Moment of inertia for a rectangular plate

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The discussion focuses on deriving the moment of inertia for a rectangular plate and a sphere. For the rectangular plate with area ab and an axis through the center, the correct moment of inertia is derived as (1/12)(M)(a^2 + b^2). The method involves integrating the distance from each point on the plate to the axis of rotation, leading to the formula after substituting the mass M in terms of area density. For the sphere, the moment of inertia is correctly calculated as (2/5)MR^2, using spherical coordinates and integrating the volume elements. The confusion arises from incorrect setups leading to an alternative result of (3/5)MR^2. The discussion emphasizes the importance of proper integration techniques and coordinate systems in deriving these formulas, clarifying where common mistakes may occur.
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can anyone help me derive the moment of inertia for a rectangular plate, area of ab, (with the axis through the center)? i know it ends up being (1/12)(M)(a^2+b^2) but when i try it, my a's and b's end up canceling... its craziness. also, when i try the sphere, instead of (2/5)MR^2, i get(3/5)MR^2...
thanks...
 
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It's hard to say where you went wrong since you don't tell us what you did but this is how I would do that problem:
The moment of inertia about a lamina (two-dimensional figure) with area density ρ is

\int\int\rho r^2 dA where r is the distance from each point to the axis of rotation and dA is the differential of area.

In this problem, take the center of the rectangle to be at (0,0) so the vertices are (a/2, b/2), etc. Then the distance from (x,y) to the axis of rotation (passing through (0,0)perpendicular to the plate) is √(x2+ y2) and the moment of inertia is
\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}\rho (x^2+ y^2)dydx
= \int_{-a/2}^{a/2}\rho(bx^2+ \frac{1}{12}b^3)dx
= \frac{1}{12}\rho(a^3b+ ab^3).

Since M= ab ρ the moment of inertia is
\frac{M}{12}(a^2+b^2).
 
Similarly, for the sphere, set up spherical coordinates so that the sphere, of radius R, has center at (0,0,0) and the axis of rotation is the z-axis. Then the distance from each point to the axis is just the projection to the xy-plane, \rho sin\phi and differential of volume is \rho^2sin\phi d\rho d\phi d\theta so the moment of inertia, with density \lambda is given by
\int_0^\pi\int_0^{2\pi}\int_0^R(\lambda\rho^4sin^3\phi)d\rho d\theta d\phi
= \lambda (\int_0^{2\pi} d\theta )(\int_0^R\rho^4d\rho)(\int_0^\pi sin^3\phi d\phi)

The first two of those can be integrated directly and give
(2\pi)(\frac{1}{5}R^5).
To integrate the third, let u= cos\phi and we get
\frac{4}{3}. The moment of inertia is given by
\frac{8}{15}\pi\lambda R^5.

Since the mass is M= \lambda\frac{4}{3}R^3, the moment of inertia is \frac{2}{5}\pi R^2.
 
thanks that helps a lot. now i know i was missing something important..
 
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