
−
h
.
61.
The gravitational potential of the centrifugal
force is
1
2
ω
2
r
2
, where
r
is the distance from the ro-
tation axis.
62.
Assume the reference frame of the large block
(which moves with acceleration
a
). Where does the
effective gravity (the net force of the gravity and the
force of inertia) have to be directed? What is
a
?
With which acceleration does the little block fall in
this reference frame? What is the tension
T
of the
thread? Having answers to these questions we can
write down the equilibrium condition for the large
block
ma
=
T
(
1
−
sin
α
)
.
63.
Let us use the displacement of the sphere (down
the inclined surface) as the generalized coordinate
ξ
.
What is the displacement of the sphere (up the other
inclined surface)? Evidently
Π
= (
m
−
M
)
g
ξ
sin
α
.
The normal force between the two bodies can be
found by projecting Newton’s second law onto the
inclined surface’s direction.
64.
Let the displacement of the large cylinder be
ξ
,
the horizontal displacement of the middle and the
leftmost cylinder, respectively,
x
and
y
. What is the
relationship between them given that the centre of
mass is at rest? What is the relationship between
them given that the length of the rods does not
change? From the two equations thus obtained we
can express
x
and
y
via
ξ
. If we assume the displace-
ment to be tiny, what is the relationship between the
vertical displacement
z
of the middle cylinder and
the horizontal projection of the rod’s length,
ξ
−
x
?
Knowing these results, applying Method no. 6 is
straightforward.
65.
Where is the small displacement
ξ
of the sphere
directed (see Idea no. 30)? What is the displacement
of the ring expressed via
ξ
? Use Method no. 6.
66.
Use Idea no. 38 along with energy conserva-
tion by projecting the force and the acceleration in
the Newton’s 2nd law radially.
67.
Let us use some ideas from kinematics to find
the acceleration of the sphere (K1, K29 and K2: by
changing into the reference frame moving with ve-
locity
v
we find the component of the sphere’s accel-
eration along the rod and by noticing that the hori-
zontal acceleration of the sphere is zero, we obtain,
using trigonometry, the magnitude of the accelera-
tion). Now use Newton’s 2nd law.
68.
Using the velocity
v
of the sphere we can ex-
press the velocity of the block at the moment being
investigated (bearing in mind that their horizontal
velocities are equal). Using Idea no. 38 we find that
the block’s (and thus the sphere’s) horizontal accel-
eration is zero; by using Newton’s 2nd law for the
sphereandthehorizontaldirectionweconcludethat
the tension in the rod is also zero. From the energy
conservation law we express
v
2
and from Newton’s
2nd law for the sphere and the axis directed along
the rod we obtain an equation wherein hides the
solution.
69.
UsingNewton’s2ndlawinvestigatewhitherthe
system’s centre of mass will move — to the left or to
the right (if the centre of mass had not move, then
the both events would have happened at the same
time).
70.
To answer the first part: show that the force per-
pendicular to velocity is zero (use Method no. 3 and
Idea no. 26). To answer the second part use Method
no. 3 and idea 54.
71.
Due to the length of the thread there are no ho-
rizontal forces, i.e. the horizontal component of mo-
mentum is conserved, and so is the energy. From
the two corresponding equation the limiting velo-
city
v
=
v
0
can be found, for which the bottom
sphere ascends exactly to the height of the top one.
Note that at that point its vertical velocity is zero, cf.
Idea no. 42.
72.
Use Idea no. 49. Options: all block keep to-
gether; everything slides; the top one slides and and
the bottom two stay together (why is it not possible
that the top two keep together and the bottom one
slides?).
73.
Which conservation law acts when the two boys
collide (during a limited time of collision) — do we
consider the collision absolutely elastic or inelastic
(can momentum be lost and where? If it is inelastic,
where does the energy go?), see Idea no. 56? After
the collision: the common acceleration of the two
boys is constant, knowing the initial and final velo-
cities finding the distance becomes an easy kinemat-
ics problem.
