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 α is defined by the relation J = –αθ , where J is the restoring couple and q the angle of twist).
Given: The mass of the circular disc is 10 kg , the period of oscillation of the disc is 1.5 s and the radius of the disc is 15 cm .
The time period of oscillation of the disc is given as,
T = 2 π I α T 2 = 4 π 2 I α α = 4 π 2 I T 2 (1)
Where, I is the moment of inertia of the disc and α is the torsional spring constant.
The moment of inertia of the disc is given as,
I = 1 2 M R 2
Where, M is the mass of the disc and R is the radius of the disc.
By substituting the expressions in equation (1), we get
α = 4 π 2 T 2 ( 1 2 M R 2 ) = 2 π 2 M R 2 T 2
By substituting the given values in the above equation, we get
α = 2 π 2 ( 10 ) ( 0.15 ) 2 ( 1.5 ) 2 ≈ 2 N-m / rad
Thus, the torsional spring constant of the wire is 2 N-m / rad .
Given: The mass of the circular disc is
The time period of oscillation of the disc is given as,
Where,
The moment of inertia of the disc is given as,
Where,
By substituting the expressions in equation (1), we get
By substituting the given values in the above equation, we get
Thus, the torsional spring constant of the wire is
