First slide

Moment of intertia

Question

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing
through the centre of the original disc. Its moment of inertia about the axis of rotation is

Moderate

A

MR²/2
B

MR²/4
C

MR²/8
D

$\sqrt{2} {MR}^{2}$

Solution

The moment of inertia of quarter disc about the centre is given by
$I_{C} = \frac{1}{4} [\frac{{MR}^{2}}{2}], where M = mass of disc$
If M' be the mass of quarter disc, then
$I_{c} = \frac{1}{4} [\frac{(4 M) R^{2}}{2}] = \frac{{MR}^{2}}{2}$

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