First slide

Moment of intertia

Question

Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of the system about an axis passing through A
and parallel to BD is

Easy

A

$2 {ml}^{2}$
B

$\sqrt{3} {ml}^{2}$
C

$3 {ml}^{2}$
D

${ml}^{2}$

Solution

$\begin{array}{l} I_{{XX}^{'}} = m (DP)^{2} + m (BQ)^{2} + m (CA)^{2} \\ = m {(\frac{\sqrt{2}}{2} l)}^{2} + m {(\frac{\sqrt{2}}{2} l)}^{2} + m (\sqrt{2} l)^{2} \\ = \frac{{ml}^{2}}{2} + \frac{{ml}^{2}}{2} + 2 {ml}^{2} = 3 {ml}^{2} \end{array}$

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