Moment of inertia of a ring about diameter

Moment of inertia of a ring about diameter uniform ring moment of inertia of the Ring about any diameter is not 2.1

talk about a 3 hour time phone definition of hearing about Z Axis passing through centre of mass of the ring and
perpendicular to its plane then ise equal to I see your type i y be the moment of inertia of the ring up of diameter along x and y


axis respectively using perpendicular axis theorem as it is equal to X + i

y but I Z equal to I see and for sharing my school to i y equation become IC equal to 2 X equal to 2 Y to i y moment of inertia of sharing about a tangent .

In its plane consider a tangent drawn to point in the plane of taking leave tangent is parallel to time towering Ashwin let IOB the moment of inertia of a ring about an Axis passing through in the Plane of the Ring literacy be the event of inertia of the Ring about diameter and distance between two equal to R a sequel to the appliance parallel axis .


theorem I was equal to 3 + him a square but I say equal to ID equal to half famous square and its equal to our equation becomes ...

I want to do after ma square + Y square equal to 3.2 m r squared let moment of inertia of writing about tangent perpendicular to its plane consider the tangent drawn to the point on IT .

The perpendicular to the plane of the ring is tangent is parallel to support the action of the Ring passing through the centre and perpendicular to the plane as shown in figure at home equal to moment of inertia


Turing about it and initial Lexus person to point and perpendicular to its plane icy cold phone definition of knowing about an Axis passing through the centre of the ring and perpendicular to its plane latest distance between two parallel Axes is equal to radius of ruling axis theorem + square what ise equal to m square and equal to or equation become a good Mr square + Y square equal to M .


moment of inertia of a disc about an Axis passing through its centre and perpendicular to its consider a thin uniform disc of mass M and radius r rotating about an Axis passing through its centre and perpendicular to its plane as shown of inertia of a disc is given by x is equal to half of them are square is in expression for moment of inertia of a disc about an Axis passing through its centre .

Perpendicular to it and I can find moment of inertia of disc about an Axis passing through diameter and joint x ray scan tangent in the plane of despise in parallel and perpendicular axis .

Theorem we have already an expression for moment of inertia of disc 5 Singh given by Shekar master about an Axis passing through its centre and perpendicular to it in solid cylinder can be considered as a combination of number of thin circular disc 51 A rather therefore moment of inertia of solid cylinder about its diameter that will be same ,
as that of a discount of inertia of a solid cylinder of mass M and radius about coronavirus reason why I could have a square .