First slide

Moment of intertia

Question

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

Moderate

A

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

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

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

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

Solution

$Here, l_{1} + l_{2} = l$
Centre of mass of the system,
$\begin{array}{l} l_{1} = \frac{m_{1} \times 0 + m_{2} \times l}{m_{1} + m_{2}} = \frac{m_{2} l}{m_{1} + m_{2}} \\ l_{2} = l - l_{1} = \frac{m_{1} l}{m_{1} + m_{2}} \end{array}$
Required moment of inertia of the system
$I = m_{1} l_{1}^{2} + m_{2} l_{2}^{2} = (m_{1} m_{2}^{2} + m_{2} m_{1}^{2}) \frac{l^{2}}{{(m_{1} + m_{2})}^{2}}$
$= \frac{m_{1} m_{2} (m_{1} + m_{2}) l^{2}}{{(m_{1} + m_{2})}^{2}} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} l^{2}$

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