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 $\frac{mr^{2}}{2}$ . 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. $\frac{55}{2}mr^{2}$
• B. $\frac{127}{2}mr^{2}$
• C. $\frac{111}{2}mr^{2}$
• D. $55 \,mr^{2}$

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

$I_{A}=\frac{mr^{2}}{2}+6\left(\frac{mr^{2}}{2}+m\left(2r\right)^{2}\right)$
$=mr^{2}\left(\frac{1}{2}+\frac{6}{2}+24\right)=\frac{55}{2}mr^{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} $
$=\frac{55}{2} mr^{2} +7m \left(2r\right)^{2}=\frac{111}{2}mr^{2}$