MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
W15D2-1 Table Problem Unwinding Tape
Two objects of equal mass M are suspended from a massless and frictionless pulley, as shown.
A is a simple weight. B is a uniform cylinder of radius R around which the tape is wrapped. The
system is released from rest. Find the tension in the tape.
\
We first apply Newton’s Second Law to each object (see figure below for force diagram and
choice of positive directions)
Mg  T  Ma A
(1)
Mg  T  MaB .
(2)
.
Comparing equations (1) and (2), the accelerations are equal
a  a A  aB .
We now apply the torque equation about the center of mass of object B finding
(3)
TR  I cm .
(4)
The constraint condition can be found as follows. Let l0 represent the initial length of the tape.
At some later time let s  R(  0 ) represent the amount of tape that has unwound, where
 (t )  0 is the angle that the cylinder has rotating. Then the length of tape at time t is given by
l(t)  l0  R( (t)  0 ) .
(5)
If we choose coordinates as shown in the figure above, then
l(t)  y A  yB   r 2 ,
(6)
where r is the radius of the pulley. Then
y A  yB   r 2  l0  R( (t)  0 ) .
(7)
We can now take two derivatives to find the constrain condition between the accelerations of the
two objects and the angular acceleration of B,
a A  aB  R .
(8)
2a  R .
(9)
Therefore using Eq. (3),
We can solve the above equation for  and substitute that into the torque equation (4) and find
that using Icm  (1`/2)MR2
T
Substitute Eq.(10) into Eq. (1) yields
Therefore Eq. (10) becomes
2aI cm
R2
 Ma .
(10)
ag/2
(11)
T  Mg / 2 .
(12)