HW 1: Physics 1B Spring 2010
C. M. Cooper, Teaching Assitant
W. Gekelman, Professor
April 7, 2010
Abstract
This hw covers all of chapter 13 in the text except section 13.6. Topics include Simple
Harmonic Motion and DampedDriven Harmonic motion. Two good problems to memorize
are problems 71 and 82 Note that problems 55, 71, and 93 of these solutions involve the
physical pendulum (section 13.6) which was not covered in section but consists of a simple one
line formula
f
=
ω
2
π
=
1
2
π
r
mgd
I
(1)
A good way to remember this formula is to consider that I used to drink a lot of Miller
Genuine Draft playing kings as a freshman and although I’d like to think I’m past that point
in my life (I am over MGD), in actuality its the opposite (MGD is over I).
Also, remember whenever we see a moment of inertia, we have to remember 3 things: 1) Use
the table on page 299 for regular bodies, 2) the parallel axis theorum, and 3) how to integrate
and ﬁnd the moment of inertia of a solid body. The parallel axis theorem is used when you are
rotating something with a standard moment of inertia a distance away from you (on the end
of a stick or string) you need to add an additional
mL
2
where
L
is the distance to the center of
mass and in many cases needs to be calculated. We do not use integration to ﬁnd the moment
of inertia of anything.
.. YET!
1
Question 13.15: Weighing an Astronaut
As mentioned in the intro, basic one and two step problems like this only require you to have the
right equation as the ﬁrst line then substitute in. These kinds of problems are easy to ace.
We start with the equation
ω
2
=
k
M
(2)
In the second case,
M
=
M
C
+
M
A
,
ω
=
ω
C
+
A
= 2
π/T
C
+
A
, but
k
and
M
C
are both unknown.
In order to solve a problem, you must have at least as many unknowns as equations!
We have two unknowns so we need a second equation.
From the ﬁrst case, have
M
=
M
C
,
ω
=
ω
C
= 2
π/T
C
, and only
k
is unknown, we solve for
k
in
the ﬁrst case to get
k
=
M
C
ω
2
C
(3)
1
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View Full DocumentWe rearrange the second case and sub in the ﬁrst case to get
M
A
=
k
ω
2
C
+
A

M
C
=
M
C
ω
2
C
ω
2
C
+
A

M
C
=
M
C
(
T
2
C
+
A
T
2
C

1) = 120
.kg
(4)
2
Question 13.33: Two masses on a spring
2.1
(a) Why does the glue fail at the lowest part?
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 Spring '10
 gekelman
 Simple Harmonic Motion, Moment Of Inertia

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