PHY126 HW Set 4 (Chap.13)
Problem 13.45
Description:
A pendulum of length
Part A
is mounted in a rocket.
What is its period if the rocket is at rest on its launch pad?
Part B
What is its period if the rocket is accelerating upward with acceleration
?
Part C
What is its period if the rocket is accelerating downward with acceleration
?
Part D
What is its period if the rocket is in free fall?
Solutions
Let a be the acceleration of the pendulum relative to the rocket, and let a0 be the acceleration of the rocket
relative to the ground (assumed to be an inertial system.) The Newton’s 2nd law is
r
r
r
∑ F = m( a + a
0
r
r
r
) → ∑ F − ma 0 = ma. Because of the acceleration the gravity constant g will be
modified to geff, , depending on the magnitude and the direction of the acceleration. For small oscillations
r
about the equilibrium position, which depends on a 0 , the period is T = 2π
L
r r where L is the
g − a0
length of the pendulum.
r
T = 2π L / g as before.
r
r
(b) a0 = (1 / 2) gˆj and g = − gˆj . This gives T = 2π L /( g + (1 / 2) g ) = 2π 2 L /(3g ) . (c) If
r
a 0 = −(1 / 2) gˆj , this gives T = 2π L /( g − (1 / 2) g ) = 2π 2 L / g .
r
(d) If a 0 = gˆj , then T = ∞.
(a) If a 0 = 0, then
Problem 13.42
Description: A mass
slides along a frictionless horizontal surface at speed
. It strikes a spring of
constant attached to a rigid wall, as shown in the figure
.
After a completely elastic encounter with the spring, the mass heads back in the direction it came from.
Part A
In terms of ,
, and
, determine how long the mass is in contact with the spring.
, and
, determine the maximum compression of the spring.
Part B
In terms of ,
Solutions
(a) While the mass in contact with the spring, the net horizontal force on it is just the spring force, so it
undergoes half a cycle of simple harmonic motion before leaving the spring with speed v0 to the left.
This takes time equal to half a period (1 / 2)T = π / ω = π m / k .
(b) v0 is the maximum speed, which is related to the maximum compression of the spring (the amplitude)
by v 0 = ωA. Thus
A = v0 / ω = v0 m / k .
Problem 13.55
Description: A thin, uniform hoop of mass
set oscillating with
small amplitude, as shown in the figure
and radius
is suspended from a thin horizontal rod and
.
Part A
Show that the period of the oscillations is
useful.
. Hint: You may find the parallel-axis theorem
Solution
Using Eq.13.13, the period of a physical pendulum is
T = 2π I /(mgL) , where I is the moment of
inertia of the hoop. We can calculate I by using the parallel-axis theorem.
I = I 0 + mh 2 = mR 2 + mR 2 = 2mR 2 . Thus, using Eq.13.13, we find the period to be
T = 2π
2mR 2
2R
I
= 2π
= 2π
.
mgL
mgR
g
Problem 13.47
Description: A mass is attached to a vertical spring, which then goes into oscillation. At the high point
of the oscillation, the spring is in the original unstretched equilibrium position it had before the mass
was attached; the low point is 5.8
below this.
Part A
What is the period of oscillation?
Express your answer using two significant figures.
Solution
We note that at the highest point, there is no spring force, since the spring is unstretched. Therefore, the
acceleration is just g (downward). This is also the maximum acceleration during the simple harmonic
motion, since a max occurs where the displacement is a maximum. Thus using Eq.13.10, we have
a max = g = ω 2 A. Once ω is known, the period of oscillation can be obtained by using Eq.13.7b. Now
2 A = 5.8 cm. Thus, using Eq.13.7b, the period is
A
1 2π
0.029 cm
= 2π
= 2π
= 0.34 s.
T= =
ω
f
g
9.8 m/s 2
Problem 13.60
Description: A solid cylinder of mass
and radius
is mounted on an axle through its center. The
axle is attached to a horizontal spring of constant , and the cylinder rolls back and forth without
slipping (see the figure
).
Part A
Write the statement of energy conservation for this system, and differentiate it to obtain an equation
.
analogous to the equation
Part B
, determine the angular frequency of the
Comparing your result with the equation
motion.
Express your answer in terms of
, , and
.
Solutions
v = ωR for rolling without slipping,
K = (1 / 2) Mv + (1 / 2) I cm ω = (1 / 2) Mv + (1 / 2)(1 / 2) MR 2 (v / R ) 2 = (3 / 4) Mv 2 . The
(a) With reference to Eq.10.20 and the condition
2
2
2
potential energy of the spring is U = (1 / 2) kx , where v = dx / dt , so
2
E = K + U = (3 / 4) M (dx / dt ) 2 + (1 / 2)kx 2 . Differentiating, we find:
2
2
dE
3
2k
⎛ dx ⎞⎛ d x ⎞ 1
⎛ dx ⎞ d x
= 0 = M 2⎜ ⎟⎜⎜ 2 ⎟⎟ + k 2 x⎜ ⎟ or 2 = −
x = −ω 2 x. So the angular frequency
dt
4
dt
2
dt
3
M
dt
dt
⎝ ⎠⎝
⎝ ⎠
⎠
is
ω=
2k
.
3M