How Do You Calculate the Moment of Inertia for a Rectangular Sheet of Steel?

[John O' Meara]

A thin rectangular sheet of steel is .3m by .4m and has mass 24kg. Find the moment of inertia about an axis (a) through the center, parallel to the long sides; (b) through the center parallel to the short sides. (c) through the center, perpendicular to the plane.
(a) We divide the sheet into N very narrow strips parallel to the axis and of width deltaXi and length L=.4m. The mass of each strip is deltaMi and which is a distance Xi from the axis. Now
deltaMi = k*L*deltaXi,
where k is the mass per unit area. Substituting this value into the expression for I (rotational inertia)
I= Sum k*L*Xi^2*deltaXi = k*L*Sum Xi^2*deltaXi
If we now pass to the limit deltaX -> 0 and N -> infinity. So the above sum can be replaced by an integral =>
I=2*k*L*[X^3/3]0,w/2 the limits of integration => I= 2*w^2, where w=.3m : I=.18 kg.m^2
(b) is similar to (a); I= .32 kg.m^2
(c) for this section I need a drawing please, on how to integrate for I. Do I divide the sheet up into N thin strips con-centric with the axis of rotation
like I would for a solid disk or cylinder? Many Thanks.

 

[Galileo]

a) and b) are correct, but your post is difficult to read. Check here:
https://www.physicsforums.com/showthread.php?t=8997

For c), the easiest way is to use the perpendicular axis theorem. Which makes it trivial. If you don't know that, a simple double integral will do. The distance of a mass element to the axis is x^2+y^2, so :

$$I_z=\sigma\int \limits_{-a/2}^{a/2}\int \limits_{-b/2}^{b/2}(x^2+y^2)dxdy$$

If you haven't had double integrals, you could also try adding the moments of inertia from each strip, like in parts a) and b), and use the parallel axis theorem to find these moments of inertia. If you don't know that theorem, brute force integration will give the same result.
Moral: Learn some 'tools of the trade' to make your life easy.

 

[kyle8921]

Yes, for part (c) you would divide the sheet into N thin strips concentric with the axis of rotation. Each strip would have a mass deltaMi and a distance Ri from the axis. The moment of inertia for each strip would be deltaIi = deltaMi*Ri^2. Then, by integrating over all the strips, the total moment of inertia would be given by the expression:
I= Sum deltaIi = Sum deltaMi*Ri^2 = k*Sum Ri^2*deltaRi, where k is the mass per unit area and deltaRi is the thickness of each strip. As before, passing to the limit of deltaRi -> 0 and N -> infinity, we can replace the sum with an integral:
I= k*Integral (0,R)[R^3]dR = k*[R^4/4]0,R = k*R^4/4, where R is the distance from the axis to the edge of the sheet. Substituting in the values for k and R, we get:
I= (24kg/(.3m*.4m))*(.4m)^4/4 = 6.4 kg.m^2
So the moment of inertia for the sheet about an axis perpendicular to the plane through its center would be 6.4 kg.m^2.