Question

Seven identical coins are rigidly arranged on a flat table in the pattern shown below, so that each coin touches its neighbours. Each coin is a thin disc of mass m and radius $r.$

Here note, that the moment of inertia of an individual coin about an axis passing through centre and perpendicular to the plane of the coin is $\frac{m{r}^2}{2}$ . The moment of inertia of the system of seven coins about an axis that passes through the point P (the centre of the coin positioned directly to the right of the central coin) and perpendicular to the plane of the coins is

• A. $\frac{55}{2}m{r}^2$
• B. $\frac{127}{2}m{r}^2$
• C. $\frac{111}{2}m{r}^2$
• D. $55m{r}^2$

Answer 1

Answer: (C)

Moment of inertia of given configuration about an axis through point A (centre of middle disc), using parallel axes theorem is

$I_A=\frac{m{r}^2}{2}+6\left(\frac{m{r}^2}{2}+m{\left(2r\right)}^2\right)$
$=m{r}^2\left(\frac{1}{2}+\frac{6}{2}+24\right)=\frac{55}{2}m{r}^2$
Now, again using parallel axes theorem, moment of inertia of configuration about $P$ is
$I_P=I_A+M_{\text{total}}\times d^2$
$=\frac{55}{2}m{r}^2+7m{\left(2r\right)}^2=\frac{111}{2}m{r}^2$