Home
Class 11
PHYSICS
For the same total mass which of the fol...

For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body.

A

a disc of radius a

B

a ring of radius oa

C

a square lamina of side `2 a`

D

four rods forming a square of side `2a`.

Text Solution

Verified by Experts

The correct Answer is:
D
For Disc `I_("disc") = (1)/(2) ma^(2)`
For Ring `I_("ring") = ma^(2)`
For square lamine of side `2a = (1)/(6) m(2a)^(2) = (2)/(3) ma^(2)`

For Rods forming a square of side `2 a`
`I_("square") = 4[(m(2a)^(2))/(12) +ma^(2)] =4[(ma^(2))/(3)+ma^(2)] = (16 ma^(2))/(3)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body

A ring and a circular disc of different materials have equal masses and equal radii. Which one will have a larger moment of inertia about an axis passing through its centre of mass perpendicular to its plane ?

The moment of inertia of a copper disc, rotating about an axis passing through its centre and perpendicular to its plane

A uniform thin bar of mass 6 m and length 12 L is bend to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is :

A square of a side a is cut from a square of side 2a as shown in the figure. Mass of this square with a hole is M. Then its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane will be

Calculate the moment of inertia of a disc of radius R and mass M, about an axis passing through its centre and perpendicular to the plane.

Derive an expression for moment of inertia of a thin circular ring about an axis passing through its centre and perpendicular to the plane of the ring.