Q.A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60 degrees and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (Use pi = 3.14)

[!TLDR] $I=\frac12MR^2=0.001$ kg m$^2$, $\omega=2\pi/T=2\pi$ rad/s, $c=I\omega^2=0.001(2\pi)^2\approx0.03948$ N m/rad. $\tau_{max}=c\theta_0=0.03948\times …

The disc has diameter 20 cm, so radius $R=0.1$ m, and mass $M=200$ g $=0.2$ kg. Suspended by the string through its centre, perpendicular to the disc, the relevant moment of inertia is that of a disc about its own central (perpendicular) axis: $I=\frac12MR^2=\frac12(0.2)(0.1)^2=0.001$ kg m$^2$. With period $T=1$ s, angular frequency of the torsional oscillation is $\omega=\frac{2\pi}{T}=2\pi$ rad/s. From $T=2\pi\sqrt{I/c}$ (section 5.13), $c=I\omega^2=0.001\times(2\pi)^2=0.001\times39.478\approx0.039478$ N m per radian. The disc is twisted through $\theta_0=60°=\frac{\pi}{3} …

Compute $I=\frac12MR^2$ for the disc, find $c=I\omega^2$ from the given period, and evaluate $\tau_{ …

Using the given 20 cm as the radius instead of the diameter (it is explicitly stated as 'a 20 cm WIDE disc'), …