A diatomic molecule is formed by two atoms which may be treated as mass points $m_{1}$ , and $m_{2}$ joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is :

Zero

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(m_{1} + m_{2}) r^{2}

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\frac{(m_{1} + m_{2})}{m_{1} m_{2}} r^{2}

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(\frac{m_{1} m_{2}}{m_{1} + m_{2}}) r^{2}

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Solution

The correct option is D $(\frac{m_{1} m_{2}}{m_{1} + m_{2}}) r^{2}$
Let point C be the centre of mass situated at distance x from atom of mass mi and at distance (r - x) from mass $m_{2}$ . By definition of centre of mass, we have
$m_{1} x = m_{2} (r - x)$
$x = \frac{m_{2}}{(m_{1} + m_{2})} r$
Moment of inertia about the centre of mass is
$I = m_{1} x^{2} + m_{2} (r - x)^{2}$
or $I = m_{1} {(\frac{m_{2} r}{m_{1} + m_{2}})}^{2} + m_{2} {(r - \frac{m_{2} r}{m_{1} + m_{2}})}^{2}$
$m_{1} {(\frac{m_{2} r}{m_{1} + m_{2}})}^{2} + m_{2} {(\frac{m_{1} r}{m_{1} + m_{2}})}^{2}$
$\frac{m_{1} m_{2}}{m_{1} + m_{2}} r^{2}$

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A diatomic molecule is formed by two atoms which may be treated as mass points m1, and m2 joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is :

A diatomic molecule is formed by two atoms which may be treated as mass points $m_{1}$ , and $m_{2}$ joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is :