How to find moment of inertia of circular plate ? show by integration. Don't fake
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Answer:
We defined the moment of inertia I of an object to be [latex] I=\sum _{i}{m}_{i}{r}_{i}^{2} [/latex] for all the point masses that make up the object. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod ((Figure)) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.
We defined the moment of inertia I of an object to be [latex] I=\sum _{i}{m}_{i}{r}_{i}^{2} [/latex] for all the point masses that make up the object. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod ((Figure)) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.In the case with the axis in the center of the barbell, each of the two masses m is a distance R away from the axis, giving a moment of inertia.
[latex] {I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}. [/latex]
[latex] {I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}. [/latex]In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is
[latex] {I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}. [/latex]In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is[latex] {I}_{2}=m{(0)}^{2}+m{(2R)}^{2}=4m{R}^{2}. [/latex]
[latex] {I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}. [/latex]In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is[latex] {I}_{2}=m{(0)}^{2}+m{(2R)}^{2}=4m{R}^{2}. [/latex]From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.