74.
Prove that for a vertical thread the velocity is
maximal (by applying Idea no. 42 for the rotation
angleoftherodshowthatitsangularvelocityiszero
in that position; use Idea no. 59). Then it only re-
mains to apply energy conservation (remember that
ω
=
0).
75.
Find the instantaneous rotation axis (make sure
that its distance from the centre of mass is
1
2
). Prove
that the centre of mass moves along a circle centred
at the corned of the wall and the floor, whereas the
polar coordinate of the centre of mass on that circle
is the same as the angle
ϕ
between the wall and the
stick. Express the kinetic energy as a function of
the derivative ˙
ϕ
of the generalized coordinate
ϕ
us-
ing the parallel-axis (Steiner’s) theorem and express
the energy conservation law as
ω
2
=
f
(
ϕ
)
; using
Method no. 6 we obtain
ε
=
˙
ω
=
1
2
f
′
(
ϕ
)
. When
the normal force against the wall reaches zero, the
acceleration of the centre of mass is vertical: present
this condition using the tangential and radial accel-
erations of the centre of mass on its circular orbit (
l
2
ε
and
l
2
ω
2
respectively) and use it as an equation to
find
ϕ
.
76.
Based on Idea no. 62 we find that
ω
=
6
v
/
l
. Us-
ing energy and momentum conservation we elimin-
ate the puck’s velocity after the collision and express
the mass ratio.
77.
The forces along the normal to the surface are
elastic forces, so the energy in vertical direction is
conserved during the collision: after the collision
the corresponding velocity component is the same
as before. To find the other two unknowns, the ho-
rizontal and angular velocities, we can obtain one
equation using Idea no. 62. The second equation
arises from
(a)
the condition that the velocity of the
ball’s surface is zero at the contact point (no sliding;
(b)
the equation arising from 58).
78.
Using the idea 49 we investigate the sliding and
rolling regimes. In the latter case the quickest way
to find the answer is to use Idea no. 63.
79.
The velocity can be found from the conservation
laws for energy and momentum (note that the hoop
is moving translationally). To find the acceleration it
is convenient to use the non-inertial reference frame
of the hoop, where the centripetal acceleration of the
block is easily found. The condition for the radial
balance of the block gives the normal force between
the block and the hoop (don’t forget the force of in-
ertia!); the horizontal balance condition for the hoop
provides an equation for finding the acceleration.
80.
Let us assume the block’s velocity to be ap-
proximately constant. For a certain time
t
l
the base
slides to the left with respect to the block and the
momentum imparted by the frictional force at that
time is also directed to the left. During the remain-
ing time
t
r
the base slides to the right with respect-
ive momentum directed to the right as well. The
equilibrium condition is that the two momenta have
equal magnitudes; hence we ding the equilibrium
value of
t
l
/
t
r
. From the graph we find the velocity
for which that ratio has the needed value.
81.
As the water flows against the paddles it ob-
tainthesameverticalvelocity
u
asthepaddlesthem-
selves. This allows to compute the momentum im-
parted to the paddle per unit time (i.e. the force),
which ends up being proportional to the difference:
F
∝
v
−
u
. From there, it is not very hard to find the
maximum of the power
Fu
.
82.
In the reference frame of the board the problem
is equivalent to the problem no. 52.
83.
Go into the (accelerated) reference frame of the
wagon, where the effective gravity
√
a
2
+
g
2
is at a
smallanglewithrespecttothevertical. Theloadwill
oscillate yet remain motionless at the end if the cable
is vertical at the stopping moment and the load’s ve-
locity is zero. It is possible when the corresponding
position is the maximal deviation during the oscil-
lation. Therefore the oscillation amplitude has to be
the same both during the acceleration and decelera-
tion, so that even when the deceleration begins the
cable has to be vertical. In that case, how are the ac-
celeration time and the oscillation period related?
84.
If the shockwave is at the point where the in-
tersection area of its wavefront and the considered
body is
S
, then what is the force acting on the body?
