PH8151 Engineering Physics Chapter 1

1.7. TORSION PENDULUM

m m m m

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,

PH8151 25 LICET