Physics 105
Solutions to Problem Set 1
By Stephen Glassman, Fall 2007
1.10
Whenever you’re asked to “show” something, there’s always a question of where
your starting point should be. Consequently, I’ll accept many different approaches
to this problem. In this solution set, we will start by (nearly) assuming equations
for circular motion in polar coordinates, and derive the cartesian form. Specifically,
we’ll assume:
˙
θ
=
±
ω
r
=
R
where the sign choice corresponds to the different directions an object can circle. In
fact, we’ll further assume the plus sign, and we’ll check that we made the correct
choice later. Integrating
˙
θ
gives us
θ
=
ωt
+
φ
. Standard trigonometry then implies:
r
= ˆ
xR
cos(
ωt
+
φ
) + ˆ
yR
sin(
ωt
+
φ
)
If we now insist that the ˆ
y
component be 0 at
t
= 0, we conclude sin(
φ
) = 0 and so
φ
∈ {
0
, π
}
. There is no sound basis for excluding
φ
=
π
, but we need to in order to
arrive at the stated result. Substituting
φ
= 0 yields:
r
= ˆ
xR
cos(
ωt
) + ˆ
yR
sin(
ωt
)
To calculate the velocity and acceleration, we take derivatives.
The calculation is
simplified by the fact, as mentioned in the text, that our cartesian basis vectors won’t
depend on time.
v
=

ˆ
xRω
sin(
ωt
) + ˆ
yRω
cos(
ωt
)
a
=

ˆ
xRω
2
cos(
ωt
)

ˆ
yRω
2
sin(
ωt
)
We take note of a few things.
r
(0) = ˆ
xR
and
v
(0) = ˆ
yRω
suggesting a counter
clockwise circle.
So, indeed, we did guess the sign correctly earlier.
We also note
that
r
·
v
= 0 and
a
=

ω
2
r
, both familiar results of circular motion.
1.12
Let’s begin by considering
b
=
c
and
v
0
= 0. The particle’s path then reduces
to the form familiar from problem 1.10.
Let’s now consider the effect of nonzero
1
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v
0
. Taking a time derivative, we see that
v
z
=
v
0
so that the particle travels with
constant speed in the ˆ
z
direction while circling in the
x

y
plane. This trajectory
describes a helix. Increasing
b
or
c
simply stretches the circle into an ellipse with a
major and minor axis. So in general, the trajectory describes a modified helix based
off an ellipse rather than a circle.
There remain several degenerate possibilities including
b
= 0 or
c
= 0 (or both), in
which case the particle can be seen to travel along a wave (or along a line). Further
possibilities include a harmonic oscillator (if only
b
= 0) or a particle at rest (if
b
=
c
=
v
0
= 0).
1.13
Let the angle between
u
and
b
be
θ
. Then, using:
v
·
w
=
vw
cos
θ
=
⇒
u
·
b
=
b
cos
θ

v
×
w

=
vw
sin
θ
=
⇒

u
×
b

=
b
sin
θ
Therefore (
u
·
b
)
2
+ (
u
×
b
)
2
= (
b
cos
θ
)
2
+ (
b
sin
θ
)
2
=
b
2
as required. In words,
u
·
b
is the length of the component of
b
along
u
, whereas

u
×
b

gives the length of
the component of
b
perpendicular to
u
. By the Pythagorean Theorem, the square
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 Fall '07
 EdgarKnobloch
 Physics, Derivative, Sin, Trigraph, DT DT DT, ijk vj vk, ijk klm

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