First slide

Moment of intertia

Question

Two particles A and B of masses m and 2m are at a separation d. PQ is a line perpendicular to the line AB. Then minimum possible moment of inertia about PQ is

Moderate

A

$\frac{3}{4} {md}^{2}$
B

$\frac{1}{4} {md}^{2}$
C

$\frac{4}{3} {md}^{2}$
D

$\frac{2}{3} {md}^{2}$

Solution

$I_{PQ}$ will be minimum when line PQ passes through the centre of mass of the system
$∴ AP = \frac{2 m \times d}{3 m} = \frac{2 d}{3} and BP = d - \frac{2 d}{3} = \frac{d}{3} \begin{array}{l} ∴ (I_{PQ})_{\min} = m {(\frac{2 d}{3})}^{2} + 2 m {(\frac{d}{3})}^{2} a \\ = \frac{2}{3} {md}^{2} \end{array}$

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