First slide

Moment of intertia

Question

A portion of a ring of radius R has been removed as shown in figure. Mass of the remaining portion is m. Centre of the ring is at origin O. Let I_A and I_O be the moments of inertia passing through points A and O are perpendicular to the plane of the ring. Then,

Moderate

A

$I_{0} = m R^{2}$
B

$I_{O} < I_{A}$
C

$I_{O} > I_{A}$
D

$B o t h (1) a n d (4) a r e c o r r e c t$

Solution

Whole mass has equal distance from the centre O. Hence, I_O = m $R^{2}$ . Further, centre of mass of the remaining portion will be to the left of point O. More the distance of axis from centre of mass, more is the moment of inertia. Hence, I_A > I_O.

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