First slide

Moment of intertia

Question

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about perpendicular axis, passing through centre?

Difficult

A

$3 \frac{{MR}^{2}}{32}$
B

$15 \frac{{MR}^{2}}{32}$
C

$13 \frac{{MR}^{2}}{32}$
D

$11 \frac{{MR}^{2}}{32}$

Solution

$I_{Total disc} = \frac{{MR}^{2}}{2}$
$M_{Removed} = \frac{M}{4} (Mass \propto area)$
$I_{removed} (about the same perpendicular axis)$
= $\frac{M}{4} \frac{{(\frac{R}{2})}^{2}}{2} + \frac{M}{4} {(\frac{R}{2})}^{2} = \frac{3 {MR}^{2}}{32}$
$_{Removed disc = I_{total} - I_{removed}}$
= $\frac{{MR}^{2}}{2} - \frac{3}{32} {MR}^{2} = \frac{13}{32} {MR}^{2}$

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