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:

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

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\sqrt{m_{1} m_{2}} l^{2}

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\frac{m_{1} m_{2}}{m_{1} + m_{2}} l^{2}

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\frac{m_{1} {+ m}_{2}}{m_{1} m_{2}} l^{2}

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Solution

The correct option is C $\frac{m_{1} m_{2}}{m_{1} + m_{2}} l^{2}$
Given: A light rod of length l has two masses $m_{1}$ and $m_{2}$ attached to its two ends.
To find the moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass
Solution:
Moment of inertia for system of particle -
$I = m_{1} r_{1}^{2} + . . . + m_{n} r_{n}^{2}$
Applied when masses are placed discretely.
Here the center of mass is
$X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} ⟹ X_{c m} = \frac{m_{1} \times 0 + m_{2} \times l}{m_{1} + m_{2}} ⟹ X_{c m} = \frac{m_{2} l}{m_{1} + m_{2}}$
And so, $r_{1} = X_{c m} = \frac{m_{2} l}{m_{1} + m_{2}}$
and $r_{2} = l - X_{c m} = l - \frac{m_{2} l}{m_{1} + m_{2}} ⟹ r_{2} = \frac{m_{1} l}{m_{1} + m_{2}}$
So the moment of inertia of the present system is
$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} ⟹ I = m_{1} {(\frac{m_{2} l}{m_{1} + m_{2}})}^{2} + m_{2} {(\frac{m_{1} l}{m_{1} + m_{2}})}^{2} ⟹ I = \frac{(m_{1} m_{2}) (m_{2} + m_{1}) l^{2}}{(m_{1} + m_{2})^{2}} ⟹ I = \frac{m_{1} m_{2} l^{2}}{m_{1} + m_{2}}$
is the moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass

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A light rod of length l has two masses m1 and m2 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 about an axis perpendicular to the rod and passing through the centre of mass is: