Solutions to Problem Set 3 (in Phys 241X)
John Hauptman
Department of Physics and Astronomy, Iowa State University, Ames, Iowa 50011, USA
(Dated: 15 September 2010)
The solutions to Phys 241X Problem Set 3 are shown here. Problems 2.14 and 2.18 are hard, but
more involved and complicated than difficult.
I.
INTRODUCTION
The assigned problems will define the content of the
course, and these problems are always open for discussion
with me or your recitation instructor. You are encouraged to talk and work with your fellow students on these
problems. Probably one of the assigned problems will be
graded in detail by your recitation instructor, while the
remaining problems will just be noted.
All problems will be taken from the text by Kleppner
and Kolenkow. You will see other problems in recitation
for group solution. Of course, be completely free to call
or email me to talk about any of these problems, or other
problems, at any time.
II.
SOLUTIONS
Prob 2.3 The horizontal force is F = F î, it pushes m1
to the right, the two masses are in contact, and m1 = 2
kg, m2 = 1, kg, and F = 3 N. Let’s do it formally: two
masses, so two “F=ma” equations. The contact force
between the blocks acts like a compressed spring that is
pushing back on the masses, so a contact force −f î acts
to the left on m1 , and the contact force +f î acts to the
right on m2 . The contact forces are equal and opposite
(3rd law). The two “F=ma” equations are
[F − f ] = m1 a1
and
[+f ] = m2 a2 ,
the constraints are that a1 = a2 since the masses move
together, both to the right. Call this common acceleration a. Then, solving for the contact force f ,
and each has gravity down, so the two equations are
[+T − M g] = M aM
The string enforces a constraint that as M accelerates
up, +ĵ, m accelerates down, am = −aM , then
T − M g = −M aM
Prob 2.5 This is Atwood’s machine, with masses M and
m on either side. The pulley is massless and frictionless,
and the tension in the string is T . There are two masses,
and one coordinate axis, so we will have two “F=ma”
equations. There is no horizontal motion, and the vertical axis is +ĵ Each mass has the force of tension +T ĵ,
and
T − mg = m(−aM ),
and these can be solved to get
T = 2(
mM
)g
m + M)
and
aM = −(
M −m
)g.
M +m
Does this make sense? If the masses are equal, M = m,
it should balance and aM = 0. OK. If m = 0, then
M is in free-fall with a = −g. OK. If M = 2m, then
T = (2/3)M g and aM = −g/3. OK.
Prob 2.6 Simplify this problem to a drum of radius R
and a mass M of cement at radius R. If the drum spins
too fast, the cement will always stick to the wall of the
drum, so what is the revolution frequency at which the
mass M is just ready to release from the wall. There is
one mass, one force (gravity), and one “F=ma” equation.
You can solve for the motion of the mass M at all times
(this is worked out in Example 2.6), then find the point
where the acceleration of the mass is zero.
A little thinking suggests that the point where the mass
will begin to fall is at the top of the drum, so let’s write
down “F=ma” at that point for vertical force and acceleration, noting that the acceleration moving on a circle
is v 2 /R,
F
m2
)=
= 1N,
f = F(
m1 + m2
3
for the given input numbers. Does this make sense? Since
the masses move together, we can consider one force, F,
acting on one mass, (m1 + m2 ), and get a = F/(m1 +
m2 ) = 3N/(1 + 2)kg = 1 m/s2 . The mass m2 must
have the same acceleration, and its force is just f , so
a2 = +f /m2 = 1N/1kg = 1 m/s2 . OK.
[+T − mg] = mam .
and
[−M g] = M (−
v2
)
R
the M ’s cancel (doesn’t matter what the mass is), so the
condition on v is that g = v 2 /R. Converting to angular frequency (ω in radians/sec), v = ωR, the frequency
above which the cement sticks to the wall, even at the
top, and g = (ωR)2 /R = ω 2 R. So the condition is
p
ω < g/R. Units OK?
Substituting values (in English units!) suggested in
2
the
q problem, g = 32 ft/s , and R = 2 ft, then ω ≈
2
(32ft/s )/(2ft) = 4 radians/sec, or (2π radians = one
circle = one rotation), this is 0.636 rotations per second,
or 38 rotations per minute.
