Physics 116, Fall 2007
Problem Set 2
due Sep 12, 2007
Reading:
Sec 1.8, 1.9, notes 1.1, 2.12.3
1. Circular motion
K&K 1.14.
SOLUTION: From rolling without slipping, we have
v
cm
=
ωR
. Taking derivative of time on both sides,
dv
cm
dt
=
d
(
ωR
)
dt
→
a
=
αR
α
=
a
R
To connect even more concretely with what we’ve done in
class, consider our starting point for the point on the rolling
tire in problem 5 of PS01. There we began with
v
p
=
v
translation
+
v
rotation
= 0
.
(1.1)
Again taking the time derivative of both sides and rearranging,
we obtain
a
translation
+
a
rotation
= 0
.
(1.2)
Now note that
a
rotation
is the tangential acceleration, and
that at the point of contact,
a
translation
=
a
ˆ
θ
. So we have
a
ˆ
θ
=

(2 ˙
r
˙
θ
+
R
¨
θ
)
ˆ
θ.
(1.3)
But, ˙
r
= 0 and
¨
θ
=

α
, so we once again have
a
=
Rα.
(1.4)
2. Rolling drum
K&K 1.15.
SOLUTION:
a) See graph.
For position vectors, the point in coordinate
system A has a position vector
r
A
, which becomes
r
B
in
system B. They are related via relation
r
A
=
r
B
+
R
. Again
taking time derivatives on both sides, we have
v
B
=
v
A

d
R
dt
.
b) See graph again.
We use the notation
v
ab
to represent
velocity of
a
relative to
b
.
v
ab
=
v
a

v
b
=
d
R
ab
dt
, and
v
a
is
0 in this case (it’s not moving in its own frame moving with
itself). So,
v
ab
=
d
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 Fall '05
 ELSER, V
 Derivative, Circular Motion

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