A circular disc of mass 10 ; kg is suspended | vedclass

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Question

(A) Mass of the circular disc,$m = 10 \; kg$. Radius of the disc,$r = 15 \; cm = 0.15 \; m$. The time period of torsional oscillations is $T = 1.5 \;

Vedclass · Accepted Answer

$1.97$

Vedclass · Answer

(A) Mass of the circular disc,$m = 10 \; kg$. Radius of the disc,$r = 15 \; cm = 0.15 \; m$. The time period of torsional oscillations is $T = 1.5 \; s$. The moment of inertia of the disc about its central axis is $I = \frac{1}{2} m r^2$. $I = \frac{1}{2} 	imes 10 	imes (0.15)^2 = 5 	imes 0.0225 = 0.1125 \; kg \; m^2$. The time period for torsional oscillations is given by $T = 2 \pi \sqrt{\frac{I}{\alpha}}$. Squaring both sides,$T^2 = 4 \pi^2 \frac{I}{\alpha}$,which gives $\alpha = \frac{4 \pi^2 I}{T^2}$. Substituting the values: $\alpha = \frac{4 	imes (3.14159)^2 	imes 0.1125}{(1.5)^2}$. $\alpha = \frac{4 	imes 9.8696 	imes 0.1125}{2.25} = \frac{4.4413}{2.25} \approx 1.974 \; N \; m \; rad^{-1}$. Thus,the torsional spring constant is approximately $1.97 \; N \; m \; rad^{-1}$.

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