1.7. TORSION PENDULUM
m m m m
d 1
d 1
d 2
d 2
(i) (ii) (iii)
Axis of cylindrical
mass
Figure 1.22: Determination of Rigidity Modulus - dynamic torsion method
suspension wire.
Now, from theparallel axes theoremof moment of inertia we have:
I 1 =I+2i+2md^21 ,I 2 =I+2i+2md^22 (1.22)
whereiis the moment of inertia of the object of mass m about its own axis (Figure3.29).
Then,
I 2 ≠I 1 =2m(d^22 ≠d^21 ) (1.23)
The periods of oscillationT 0 ,T 1 ,T 2 are related to the respective moments of inertia by
the following relations:
T 02 =4fi^2
I
c
T 12 =4fi^2
I 1
c
T 22 =4fi^2
I 2
c
Therefore we can write,
T 22 ≠T 12 =
4 fi^2
c
(I 2 ≠I 1 )
and using the relation(1.23) we obtain,
T 02
T 22 ≠T 12
=
I
I 2 ≠I 1
=
I
2 m(d^22 ≠d^21 )
Hence,
I=
2 m(d^22 ≠d^21 )T 02
T 22 ≠T 12
(1.24)
Substituting from equation (1.24) into the expression (1.19) for Rigidity Modulus, we
arrive at,