Unit 1 Engineering Physics

24 CHAPTER 1. PROPERTIES OF MATTER

The moment of inertia of a circular disc of massMand radiusR, about an axis passing
through its centre and perpendicular to the circular face, is

I=

1

2

MR^2 (1.19)

Substituting from equation (5.294) into (5.293) we obtain:

n=

4 filMR^2

T 02 r^4

(1.20)

Hence by experimentally measuring the length (l) and radius (r) of suspension wire ,
mass (M) and radius (R) of the suspended disc and the period of oscillation (T 0 ) of the
pendulum, the Rigidity Modulus of the wire can be obtained from (5.295).

Rigidity Modulus - Determination by Dynamic Torsion Method

It is possible to obtain an expression for the Rigidity Modulus of the suspension wire
which does not explicitly require the expression (5.294) for the moment of inertia of the
disc. For this we adopt an experimental procedure known as dynamic torsion method
which is explained in the following.

The periods of oscillation of the given torsion pendulum are determined in three dier-
ent configurations as shown in Figure5.66by adding extra weights (usually of cylindrical
shape) to the suspended disc. In configuration (i), the period of oscillationT 0 is measured
for the bare disc with no extra masses. In configuration (ii), the period of oscillationT 1
is measured with two identical bodies (each having massm) placed symmetrically on
the disc at distancesd 1 from the suspension wire. In configuration (iii), the period of
oscillationT 2 is measured with the identical bodies placed symmetrically on the disc at
distancesd 2 from the suspension wire (withd 2 >d 1 ).
As mentioned before,I= the moment of inertia of the suspended disc about an axis
passing through its centre and perpendicular to the circular face (i.e., about the axis co-
inciding with the suspension wire).
LetI 1 = the moment of inertia of the configuration-ii
andI 2 = the moment of inertia of the configuration-iiiabout the axis coinciding with the
suspension wire.
Now, from theparallel axes theoremof moment of inertia we have:

I 1 =I+2i+2md^21 ,I 2 =I+2i+2md^22 (1.21)

whereiis the moment of inertia of the object of mass m about its own axis (Figure5.66).
Then,
I 2 ≠I 1 =2m(d^22 ≠d^21 ) (1.22)

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 )

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