Download MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC-W11D3-2 Group Problem 2
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
d 2z
the length of the tape that is unwound from the upper drum. Then, ! B R = 2 . The
dt
dz
linear velocity v A may then be expressed as the sum of two terms, the rate
at which
dt
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
aA =
d 2z
+ ! AR = !B R + ! AR .
dt 2
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
falling drum about its center is also ! A = TR , also directed clockwise. Thus,
! B = TR / I = 2T / MR , ! A = TR / I = 2T / MR . Where we have assumed that the moment
of inertia of the drum and unwinding tape is I = (1/ 2) MR 2 . Newton’s Second Law,
applied to the falling drum, with the positive direction downward, is Mg ! T = Ma A . We
now have five equations,
d 2z
d 2z
2T
2T
! B R = 2 , a A = 2 + ! A R, ! B =
, !A =
, Mg " T = Ma A ,
dt
dt
MR
MR
in the five unknowns ! A , ! B , a A ,
d 2z
and T .
dt 2
It’s easy to see that
! A = !B .
Therefore
a A = ! B R + ! A R = 2! A R .
The tension in the tape is then
T=
! A MR a A MR Ma A
=
=
2
4R
4
Newton’s Second Law then becomes
Mg !
Ma A
= Ma A .
4
Therefore solving for the acceleration yields
aA =
4
g
5
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 .