SHMMedium
Question
A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is $\_\_\_\_$ . (g is acceleration due to gravity)
Options
A.$2\pi\sqrt{\frac{5R}{4g}}$
B.$2\pi\sqrt{\frac{2R}{3g}}$
C.$2\pi\sqrt{\frac{3R}{2g}}$
D.$2\pi\sqrt{\frac{3R}{g}}$
Solution
Sol. [Key Concept: About axis A, $I = \frac{{mR}^{2}}{4} + {mR}^{2} = \frac{5{mR}^{2}}{4}$; then $T = 2\pi\sqrt{I/mgR} = 2\pi\sqrt{5R/4g}$.]
$$I_{\text{disc}\text{~}} = \frac{{mR}^{2}}{4} + {mR}^{2} = \frac{5{mR}^{2}}{4} $$∴ For small displacement
$${\tau = - mgR\theta }{\therefore\alpha = - \frac{mgR}{\frac{5{mR}^{2}}{4\text{ }g}}\theta }{\therefore\omega = \sqrt{\frac{4\text{ }g}{5R}} }{\therefore T = 2\pi\sqrt{\frac{5R}{4\text{ }g}} }$$
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