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 0.15m. Determine the torsional spring constant of the wire. (Note : Torsional spring constant 'alpha' is defined by the relation J=-alpha theta where J is the restoring couple is torque).

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 0.15m. Determine the torsional spring constant of the wire. (Note : Torsional spring constant 'alpha' is defined by the relation J=-alpha theta where J is the restoring couple is torque).
Answer» <html><body><p></p>Solution :`T=2pisqrt((I)/(<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>))or alpha=(4pi^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)I)/(T^(2))` <br/> `"i.e."alpha=(4pi^(2)((MR^(2))/(2)))/(T^(2))=(2pi^(2)MR^(2))/(T^(2))""<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> alpha=(<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a>(3.142)^(2)xx10xx(0.15)^(2))/((1.5)^(2))=(4.4425)/(2.25)` <br/> `alpha="1.974 Nm rad"^(-1)`</body></html>