Let us assume that the body stays (almost) at the
same place as the shockwave passes it. Then the mo-
mentum imparted during the time
dt
can be found
using the cross-sectional area
S
and the distance
dx
=
c
s
·
dt
covered by the wavefront. Note that
S
·
dx
is the volume element. Finally we sum over
all imparted momenta.
85.
The rod will act like a spring (since the rod is
thin and made out of steel, while steel is elastic).
After the left sphere has collided with the station-
ary sphere, the latter will acquire velocity
v
0
and
the former will stay at rest. Then the dumbbell, as
a system of spheres and springs, will begin oscillat-
ing around its centre of mass. What is the velocity of
the centre of mass? Convince yourself that after half
a period the single sphere is already far enough that
the left sphere is not going to collide with it again.
The oscillations of the dumbbell will decay little by
little — so some energy will be lost there.
1.
arcsin
R
µ
(
R
+
l
)
√
µ
2
+
1
.
2.
arcsin
m
M
+
m
µ
√
µ
2
+
1
.
3.
mg
/2.
4.
a)
µ
mg
/
√
1
+
µ
2
; b)
mg
sin
(
arctan
µ
−
α
)
.
5.
µ
≥
|
g
sin
α
−
a
cos
α
|
g
cos
α
+
a
sin
α
, if
g
+
a
tan
α
>
0.
6.
a)
ω
2
R
≥
g
√
1
+
µ
−
2
;
b)
ω
2
R
≥
g
√
1
+
µ
−
2
, if
µ
<
cot
α
and
ω
2
R
≥
g
(
cos
α
+
µ
−
1
sin
α
)
if
µ
>
cot
α
7.
v
/2.
8.
tan2
α
=
h
/
a
9.
µ
1
≥
√
l
2
−
h
2
/
h
10.
3
mg
11.
2arctan
[(
1
+
m
M
)
cot
α
]
12.
√
2
HL
µ
+
µ
2
H
2
−
µ
H
≈
√
2
HL
µ
−
µ
H
≈
7,2m.
13.
a)
ω
2
<
g
/
l
; b)
(
2
−
√
2
)
g
/
l
14.
1
2
(
1
−
3
−
1/2
)
ρ
v
≈
211kg/m
3
15.
π
3
ρ
R
3
16.
v
/
µ
2
cot
2
α
−
1
17.
4
3
π
Gr
3
∆
ρ
/
g
(
r
+
h
)
≈
0,95cm
18.
−
ω
19.
µ
mgv
/
ω
R
20.
cos
ϕ
tan
α
<
tan30
◦
21.
L
−
π
R
/2cos
α
; 2
π
√
L
/
g
22.
1
12
mg
,
1
3
mg
,
7
12
mg
23.
mg
/
(
2
M
+
m
)
24.
m
<
M
cos2
α
.
25.
mg
sin
α
/
[
M
+
2
m
(
1
−
cos
α
)] =
mg
sin
α
/
[
M
+
4
m
sin
2
α
2
]
.
26.
g
(
m
1
sin
α
1
−
m
2
sin
α
2
)(
m
1
cos
α
1
+
m
2
cos
α
2
)
(
m
1
+
m
2
+
M
)(
m
1
+
m
2
)
−
(
m
1
cos
α
1
+
m
2
cos
α
2
)
2
.
27.
mg
(
5
√
2
−
4
)
/6
)
; Simultaneously.
28.
cos
α
≥
1
3
(
2
+
v
2
/
gR
)
29.
2
m
M
+
m
√
2
gR
30.
mMg
/
(
m
+
4
M
)
31.
F
x
=
2
Ra
ρ
,
F
y
=
ρ
(
m
+
ρ
L
)
g
−
(
L
−
π
R
−
2
l
)
a
, where
a
=
ρ
g
(
L
−
π
R
−
2
l
)
/
(
m
+
ρ
L
)
.
32.
The one that had not been pushed.
33.
