Physics Challenges for Teachers and Students - Scitation

Physics Challenges for
Teachers and Students
◗ A New Kid on the Block
Solutions to November 2004 Challenges
Challenge: Four blocks are arranged on a smooth horizontal
surface as shown. The masses of the blocks are given
(see the diagram). The coefficient of static friction between
the top and the bottom blocks is s . What is the maximum
value of the horizontal force F, applied to one of the bottom
blocks as shown, that makes all four blocks move with the
same acceleration?
Solution: The maximum value of the friction
force comes from the coefficient of friction
times the normal force between the blocks.
The normal force, in this case, equals the
weight of the top block, so fmax = µ smg. Where
does friction need to be the greatest? The role
of friction between the back two blocks is to
accelerate the bottom block (M). On the other
hand, the role of friction between the front two
blocks is to accelerate the top block and everything
behind it (m+m+M). (Since all objects
move as one, we can think of any combination
as one object.) Clearly, friction will reach its
maximum value between the front two blocks
first. Based on the maximum value of friction
and the mass that needs to be accelerated by
this force, we can find the maximum acceleration
of these blocks:
ΣFmax
f max µ smg
amax
= =
=
m ( m + m + M ) 2m
+ M
total
f max
Now, looking at the entire system as a whole, F
acts to accelerate all the masses (m+m+M+M),
so:
µ smg
2µ
F = ΣFmax = mtotalamax = ( m + m + M + M ) =
2m
+ M
2µ
smg( m + M )
2m
+ M
(Contributed by H. Scott Wiley, Science Academy
of South Texas, Weslaco, TX)
◗ Breaking Uneven
Challenge: A straight metal rod of length 3l is bent through
the right angle as shown. The bent rod is then placed on a
rough horizontal table. A light string is attached to the vertex
of the right angle. The string is then pulled horizontally
so that the rod slides at a constant velocity. Find the angle α
that side 2l would make with the string.
Solution: The rod is in equilibrium, so the sum
of the torques about the axis through the vertex
and perpendicular to the plane of the rod
is zero. Let the frictional force on the short leg
be f and the moment arm about the axis be a.
Then the frictional torque about the longer leg
is 2f and the moment arm of this force is a/2.
Then
S1 THE PHYSICS TEACHER ◆ Vol. 43, 2005
2l
90˚
l
- ?
F
mg( m + M )
2m
+ M
s

sinθ = θ
a 2
a
and cos =
l
l 2
where θ and α are supplementary angles.
Consequently,
tanθ = so
1
4
α = 108o –tan-1 0.25 = 166o .
(Contributed by Eugene P. Mosca, U.S. Naval
Academy, Annapolis, MD)
◗ A Vicious Semicircle
Challenge: A block of mass M can slide without friction
along a horizontal table. A simple pendulum of mass m and
length l is mounted on the block so that it can swing freely
in the vertical plane. The pendulum is released from the
horizontal position as shown. Find the maximum tension of
the string during the subsequent motion of the pendulum.
Assume that the string is light and that the pendulum bob
is very small.
M
Solution: Let the counterclockwise angle that
the string makes with the vertical be θ, so that
the bob starts out at +90, swings around to
–90 where every part of the system again
comes instantaneously to rest, then oscillates
back and forth in this manner ad infinitum.
It turns out, as I shall prove below, that the
maximum tension occurs at 0 when the string
is vertical. I shall find an expression for the tension
at any angle and then maximize it.
I will consider the fourth quarter of a period
of oscillation, in which the bob is traveling up-
m
ward to the right with decreasing speed υ relative
to the mount, and the block in reaction is traveling
to the left with speed V and slowing down so
that its acceleration A is to the right. Consider
the free-body diagram for the bob and the base
with mounting pole.
V
M
l
T
V
ac v
at mg
The two forces on the bob are the tension T
and the weight mg. The acceleration of the bob
relative to the ground is the vector sum of the
acceleration A of the mount plus the acceleration
a of the bob relative to the mount. I have decomposed
a into two components: the centripetal
acceleration a c = υ 2 /l and the tangential acceleration
a t , whose directions at the present instant
are as drawn. Newton’s second law for the bob in
the radial direction is
2
υ
T − mg cosθ = m −mA
sin θ,
(1)
l
while for the base in the horizontal direction we
have
T sin θ = MA . (2)
Solve Eq. (2) for A and substitute that into
Eq. (1) to get
2
⎛ m 2 ⎞
υ
T ⎜1+
mg m . (3)
⎝⎜
sin θ
M ⎠⎟
= cosθ
+
l
Next we use the conservation laws to find the
speed υ, noting that the bob’s velocity relative
to the ground is the vector sum of υ and V.
Suppose the zero level of gravitational potential
energy is chosen to lie at the initial position of
the bob, so that the total energy E of the system
THE PHYSICS TEACHER ◆ Vol. 43, 2005 S2
T
A
A

