Oscillations

© examsnet.com
Question : 23
Total: 25
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J=αθ, where J is the restoring couple and q the angle of twist)
Solution:  
Here, m=10kg; R=15cm ; T=1.5s Moment of inertia of the disc about wire,
I=
1
2
M
R2
=
1
2
×10
×(0.15)2
=0.1125kgm2
When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If α is angular acceleration produced, then restoring torque
τ=Iα...(i)
The torsional spring constant k of the wire is defined by the relation
τ=kα...(ii)
where θ is the angle of twist
From the equations (i) and (ii) we have
Iα=kθ or α=
k
I
θ

Since
k
I
is constant, it follows that angular acceleration is directly proportional to the angle of twist (angular displacement). Hence, the motion executed by the disc is simple harmonic in nature and the period of torsional oscillation is given by
T=2π
θ
α
=2π
θ
kθI
=2π
I
k

or k=
4π2I
T2
=
4π2×0.1125
1.52
=1.97Nmrad1
© examsnet.com
Go to Question: