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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 alpha is defined by the relation J=-alpha theta, whereJ is the restoring couple and theta the angle of twist). |
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Answer» Solution :Here , m=10 kg , R=15 cm , T=1.5 s MOMENT of inertia of disc, `I= 1/2 mR^2= 1/2 xx 10 xx (0.15)^2 kgm^2` Now, `T= 2pi SQRT(I/alpha) ` so, `alpha = (4pi^2 I)/(T^2) = 4 xx ((22)/(7))^2 xx 1/2 xx (10 xx (0.15)^2)/((1.5)^2) == 1.97 N` m/rad |
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