10.56. A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is
supported from the ceiling (see figure). Suddenly one of the wires breaks and the remaining wire
does not slip as the disk rolls down. Use energy conservation to find the speed of the center of
this disk after it has fallen a distance of 1.20 m.
Identify: The kinetic energy of the disk is K  2 Mvcm  2 I . As it falls its gravitational potential
energy decreases and its kinetic energy increases. The only work done on the disk is the
2
1
1
2
work done by gravity, so K1  U1  K2  U 2 .
I cm  12 M ( R22  R12 )
Set Up:
y2  1.20 m
.
K1  U1  K2  U 2
Execute:
, where R1  0.300 m and R2  0.500 m . vcm  R2 . Take y1  0 , so
K1  0
.
1
2
Mv  I cm  Mgy2
1
2
2
2
I cm  M 1  [ R1 / R2 ]  vcm
 0.340Mvcm
2
cm
2
vcm 
Evaluate:
1
2
1
4
U1  0
,
.
K2  U 2
.
2
.
2
.
Then
2
0.840Mvcm
 Mgy2
and
 gy2
(9.80 m/s )(1.20 m)

 3.74 m/s
0.840
0.840
2
A point mass in free-fall acquires a speed of 4.85 m/s after falling 1.20 m. The disk
has a value of vcm that is less than this, because some of the original gravitational potential
energy has been converted to rotational kinetic energy.