Gravitational Torsion Balance
012–06802B
14
the two equilibrium positions and record the result as
Δ
S
.
3.
Determine the period of the oscillations of the small mass
system by analyzing the two graphs. Each graph will
produce a slightly different result. Average these results
and record the answer as
T
.
4.
Use your results and equation 1.9 to determine the value
of
G
.
5.
The value calculated in step 4 is subject to the same
systematic error as described in Method I. Perform the
correction procedure described in that section (
Analysis,
step 3
) to find the value of
G
0
.
M
E
T
H
O
D
III
:
M
ea
su
r
e
m
e
n
t
by
A
cce
l
e
r
a
t
i
on
Observation Time: ~ 5 minutes
Accuracy: ~ 15%
Theory
With the large masses in Position I, the gravitational attraction,
F
, between each small mass (
m
2
) and its neighboring large
mass (
m
1
) is given by the law of universal gravitation:
F = Gm
1
m
2
/b
2
(3.1)
This force is balanced by a torque from the twisted torsion
ribbon, so that the system is in equilibrium. The angle of twist,
θ
, is measured by noting the position of the light spot where
the reflected beam strikes the scale. This position is carefully
noted, and then the large masses are moved to Position II. The
position change of the large masses disturbs the equilibrium of
the system, which will now oscillate until friction slows it
down to a new equilibrium position.
Since the period of oscillation of the small masses is long
(approximately 10 minutes), they do not move significantly
when the large masses are first moved from Position I to
Position II. Because of the symmetry of the setup, the large
masses exert the same gravitational force on the small masses
as they did in Position I, but now in the opposite direction.
Since the equilibrating force from the torsion band has not
Содержание AP-8215
Страница 24: ......