If
F
≤
2
µ
mg
m
+
M
2
m
+
M
:
a
1
=
a
2
=
1
2
F
M
+
m
;
otherwise
a
1
=
F
M
−
µ
g
m
M
,
a
2
=
µ
g
m
2
m
+
M
.
34.
On a half-circle.
35.
(a) v
/5;
(b) v
/4.
36.
n
(
n
−
1
)
/2
37.
2
µ
gL
(
1
+
m
M
)
38.
3,5; was coming from below right.
39.
A:
√
2
gh
;
√
gh
.
40.
2
R
µ
√
gl
sin
α
,
√
gl
sin
α
.
41.
u
−
µ
√
2
gh
.
42.
mg
(
h
+
µ
a
)
.
43.
arctan
2
5
≈
21
◦
48
′
.
44.
(a)
(
ω
l
+
3
v
)
/4;
(b)
(
ω
l
+
v
)
/2.
45.
At a distance 2
l
/3 from the holding
hand, where
l
is the length of the bat.
46.
2
3
F
M
a
R
47.
(
v
x
0
,
v
y
0
−
5
7
u
)
48.
L
/
v
0
+
π
√
m
/2
k
49.
1
2
π
2
(
n
+
1
2
)
2
R
tan
α
50.
1,03s
51.
2,0g
52.
v
1
=
v
2
=
v
; cot
2
α
2
53.
√
gH
.
54.
5m/s.
55.
(a)
tan
≤
2
µ
;
(a)
impossible.
56.
g
(
1
−
x
l
)
sin
−
1
α
;
π
2
√
l
sin
α
/
g
57.
µ
<
cot
α
.
58.
µ
1
<
tan
α
2
and
µ
2
<
tan
α
2
.
59.
R
>
h
/2
60.
3
√
3
m
/
πρ
61.
ω
2
R
2
/2
g
62.
M
/
m
=
cot
α
−
1.
63.
2
mM
M
+
m
g
tan
α
64.
g
/9.
65.
g
m
+
M
m
+
M
sin
2
α
sin
2
α
.
66.
2/3
R
67.
m
[
g
−
v
2
(
2
l
−
x
)
/
√
2
l
2
]
68.
M
/
m
=
4,
u
=
√
gl
/8.
69.
The first one arrives first
70.
A straight line; if
ω
̸
=
0
71.
√
2
gl
(
1
+
m
/
M
)
72.
F
3
m
, if
F
m
µ
g
<
6;
F
4
m
+
1
2
µ
g
, if
6
<
F
m
µ
g
<
10; 3
µ
g
, if
F
m
µ
g
>
10
73.
m
2
v
2
/2
(
M
2
−
m
2
)
µ
g
74.
(
l
−
H
2
)
g
75.
arccos
2
3
≈
48
◦
12
′
76.
M
/
m
=
4.
77.
(a)
ω
=
5
v
0
/7
R
,
v
x
=
5
v
0
/7,
v
y
=
√
2
gh
;
(b) v
y
=
√
2
gh
,
v
x
=
v
0
−
2
µ
v
y
,
ω
=
5
√
2
gh
µ
/
R
.
78.
5
7
g
sin
α
, if
µ
>
2
7
tan
α
, otherwise
g
sin
α
−
µ
g
cos
α
79.
2
gr
m
+
M
1
+
cos
ϕ
m
sin
2
ϕ
+
M
m
cos
ϕ
;
gm
sin2
ϕ
m
sin
2
ϕ
+
M
[
1
2
+
m
2
cos
ϕ
(
1
+
cos
ϕ
)
(
m
sin
2
ϕ
+
M
)(
m
+
M
)
]
80.
0,6m/s
81.
1
4
µ
v
2
82.
v
/cos
α
83.
n
−
2
Lg
/4
π
2
l
,
n
=
1,2,...
84.
(a),(b)
(
p
1
−
p
0
)
V
/
mc
s
.
85.
1
2
v
0
; no, a fraction goes into the longit-
udinal oscillations of the rod and then (as
the oscillations die) into heat
