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A uniform disc of radius *r* is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period?

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#### Solution

Let *m* be the mass of the disc and* r *be its radius.

Consider a point at a distance *x** *from the centre of gravity.

Thus, *l* = *x*

Moment of intertia \[\left( I \right)\] about the point x will be,

I = I_{C.G} +mx^{2}

\[= \frac{m r^2}{2} + m x^2 \]

\[ = m\left( \frac{r^2}{2} + x^2 \right)\]

Time period(*T*) is given as,

\[T = 2\pi\sqrt{\frac{I}{mgl}}\]

\[\text { On substituting the respective values in the above equation, we get: }\]

\[T = 2\pi\sqrt{\frac{m\left( \frac{r^2}{2} + x^2 \right)}{mgx}} (l = x)\]

\[ = 2\pi\sqrt{\frac{m\left( r^2 + 2 x^2 \right)}{2mgx}}\]

\[ = 2\pi\sqrt{\frac{r {}^2 + 2 x^2}{2gx}} \ldots(1) \]

To determine the minimum value of *T,*

\[T = 2\pi\sqrt{\frac{I}{mgl}}\]

\[\text { On substituting the respective values in the above equation, we get: }\]

\[T = 2\pi\sqrt{\frac{m\left( \frac{r^2}{2} + x^2 \right)}{mgx}} (l = x)\]

\[ = 2\pi\sqrt{\frac{m\left( r^2 + 2 x^2 \right)}{2mgx}}\]

\[ = 2\pi\sqrt{\frac{r {}^2 + 2 x^2}{2gx}} \ldots(1) \]

To determine the minimum value of *T,*

\[\frac{d^2 T}{d x^2} = 0\]

\[\text { Now }, \]

\[\frac{d^2 T}{d x^2} = \frac{d}{dx}\left( \frac{4 \pi^2 r^2}{2gx} + \frac{4 \pi^2 2 x^2}{2gx} \right)\]

\[ \Rightarrow \frac{2 \pi^2 r^2}{g}\left( - \frac{1}{x^2} \right) + \frac{4 \pi^2}{g} = 0\]

\[ \Rightarrow - \frac{\pi^2 r^2}{g x^2} + \frac{2 \pi^2}{g} = 0\]

\[ \Rightarrow \frac{\pi^2 r^2}{g x^2} = \frac{2 \pi^2}{g}\]

\[ \Rightarrow 2 x^2 = r^2 \]

\[ \Rightarrow x = \frac{r}{\sqrt{2}}\]

Substituting this value of *x *in equation (1), we get:

\[T = 2\pi\sqrt{\frac{r^2 + 2\left( \frac{r^2}{2} \right)}{2gx}}\]

\[ = 2\pi\sqrt{\frac{2 r^2}{2gx}} = 2\pi\sqrt{\frac{r^2}{g\frac{r}{\sqrt{2}}}} \]

\[ = 2\pi\sqrt{\frac{\sqrt{2} r^2}{gr}} = = 2\pi\sqrt{\frac{\sqrt{2}r}{g}}\]

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