1
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall Term 2010
Problem Set 11: Two Dimensional Rotational and Translational Motion
Solution
Problem 1:
A drum
A
of mass
m
and radius
R
is suspended from a drum
B
also of mass
m
and
radius
R
, which is free to rotate about its axis. The suspension is in the form of a
massless metal tape wound around the outside of each drum, and free to unwind. Gravity
is directed downwards. Both drums are initially at rest. Find the initial acceleration of
drum
A
, assuming that it moves straight down.
Solution:
The key to solving this problem is to determine the relation between the three kinematic
quantities
A
!
,
B
!
and
A
a
, the angular accelerations of the two drums and the linear
acceleration of drum
A
.
One way to do this is to introduce the auxiliary variable
z
for
the length of the tape that is unwound from the upper drum.
Then,
2
2
B
d z
R
dt
!
=
.
The
linear velocity
A
v
may then be expressed as the sum of two terms, the rate
dz
dt
at which
the tape is unwinding from the upper drum and the rate
A
R
!
at which the falling drum is
moving relative to the lower end of the tape.
Taking derivatives, we obtain
2
2
A
A
B
A
d z
a
R
R
R
dt
!
!
!
=
+
=
+
.
Denote the tension in the tape as (what else)
T
.
The net torque on the upper drum about
its center is then
B
TR
!
=
, directed clockwise in the figure, and the net torque on the

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falling drum about its center is also
A
TR
!
=
, also directed clockwise.
Thus,
/
2
/
B
TR
I
T
MR
!
=
=
,
/
2
/
A
TR
I
T
MR
!
=
=
.
Where we have assumed that the moment
of inertia of the drum and unwinding tape is
2
(1/ 2)
I
MR
=
. Newton’s Second Law,
applied to the falling drum, with the positive direction downward, is
A
Mg
T
Ma
!
=
.
We
now have five equations,
2
2
2
2
2
2
,
,
,
,
,
B
A
A
B
A
A
d z
d z
T
T
R
a
R
Mg
T
Ma
dt
dt
MR
MR
!
!
!
!
=
=
+
=
=
"
=
in the five unknowns
A
!
,
B
!
,
A
a
,
2
2
d z
dt
and
T
.
It’s easy to see that
A
B
!
!
=
.
Therefore
2
A
B
A
A
a
R
R
R
!
!
!
=
+
=
.
The tension in the tape is then
2
4
4
A
A
A
MR
a MR
Ma
T
R
!
=
=
=
Newton’s Second Law then becomes
4
A
A
Ma
Mg
Ma
!
=
.
Therefore solving for the acceleration yields
a
A
=
4
5
g
This result is certainly plausible.
We expect
A
a
g
<
, and we also expect that with both
drums free to rotate, the acceleration will be almost but not quite
g
.