The moment of inertia of a rectangular plate about its diagonal

[Outrageous]

Homework Statement

Find the moment of inertia of a rectangular plate of mass m, sides 2h and 2k , rotates about its diagonal.

Homework Equations

I= Ʃmr^2

The Attempt at a Solution

I=Ʃ m(x^2 +y^2) , let 2h parallel to x-axis, 2k parallel to y-axis. I= m(x cosθ)^2 + m (y cos ∅)^2 where angle θ is the angle between diagonal and x-axis... But this is not parallel theorem case ...

No idea how to start , please give some hint.
Thanks

 

[voko]

Look at the picture. The rectangle is ABCD, the axis of rotation is AC. Clearly the moment of ABCD is the sum of the moments of ABC and ACD, and these two moments are obviously equal.

The moment of ACD is likewise the sum of the moments of ADF and CDF.

Now, let AC be the x-axis, and FD be the y-axis. Then y is the distance from the axis of rotation. Can you compute the moments of ADF and CDF in these coordinates?

 

[Filip Larsen]

Another approach, that also works for a general plate (or body when using three dimension), would be to start with the moment of inertia, I, around the center of the plate expressed as a 2x2 matrix, which for a rectangular plate is a nice simple diagonal matrix. From this matrix you can find the moment of inertia around any axis n as I_n = n^T I n where n^T is the transpose of n.

 

[Outrageous]

Look at the picture. The rectangle is ABCD, the axis of rotation is AC. Clearly the moment of ABCD is the sum of the moments of ABC and ACD, and these two moments are obviously equal.

The moment of ACD is likewise the sum of the moments of ADF and CDF.

Now, let AC be the x-axis, and FD be the y-axis. Then y is the distance from the axis of rotation. Can you compute the moments of ADF and CDF in these coordinates?

I tried to find the moment of inertia of a triangle, but very complicated... Can you tell how do you calculate ?

 

[Outrageous]

The moment of inertia of the right angle triangle only. Not ADF .

 

[voko]

Look at the picture. Find the moment of inertia of the triangle about the x-axis. The equation for the outer edge of the triangle is $y = \frac b a x$. Its moment of inertia is then given by $$

\rho \int_{x = 0}^{x = a} \int_{y = 0}^{y = \frac b a x} y^2 dx dy

= \rho \int_{x = 0}^{x = a} \left[ \frac {y^3} {3} \right]_{y = 0}^{y = \frac b a x} dx

= \rho \frac {b^3} {3 a^3} \int_{x = 0}^{x = a} x^3 dx

= \rho \frac {b^3} {3 a^3} \left[ \frac {x^4} {4} \right]_{x = 0}^{x = a}

= \rho \frac {ab^3} {12}

$$

You can use the above to compute the moment of inertia of any right-angled triangle about one of its two legs. All you need is the parameters $a$ and $b$. I hope you can continue from here.

 

[Outrageous]

Look at the picture. Find the moment of inertia of the triangle about the x-axis. The equation for the outer edge of the triangle is $y = \frac b a x$. Its moment of inertia is then given by $$

\rho \int_{x = 0}^{x = a} \int_{y = 0}^{y = \frac b a x} y^2 dx dy

= \rho \int_{x = 0}^{x = a} \left[ \frac {y^3} {3} \right]_{y = 0}^{y = \frac b a x} dx

= \rho \frac {b^3} {3 a^3} \int_{x = 0}^{x = a} x^3 dx

= \rho \frac {b^3} {3 a^3} \left[ \frac {x^4} {4} \right]_{x = 0}^{x = a}

= \rho \frac {ab^3} {12}

$$

You can use the above to compute the moment of inertia of any right-angled triangle about one of its two legs. All you need is the parameters $a$ and $b$. I hope you can continue from here.

I got the answer.Thanks for making it simple.

 

[Outrageous]

Another approach, that also works for a general plate (or body when using three dimension), would be to start with the moment of inertia, I, around the center of the plate expressed as a 2x2 matrix,

The axis of rotation perpendicular to the plate?
What do you mean by 2x2?

 

[Filip Larsen]

The axis of rotation perpendicular to the plate?
What do you mean by 2x2?

I am referring to the concept of expressing the "total" moment of inertia of a body around a reference point as a tensor, that is, expressing it as a matrix in a particular choice of coordinate system. This matrix can be defined as integral over the whole body, like show by Weisstein [1] equation 4. The diagonal elements of this matrix is just the moment of inertia of the body around the x, y, and z axis. The off-diagonal elements can be considered a sort of area moment of inertia [2] and they may be new to you.

Before you start to worry that you now have to make nine integrals to find all the elements of the moment of inertia matrix for your rectangular plate just so you can calculated the moment of inertia around your selected axis, let's see if there is something that will make it simpler. First, you are only interested in an axis that lies in the xy-plane, so we can forget about the z-dimension reducing the moment of inertia matrix to two by two instead three by three. Secondly, notice that the area moment in general is zero if the body has its mass symmetrically distributed around either x- og y-axis. Since the mass in your rectangle is symmetrically placed around both the x- and y-axis the off-diagonal elements are zero, leaving you with a two by two matrix where only two of the elements are non-zero.

So for your rectangle the moment of inertia around origo in the xy-plane can be written as
<br /> \mathbf I = \left[\begin{matrix} I_x &amp; 0 \\ 0 &amp; I_y \end{matrix}\right]<br />
where I_x = \frac 1 3 m h^2 and I_y = \frac 1 3 m k^2 are the moments of inertia around the x- and y-axis respectively, which I assume you already know. If not, they can easily be found by direct integration and are also given in most standard reference on moment of inertia.

To find the moment of inertia of any other axis \vec n in the xy-plane that goes through origo you can now calculate
<br /> I_n = \vec n^T \mathbf I \vec n<br />

Now you only need to express the $\vec n$ vector as a function of your angle θ and you are good to go. At some point you will probably want to replace $\cos θ$ and $\sin θ$ in your result with expressions involving $h$ and $b$ in order to get a nice result.

If this whole thing didn't make sense to you, then don't worry about it. Just stick with direct integration of the moment of inertia, as described by voko. The approach I describe here is useful because it is general and, for bodies with symmetries, can replace integrals over complicated geometric shapes (in your case triangles) with simple shapes (in your case rectangles). If you ever get to study motion of complex bodies like aircrafts and rotating machinery you will probably find that the moment of inertia tensor is extremely handy.


[1] http://mathworld.wolfram.com/MomentofInertia.html
[2] http://mathworld.wolfram.com/AreaMomentofInertia.html