is zero. This must also equal the sum of the
kinetic and potential energies of the system at
the current instant,
1
2 2 1 2
m ⎡(
υ cos θ− V ) + ( υ sin θ) ⎤ MV mgl cosθ
0
2 ⎣⎢
⎦⎥ + − =
2
(4)
Meanwhile, conservation of linear momentum
in the horizontal direction implies that
m(υ cos θ – V ) = MV . (5)
Solve Eq. (5) for V and substitute that into
Eq. (4) to prove that
2
υ
M + m
m = 2mg
cosθ
. (6)
l
2
M + m sin θ
Substituting Eq. (6) into (3) leads to the
desired expression for T as a function of θ,
m
3
3cos θ + ( 3cos
θ −cos
θ)
T
M
(7)
=
mg
2 .
m 2 ( 1+
sin θ
M )
To maximize this, we should make the numerator
as big as possible and the denominator as
small as possible. Both requirements are satisfied
when θ is 0. Therefore,
Tmax
m
= 3+ 2 . (8)
mg M
It is interesting to explore the limiting cases
of this expression. If M / m → ∞ then Tmax =
3mg, which is the tension at the bottom of the
swing (corresponding to a centripetal acceleration
of 2g) of a pendulum bob traveling in a
semicircular arc at the end of a mount fixed
to the Earth (acting as a really massive block).
On the other hand, consider what happens if
M / m → 0. Since there are no external horizontal
forces on the system, the center of mass
cannot move sideways. If M = 0 this means
the bob does not move horizontally relative to
the Earth, but simply free-falls straight down,
and the massless mounting block slides over
in exactly the right manner to always keep the
string just barely taut. The tension in the string
is zero (since there cannot be a net horizontal
force exerted on a massless block) until the
angle reaches zero, at which point the tension
suddenly (but smoothly) diverges to provide
the impulse needed to reverse the bob’s vertical
motion.
(Contributed by Carl E. Mungan, U.S. Naval
Academy, Annapolis, MD)
Column Editor's Note: If one does assume
a priori that the maximum tension corresponds
to the moment when the bob passes through
the lowest point, the solution path may be different,
for example, this solution:
The maximum tension occurs when the string
is vertical. In that position (which I call “final”
position), both the tension T of the string and
the weight mg of the mass m are vertical, so (according
to Newton’s second law)
T – mg = ma my ,
where amy is the vertical acceleration of the
mass m at that instant. This acceleration can be
written as amy = vrel
l
2 (centripetal acceleration),
where vrel is the velocity of the mass m relative
to the pivot. Then,
2
vrel
T = m( g + ).
l
Let us calculate vrel using energy conservation.
The energy of the system at the initial position
is Ei = mgl (I put the zero level of the gravitational
potential energy at the final position).
The total energy at the final position is
1 2 1 2
E mv Mv Then,
f = m + M .
2 2
1 2 1 2
mvm + MvM = mgl.
2 2
Note that, at the final position, both masses m
S3 THE PHYSICS TEACHER ◆ Vol. 43 2005

and M are moving, because the resultant force
on the system (m+M) is zero (linear momentum
conservation). For this reason,
m
m
mv Mv v v v
M M v
2
2
2
m + M = 0 ⇒ M = − m ⇒ M = 2 m.
Then,
1 1
1
2 2
2 1
2
2
2 2
mv M 2
2 2
1
m
m
v mgl
M M v
m + m = ⇒ ( + ) m =
gl
gl ⇒ vm
= .
m
+
M
Now as
m
v v v v
,
M v
m
M v
rel = m − M = m + m = ( + ) m 1
we have
v
m
M v
m
M
gl
2
2 2
v m
= ( 1+ ) m = ( 1+ )( 2 ) ⇒ = 2g ( 1+
).
l M
2 rel
rel
Putting this in the expression for T, we obtain
T m g g g m
M
mg
m
= ( + 2 + 2 ) = ( 3 + 2 ).
M
The maximum tension of the string is
m
mg(
3+ 2 ).
M
Note that if M >> m, T
3mg, the “standard” result (for fixed mass M).
If, instead, M = m, T = 5mg.
(Contributed by Fernando Ferreira, Universidade
da Beira Interior, Covilhã, Portugal)
Several other readers also sent us correct solutions
to the November Challenges. We would
like to recognize the following contributors:
Michael C. Faleski (Delta College, Midland, MI)
John F. Goehl Jr. (Barry University, Miami Shores,
FL)
Art Hovey (Milford, CT)
Mark Lenfestey (Homestead HS, Wayne, IN)
Adam Plana (The Wheatley School, Old Westbury,
NY)
Bayani I. Ramirez (San Jacinto College South,
Houston, TX)
Ben Stam and Tim Thomas, students (Oklahoma
School of Science and Mathematics Regional
Center, Enid, OK)
Leo H. van den Raadt (Heemstede, The Netherlands)
We appreciate your submissions and hope to
receive more solutions in the future.
Note to contributors:
For the duration of the WYP contest, the submission
guidelines have been slightly changed.
Please read carefully—especially if you are a
regular contributor.
– only email submissions will be considered;
– email your solutions to Boris Korsunsky at
korsunbo@post.harvard.edu;
– please email the solutions as Word files;
– please email each solution as a separate file;
– note that each problem, in addition to a
very clever title, has a code such as F1. Please
name each file as “problem code-first initiallast
name.” For instance, “F1DVader” if your
name is Darth Vader and you are sending the
solution to problem F1;
– please state your name, hometown, and professional
affiliation in each file.
We look forward to your (and your students’)
participation.
Please send correspondence to:
Boris Korsunsky
korsunbo@post.harvard.edu
THE PHYSICS TEACHER ◆ Vol. 43, 2005 S4