Question

An annular ring of internal and outer radii r and R respectively oscillates in a vertical plane about a horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate its time period and show that the length of an equivalent simple pendulum is $\frac{3R}{2}$ as $r\to 0$ and 2R as $r\to R$.

Answer 1

Mass per unit area$=\frac{m}{\pi (R^2-r^2)}=\sigma$ (say)
Whole mass $m_1=\left(\pi R^2\right)\sigma$
$=\left(\frac{R^2}{R^2-r^2}\right)m$
Mass of cavity $m_2=\left(\pi R^2\right)\sigma$
$=\left(\frac{r^2}{R^2-r^2}\right)m$
$I=\frac{3}{2}{m}_1{R}^2-[\frac{1}{2}{m}_2{r}^2+m_2{R}^2]$
$=\frac{3}{2}\left[\frac{R^2}{R^2-r^2}\right]m{R}^2-[\frac{1}{2}\left(\frac{r^2}{R^2-r^2}\right)m{r}^2+\left(\frac{r^2}{R^2-r^2}\right)m{R}^2]$
$=\frac{m}{2(R^2-r^2)}[3R^4-r^4-2r^2{R}^2]$
$=\frac{m(3R^2+r^2)}{2}$
Now, $T=2\pi \sqrt{\frac{I}{mgl}}$ (i)
Here, l=R
$\therefore$ $\frac{I}{mR}=\frac{3R}{2}+\frac{r^2}{2R}$
Substituting in Eq. (i) we have,
$T=2\pi \sqrt{\frac{\frac{3R}{2}+\frac{r^2}{2R}}{g}}$
Comparing with
$T=2\pi \sqrt{\frac{l}{g}}$
l of pendulum$=\frac{3R}{2}+\frac{r^2}{2R}$
$l=\frac{3R}{2}$ as $r\to 0$
and $l=2R$ as $r\to R$