Moment of Inertia of Plate: Integration Proves 1/3ma^2+b^2

[astr0]

Show by integration that the moment of inertia of the plate about an axis that is perpendicular to the plate and passes through one corner is $$\frac{1}{3}m(a^{2}+b^{2})$$

I'm not sure at all how to approach this problem. I know that the moment of inertia is $$\int r^{2}dm$$ but how do I use that in this instance?

Any help is appreciated.

 

[ehild]

Write up both r² and dm in terms of x,y and integrate with respect to x and y.

ehild

 

[astr0]

Should I use two separate integrals, then add them together?

 

[ehild]

It is a double integral of ρr²=ρ(x²+y²).
First integrate both terms with respect to x from 0 to a, taking y as constant. Then integrate the result with respect to y from 0 to b.

ehild

 

[rwisz]

The moment of inertia of a rigid body is defined as the sum of the products of the mass of each particle in the body and the square of its distance from the axis of rotation. In the case of a plate, we can consider it as a collection of infinitesimal particles with mass dm distributed over its surface.

To calculate the moment of inertia of the plate about an axis perpendicular to the plate and passing through one corner, we can use the parallel axis theorem. This theorem states that the moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass of the body plus the product of the mass of the body and the square of the distance between the two axes.

In this case, we can choose the axis passing through the center of mass of the plate as our reference axis. The distance between this axis and the axis passing through the corner can be expressed as the sum of the distances a and b, where a is the distance from the center of mass to the edge of the plate and b is the distance from the edge of the plate to the corner.

Using the parallel axis theorem, the moment of inertia of the plate about the axis passing through the corner can be written as:

I = Icm + md^2

Where Icm is the moment of inertia about the center of mass and d is the distance between the two axes, which is equal to a+b.

Substituting the expression for the moment of inertia about the center of mass (which can be calculated using the integration formula given in the question) and simplifying, we get:

I = \frac{1}{12}ma^2 + \frac{1}{12}mb^2 + ma^2 + mb^2

= \frac{1}{3}ma^2 + \frac{1}{3}mb^2

= \frac{1}{3}m(a^2 + b^2)

Therefore, we have shown by integration that the moment of inertia of the plate about an axis perpendicular to the plate and passing through one corner is \frac{1}{3}m(a^2 + b^2), which is the same result as the one given in the question.