2
Prob 2.10 A satellite in geosynchronous (“earth-sametime”) orbit has a revolution frequency of one orbit per
day. There is one force, one mass, and the acceleration
is a = −v 2 /r, where v is the orbital velocity and r is the
orbital radius. M is the mass of Earth (≈ 6.0 × 1024 kg)
and Earth’s radius is R = 6.4 × 106 m. The “F=ma”
equations is
v2
Mm
−G 2 = m(− ).
r
r
with the period constrained to be T = 2πr/v = 1day ≈
(π × 107 sec/year) / (365 days/year) ≈ 8.6 × 104 s. You
can now directly calculate r as
r = [GM (T /2π)2 ]1/3
Secondly, when MC moves down by 1 cm, that is, δx =
+1 cm, both masses MA , MB move by +1 cm, δxA = +1
cm and δxB = +1 cm (in the case of MA = MB , for
example). We can write this as
δxC = (δxA + δxB )/2 = 1 cm,
and the same has to be true of the derivative, x˙C =
(x˙A + x˙B )/2 and the accelerations,
x¨C = (x¨A + x¨B )/2
and
T =
2πρR
= 8.6 × 104 s.
v
After some manipulations, including isolating the term
GM/R2 = g ≈ 10 m/s2 , I get
g/R = (
2π 2 3
) ρ ,
T
so
+TAB = MA aA ,
+TAB = MB aB ,
and
[−TC + MC g] = MC aC ,
with the understanding from our “convenient” coordinate
system that aC is positive for acceleration downwards.
Overview: there are five unknowns (aA , aB , aC , TAB
and TC ), and five equations (three “F=ma” and two constraints). This problem ought to be solvable. First, I
eliminated TAB and TC , and after some algebra, got
ρ = [(g/R)(T /2π)2 ]1/3 ≈ 6.63 Earth radii.
aA =
This scaling by R was not so clean, but at least it simplified the numerical computation.
Prob 2.14 Using the figure in the text, label the accelerations of MA as aA , of MB as aB , and of MC as
aC . There are two strings and therefore two tensions,
label the tension between masses MA and MB with TAB
and between the upper pulley and MC as TC . There are
three masses and therefore three “F=ma” equations. In
addition, there are two constraints arising from the two
strings connecting the masses.
I would like to define an unconventional, but very convenient, coordinate system: the +î axis points to the
right for MA and MB , “turns the corner” (following the
string with tension TC ), and points down at MC . This
problem is “simple enough” so we can get away without defining a separate ĵ axis, then specifying how ymovement relates to x-movement. That would be another constraint equation.
First, the tension TAB around the upper pulley is the
same on both sides of the pulley, and the only force on
MA is +TAB the only force on MB is +TAB . At the
pulley, both lines pull with tension TAB to the left, and
this total tension of 2TAB is balanced by tension TC , so
TC = 2TAB .
aC = (aA + aB )/2.
This is another constraint that comes from the fact that
these masses are connect by a “contact force”, the string.
(I guessed at this by going to a special case.)
The “F=ma” equations are
The problem suggests that the “simplest way” is to find
the answer in units of Earth’s radius, that is, use a scaled
radius (call it ρ, which will have no units) of ρ = r/R.
The “F=ma” equation and the period T are
v2
GM
=
ρ2 R 2
ρR
or
aB = (
2MC g
,
MC + MC (MA /MB ) + 4MA
2MC g
MA
)[
],
MB MC + MC (MA /MB ) + 4MA
and
aC = 2(
MA
2MC g
)[
] − g.
MC MC + MC (MA /MB ) + 4MA
This is a complicated answer, so let’s check it in some
simple cases. First we should check units (for algebraic
mistakes).
Check: when MA = 0, MC should drop in free fall, and
MB should not move (like it is infinite mass compared to
MA ). Plugging in, we get aC = −g, aB = 0, and aA = 2g
(which is right).
Check: when MA = MB = MC , this is like a mass of 3M
and force of F = M g, so we expect a common acceleration of all masses to be a = M g/3M or a = g/3. OK.
This was not a simple problem.
Prob 2.18 With reference to the figure in the text, the
painter exerts a force F in both lines, and both lines
have tension F = T everywhere (because all the pulleys
are frictionless and massless). Like a stretched string,
3
tension T pulls in from both ends. So, 2T pulls up on
the painter, and 2T pulls up on the plank.
There is a contact force between the painter and the
plank, which we label as +f on the painter and −f on
the plank. The plank is being compressed, and springs
back, making +f on the painter.
The painter and plank move together with common
acceleration a.
We have two masses, M painter and m plank. Now we
just have to write down the two “F=ma” equations,
(painter)
[−M g + f + 2T ] = M a
(plank)
[+2T − f − mg] = ma.
and
Solving for a, I get
a=
4T
− g.
(M + m)
Note that because of the pulleys, the tension is pulling
both the painter and the plank up with total force 4T ,
and the painter and the plank (M + m) are a common mass, so the acceleration due to the force F is
a = 4T /(M + m). In additon, there is gravity, −g.
The statement of the problem is a little unrealistic,
since the painter provides a tension equal to his own
weight, F = M g, on two lines, so the painter is lifting twice his own weight. Of course, this accelerates the
whole thing.
Check: If M = m and F = M g, then a = 4M g/(2M ) −
g = 2g − g = g.
Check: If the painter pulled with F = M g/2 and M = m,
then we have a = 0, and nothing moves, he just supports
himself and the plank against gravity.