Oscillations
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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 = – α θ , where J is the restoring couple and q the angle of twist)
Solution:
Here, m = 10 k g ; R = 15 c m ; T = 1.5 s Moment of inertia of the disc about wire,
I =
M R 2 =
× 10 × ( 0.15 ) 2 = 0.1125 kg m 2
When the wire is twisted by rotating the disc and then released, restoring torque will be set up. Ifα is angular acceleration produced, then restoring torque
τ = I α ...(i)
The torsional spring constant k of the wire is defined by the relation
τ = − k α ...(ii)
where θ is the angle of twist
From the equations (i) and (ii) we have
I α = − k θ or α = −
θ
Since
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 = 2 π √
= 2 π √
= 2 π √
ork =
=
= 1.97 N m rad − 1
When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If
The torsional spring constant k of the wire is defined by the relation
where θ is the angle of twist
From the equations (i) and (ii) we have
Since
or
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