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 1.5s. The radius of the disc is 15 cm. Determine thetorsional spring constant of the wire. (Torsional spring constant alpha is defined by the relation J= -alpha theta, where J is the restoring couple and theta the angle of twist).

A circular disc of mass 10 kg is suspended by a wire attached to its c

1.	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 1.5s. The radius of the disc is 15 cm. Determine thetorsional spring constant of the wire. (Torsional spring constant alpha is defined by the relation J= -alpha theta, where J is the restoring couple and theta the angle of twist).
Answer» Solution :This QUESTION is of torsional pendulum. Its time period, `T= 2PI sqrt((I)/(alpha))` where I is the moment of inertia For disc `I = (1)/(2) mr^(2)` where m = mass of disc and r = radius of disc and `alpha =` torsional spring constant. `T^(2)= (4pi^(2)I)/(alpha)` `therefore alpha = (4pi^(2)I)/(T^2) = (4pi^(2))/(T^2)XX (1)/(2) mr^(2)` `= (2pi^(2)mr^(2))/(T^2)` `=(2XX (3.14)^(2)xx10xx(0.15)^(2))/((1.5)^(2)) = 1.97192` `therefore APPROX 1.97 (Nm)/(rad)" or "alpha approx 2.0 (Nm)/(rad)`