Given: R = 10cm = 0.1 m, M = 0.2 kg, θm=600 = p/3 rad, T=1 s, π3 = 31
The MI of the disc about the rotation axis (perpendicular through its centre) is
I = ½ MR2 = ½ (0.2)(0.1)2 = 10-3 kg-m2
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4π2\(\frac{I}{T^2}\)
The magnitude of the maximum restoring torque,
τmax = cθm = 4π2\(\frac{I}{T^2}(\frac{π}{3})\)
= \(\frac{4}{3}π^3\frac{I}{T^2}=\frac{4}{3}(31)(\frac{10^{-3}}{1^2})\)
= 41.33 x 10-3 = 0.04133 N-m