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A solid thin metallic disc of radius R is hinged at point O on its rim and freely suspended as shown in the figure. The lower endge of the disc is slightly pulled to the right and released. Then time period of oscillation of the disc is

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$2 π \sqrt{\frac{2 R}{g}}$

$2 π \sqrt{\frac{R}{g}}$

$2 π \sqrt{\frac{3 R}{2 g}}$

$\frac{π}{2} \sqrt{\frac{2 R}{3 g}}$

answer is C.

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Detailed Solution

$T = 2 π \sqrt{\frac{I_{0}}{{mgr}_{C}}}$

Where $I_{0} =$ moment of inertia of the disc about a horizontal axis through $O = \frac{1}{2} {mR}^{2} + {mR}^{2} = \frac{3}{2} {mR}^{2}$

$r_{C} =$ Distance of centre of mass from the hinge point O = R

$∴ T = 2 π \sqrt{\frac{\frac{3}{2} {mR}^{2}}{mg . R}} = 2 π \sqrt{\frac{3 R}{2 g}}$

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