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?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let's say from the centre ,the holes is at a distance of x (point A)

How$l_A=l_c+m{x}^2$

$=\frac{M{r}^2}{2}+m{x}^2$

${\gamma }_A=l_A\alpha$

$-mg\times sin\theta =(\frac{m{r}^2}{2}+m{x}^2)\frac{d^2\theta }{d{t}^2}$

For small $\theta$

$\frac{(-mg\times \theta )}{(\frac{m{r}^2}{2}+m{x}^2)}=\frac{d^2\theta }{d{t}^2}$

Comparing with standard equation

$T=2\pi \sqrt{\frac{r^2+2(x{)}^2}{2gx}}$

To find minimum value of T

$\frac{dT}{dx}=0$

$\Rightarrow x=\frac{r}{\sqrt{2}}$$T_{min}=2\pi \sqrt{\frac{r^2+2{\left(\frac{r}{\sqrt{2}}\right)}^2}{2g\frac{r}{2}}}2\pi \sqrt{\frac{\sqrt{2}r}{g}}$