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Physics 205—Final Exam Fall 2003
Solutions
x
M
m
g
l
- (40 pts) A plane pendulum consists of
a bob of mass m suspended by a massless
rigid rod of length l that is hinged to a sled
of mass M. The sled slides without friction
on a horizontal rail. Gravity acts with the
usual downward acceleration g.
a) Taking x and θ as generalized coordinates write the Lagrangian for the system.
Solution: We start by computing the Cartesian coordinates of the bob
x 1 = x + l sin θ =⇒ x˙ 1 = ˙x +
θl cos θ
y 1
= y + l cos θ =⇒ y˙ 1
= ˙y +
θl sin θ
thus
T =
m
x˙
2
1
2
1
M
x˙
2
= (m + M )/over2 +
m
θ xl˙ cos θ +
θ
2
l
2
and
V = mgl (1 − cos θ)
hence
L = T − V =
(m + M )
m
θ xl˙ cos θ +
θ
2
l
2
where uninteresting constants have been dropped.
b) Use Lagrange’s equations to derive the equations of motion for the system.
Solution: For x we have
d
dt
∂L
∂ x˙
∂L
∂x
= (m + M )¨x + ml
θ cos θ − m
θ
2
l sin θ = 0
and for θ
d
dt
∂L
θ
∂L
∂θ
= mxl¨ cos θ + m
θl
2
c) Use the equations from part (b) to find the frequency ω for small oscillations of the
bob about the vertical. (Hint: You will need to make some approximations.)
Solution: For small oscillations, θ, x, and their derivatives will be small, so we can neglect
terms containing
θ
2 and ˙x
θ. Also cos θ ' 1 and sin θ ' θ. With this we obtain
(m + M )¨x + ml
θ ' 0 =⇒ x¨ ' −
m
m + M
l
θ
and
x¨ +
θl + gθ ' 0 =⇒
m
m + M
l
θ ' −gθ
yielding
ωosc '
g
m
m+M
l
g(m + M )
lM
d) At time t = 0 the bob and the sled, which had previously been at rest, are set in
motion by a sharp tap delivered to the bob. The tap imparts a horizontal impulse
∆P = F ∆t to the bob. Find expressions for the values of
θ and ˙x just after the
impulse.
Solution: Here we conserve linear and angular momentum.
∆P =
∆p i
= m( ˙x +
θl) + M x˙
and
∆L = ∆P l =
Li = ml
θ =⇒
θ =
∆P
ml
Note that since
∆P = ml
θ =⇒ x˙ = 0
a
m
g
θ
Ω
- (45 pts) A bead of mass m slides with-
out friction on a rotating circular hoop. The
hoop is forced to rotate about a vertical axis
along its diameter at a constant angular ve-
locity Ω. The position of the bead can be
described by the angle θ, which is the angle
that a line running between the center of the
hoop and the bead makes with the vertical.
a) What is the Lagrangian for the system?
Solution:
L = T − V =
ma
2
θ
2
2
sin
2
θ
− mga(1 − cos θ)
as shown in the sketch. Calculate the displacement of the oscillator for times t >
T /2.
Solution: Starting with the general Green function solution
x(t) =
t
−∞
dt
′
F (t
′ )
mω 1
e
−γ(t−t
′ )
sin [ω 1
(t − t
′
)]
where since there is no damping γ = 0 and ω 1
= ω 0
. We thus have
x(t) =
t
−∞
dt
′
F (t
′ )
mω 0
sin [ω 1
(t − t
′
)]
since the force is zero prior to −T /2 and after T /2 and we only want to know x(t) for
t > T /2, we can write
x(t) =
T / 2
−T / 2
dt
′
F 0 sin(ω 0 t
′
)
mω 0
sin [ω 0
(t − t
′
)]
F
0
mω 0
[
sin ω 0
t
T / 2
−T / 2
sin ω 0
t
′
cos ω 0
t
′
dt
′
− cos ω 0
t
T / 2
−T / 2
sin
2
ω 0
t
′
dt
′
]
where we have used the standard trig. ID for the sine of a difference. The first term in
the integral vanishes and the second can be evaluated by noting that the average of sin
2
ωt
over one period is 1/2. This yields
x(t) = −
F 0
mω 0
cos ω 0 t
T
F 0 π
mω
2
0
cos ω 0 t
e
3
- (40 pts) A quarterback in American foot-
ball throws the ball in such a way that it ap-
pears to “wobble” rather than spinning smoothly
as it flies. For this problem assume that the foot-
ball can be treated as an axisymmetric top having
I
1
= I
2
= 2I
3
, where the ˆe 3
axis is the long axis of
the football. Assume further that the football is
released with a rapid spin, i.e., ω 3 6 = 0, where ω 3
is the component of the football’s angular velocity
along the ˆe 3
body axis.
a) What does the wobbling motion indicate? In particular what does it imply about
the other (ˆe 1 and ˆe 2 ) components of ω? (No calculation is required here.)
Solution: This worked out to be a little bit more open ended than I had intended. The
answer I was looking for was that it means that the angular velocity was not aligned with
a body axis, meaning that it would also not be aligned with the angular momentum, which
is a vector fixed in space (conservation of angular momentum with no external torques). I
accepted most reasonable variations on this theme.
b) Expressing vectors with respect to the body fixed axes, ˆe 1 , ˆe 2 , eˆ 3 , derive three equa-
tions that relate the components of ~ω to one another and to the moments of inertia
I
1
, I
2
, I
3
Solution: Euler’s equation is
d
L
dt
inertial
d
L
dt
body
L =
Γexternal = 0
If we choose body axes that are principal axes then
dL i
dt
body
d
dt
I
ii
ω i
body
= I
i
ω˙ i
combining this with Euler’s equation and writing it by components yields
I
1
ω˙ 1
= ω 2
ω 3
(I
2
− I
3
I
2
ω˙ 2
= ω 3
ω 1
(I
3
− I
1
I
3
ω˙ 3
= ω 1
ω 2
(I
1
− I
2
c) If the football is spinning rapidly, then one can assume that ω 3
¿ ω 1 , 2
. Use this
assumption to derive an approximate expression for ~ω as a function of time. Find
the period of the wobble in terms of I 1 = I 2 , I 3 , and ω 3.
Solution: Since the football is axisymmetric I 1 = I 2 = I and ω˙ 3 = 0. Defining Ω ≡
(I
3
− I)/Iω 3
yields
ω˙ 1
= −Ωω 2
and ω˙ 2
= +Ωω 1
which, when differentiated, can be solved to yield
ω¨ 1
2
ω 1
and ω¨ 2
2
ω 2
Thus both ω 1 and ω 2 vary harmonically with frequency
ω 3
and the period of the wobble is
T =
2 π
4 π
ω 3
