Question

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?

Answer 1

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 +mx²

\[= \frac{m r^2}{2} + m x^2 \] 

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

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:

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

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

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