A light rod of length, l has two masses, m _ 1 and m _ 2 attached...

A light rod of length,

l

has two masses,

{m}_{1}

and

{m}_{2}

attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is A light rod of length, l has two masses, m _ 1 and m _ 2 attached to its two ends. The moment of inertia of the system

A light rod of length, l has two masses, m _ 1 and m _ 2 attached to its two ends. The moment of inertia of the system

Options

A $\dfrac{{m}_{1}{m}_{2}}{{m}_{1} + {m}_{2}}{l}^{2}$
B $\dfrac{{m}_{1} + {m}_{2}}{{m}_{1}{m}_{2}}{l}^{2}$
C $\left({m}_{1} + {m}_{2}\right){l}^{2}$
D $\sqrt{{m}_{1}{m}_{2}}{l}^{2}$

Correct answer

A. $\dfrac{{m}_{1}{m}_{2}}{{m}_{1} + {m}_{2}}{l}^{2}$

Step-by-step solution

A light rod of length, l has two masses, m _ 1 and m _ 2 attached to its two ends. The moment of inertia of the system

${x}_{1} = \dfrac{{m}_{2}l}{{m}_{1} + {m}_{2}}$ and ${x}_{2} = \dfrac{{m}_{1}l}{{m}_{1} + {m}_{2}}$ .

$I = {m}_{1}{x}_{1}^{2} + {m}_{2}{x}_{2}^{2} = \dfrac{{m}_{1}{m}_{2}}{{m}_{1} + {m}_{2}}{l}^{2}$ .

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