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It can be done, but it would require taking an integral for the moment of inertia of each particle of the disc... something i don't have the time to do right now ^^;

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The moment of inertia of a hollow cylinder can be calculated with the help of the formula, I=0.5xMxRxR. Here I is inertia, M is mass and R is the radius.

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 for a solid sphere: I = (2 /5) * mass * radius2 (mass in kg, radius in metres)

I believe it is I = mk^2 where k is radius of gyration and m is mass.

(1/2)mr^2 where m=mass r=radius

The answer will depend on whether the axis isthrough the centre of the disk and perpendicular to its plane,a diameter of the disk, orsome other axis.Unless that information is provided, the answer is meaningless.

It is the square root of ratio moment of inertia of the given axis to its mass.

.00288 kgm^2

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

Its diameter is 2R, whatever the mass.

radius of gyration = sqrt(Moment of inertia/cross section area) Regards, Sumit

Radius of gyration is the distance from the centre of gravity to the axis of rotation to which the weight of the rigid body will concentrate without altering the moment of inertia of that particular body.

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.

Depends on the axis considered. For the axis through the center perpendicular to the plane of the disk: I = m*r^4 /2, where r is the radius of the disk. For an axis through the center on the plane of the disk: I = m*r^4 /4

One way would be to increase the mass of the object. Another would be to increase the radius of the arc.

If we are considering the semi circular wire of mass m to have a relatively small cross sectional radius, r compared to radius of the semi circular arc, R And that we are considering the semi circular wire to lie in a xy plane. The mass moment of inertia taken about the z-axis (perpendicular to the plane) located at the cg position will be. Izz=m*(R^2)*(PI-(4/PI))/PI I did a check of this calculation with the way Catia calculated the same property. And the difference was found to be 0.03% When using the following geometry. density=0.000001kgmm^3 Semi circular wire Radius R =60mm cross sectional radius of wire = 1mm results from Catia V=592.176mm^3 m=0.000592176kg Izz=1.268159kgmm^2 hand calculated results V=592.176mm^3 m=0.000592176kg Izz=1.267835kgmm^2 % difference [((1.268159)-(1.267835))/(1.267835)]*100=0.03% Please carry out a check on the Izz calc. I determined it from 1st principles, so I may of made a mistake

Radius

find the strength of the member subject to bending or shear. Moment of inertia is used to find radius of gyratia or flexural regidity so that member strength flexural stress is found

(pi/2)*(ro^4-ri^4)

Mass and radius

Ans : By the formula of moment of inertia , I=mr2 (2=square) As by the formula it is clear that the moment of inertia depends on the mass and the radius of a particular body , so as the mass increases moment of inertia will considerably increase , So as the water drips into the beaker the mass of the beaker will increase By the law of conservation of angular momentum Moment of inertia is inversely proportional to the angular velocity ( omega ) , Since in the above case of coasting rotating system moment of inertia is increased so the angular velocity (omega) will also decrease and hence the coasting rotating system will now rotate slowly as compared to its rotation before dripping water !

The radius of a cylinder is half the thickness of its circular cross section.

A football field is rectangular, therefore there is no radius. If it were a circular shape then it would have a radius.

I=mr2 therefore r = root (I/m) = root (5/45) = 0.333 recuring

mass and radius