First slide

Moment of intertia

Question

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

Moderate

A

Zero
B

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

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

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

Solution

The moment of inertia about centre of mass
$I = m_{1} x^{2} + m_{2} (r - x)^{2}$
The moments of masses about centre of mass
$m_{1} X = m_{2} (r - X)$
or $x = (\frac{m_{2}}{m_{1} + m_{2}}) r$
$\begin{array}{l} ∴ I = m_{1} {(\frac{m_{2} r}{m_{1} + m_{2}})}^{2} + m_{2} {(\frac{m_{1} r}{m_{1} + m_{2}})}^{2} \\ = (\frac{m_{1} m_{2}}{m + m_{2}}) r^{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App