This is known as parallel axes theorem.
Statement:
If IG be the moment of inertia of a body of mass M about an axis passing through its centre of gravity, then MI (I) of the same body about a parallel axis at a distance 'a' from the previous axis will be given as I = IG + M a2
the moment of inertia of a solid cylinder about an axis passing through its COM and parallel to its length is mr2/2 where r is the radius.
A rotating body that spins about an external or internal axis (either fixed or unfixed) increase the moment of inertia.
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(1/2) mr2, assuming the axis of rotation goes through the center, and along the axis of symmetry.
It is the square root of ratio moment of inertia of the given axis to its mass.
the moment of inertia of a body about a given axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass and the product of its mass and square of perpendicular distance between two axis Iz=Ix+Iy
the moment of inertia of a solid cylinder about an axis passing through its COM and parallel to its length is mr2/2 where r is the radius.
mass moment of inertia is the property of the body to resist rotation about the given axis where as the area moment of inertia is the resistance to bending about the given axis
The Radius of Gyration of an Area about a given axis is a distance k from the axis. At this distance k an equivalent area is thought of as a line Area parallel to the original axis. The moment of inertia of this Line Area about the original axis is unchanged.
The moment of inertia of a cube depends on what its axis of rotation is. About an axis perpendicular to one of its sides and through the centre of the cube is (ML2)/6. Where M is the Mass of the Cube and L the length of its side. Due to the symmetry of the cube, you can find the Moment of Inertia about almost any other axis by using Parallel and Perpendicular Axis Theorems.
If an ellipse has a radius A long the x-axis and B along the y-axis (A > B) then the moment of inertia about the x-axis is 0.25*pi*ab^3
A rotating body that spins about an external or internal axis (either fixed or unfixed) increase the moment of inertia.
The moment of inertia formula isIxx= bh3 / 12B= base H= height and Ixx = moment of inertia of a rectagular section about x-x axis.
Think of it as the difference in moment of inertias for two solid cubes. Calculate the moment of inertia of a solid cube with dimensions equal to the inner dimensions of your hollow cube. Then calculate the moment of inertia of a solid cube with dimensions equal to the outer dimensions of your hollow cube. Subtract the moment of inertia of the inner dimensions from the moment of inertia of the outer dimensions to get the moment of inertia of what's left. Same concept applies to finding the area of a thin-walled circle. Outer area - inner area = total area. Outer moment of inertia - inner moment of inertia = total moment of inertia. This approach won't work however if you're considering hollow shell - a cube with walls of zero thickness. If the axis of rotation goes through the cube center, perpendicular to one of its walls, first calculate moment of inertia of the wall that the axis passes through (let's call it Ia). For all equations below d equals surface density(mass per unit of area) and a is length of cube's side. Ia= d * a4 / 6 Then you have to calculate moments of inertia of four walls parallel to the axis. This will be Ib=4 * Iwall=4*d*a4/3. Total moment of the shell will be then: I=2*Ia+Ib=1.5*d*a4. If the axis is through the center and ┴ one face, I = (m/6)*[a² - (a-t)²], or I = (m/6)(2at - t²) for any value of t, however small. Source: CRC Std Math Tables
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metre4
Just moment of inertia is incomplete requirement as the axis about which it is to be measured is also very important