PH8151 Engineering Physics Chapter 1

26 CHAPTER 1. PROPERTIES OF MATTER

n=

8 fiIl

T 02 r^4

=

8 fil

T 02 r^4

A

2 m(d^22 ≠d^21 )T 02

T 22 ≠T 12

B

Hence, the rigidity modulus is given by:

n=

16 fiml(d^22 ≠d^21 )

r^4 (T 22 ≠T 12 )

(1.25)

Therefore, from the experimental data obtained by measuring the massmof one of the
bodies placed on the disc, the distancesd 1 andd 2 at which they are placed from the disc
centre, the periodsT 1 andT 2 of oscillation and the radiusrof the suspension wire, the
rigidity modulus of the wire can be calculated.

Worked out Example 1.7.1

Data obtained in a torsional pendulum experiment are given below.

Mean radius of the suspension wire (r) = 0.3 mm = 0. 3 ◊ 10 ≠^3 m

Length of the wire(l) = 70.4 cm = 70. 4 ◊ 10 ≠^2 m

Mass of one of the identical cylinders (m)=100g= 100 ◊ 10 ≠^3 kg

Distancesd 1 andd 2 between suspension wire & massmare:

d 1 = 2.5 cm = 2. 5 ◊ 10 ≠^2 m,d 2 = 4.5 cm = 4. 5 ◊ 10 ≠^2 m

Periods of oscillations: T 0 =10. 6 s,T 1 =11. 2 s,T 2 =11. 7 s

Calculate the moment of inertia of the disc and the rigidity modulus of the suspension

wire.

Solution:

Moment of inertia of the disc

I =

2 m(d^22 ≠d^21 )T 02

T 22 ≠T 12

=

(2)(100◊ 10 ≠^3 kg) [(4. 5 ◊ 10 ≠^2 m)^2 ≠(2. 5 ◊ 10 ≠^2 m)^2 ](10.6s)^2

(11.7s)^2 ≠(11.2s)^2

=2. 7 ◊ 10 ≠^3 kg.m^2

Rigidity Modulus of the suspension wire

n =

8 fiIl

T 02 r^4

=

(8)(3.14)(2. 747 ◊ 10 ≠^3 kg.m^2 )(70. 4 ◊ 10 ≠^2 m)

(10.6s)^2 (0. 3 ◊ 10 ≠^3 m)^4

=4. 8 ◊ 1010 Nm≠^2

26