Moment of inertia for a rectangular plate

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Discussion Overview

The discussion revolves around deriving the moment of inertia for a rectangular plate and a sphere, focusing on the mathematical approaches and integration techniques involved in these calculations. Participants explore the theoretical underpinnings and practical applications of these concepts in physics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks help to derive the moment of inertia for a rectangular plate, expressing confusion over their calculations and results.
  • Another participant provides a detailed method for calculating the moment of inertia of the rectangular plate using double integration, specifying the setup and the resulting formula.
  • A third participant outlines a method for calculating the moment of inertia of a sphere using spherical coordinates, detailing the integration process and the resulting expression.
  • The initial poster acknowledges the assistance received and indicates they were missing an important aspect in their approach.

Areas of Agreement / Disagreement

Participants present different approaches and methods for calculating the moment of inertia, but there is no explicit consensus on the initial poster's confusion or errors. The discussion remains open with various perspectives on the calculations.

Contextual Notes

The discussion includes assumptions about the density and dimensions of the objects involved, as well as the integration limits used in the calculations. Specific steps in the integration process for the sphere are not fully resolved, and the initial poster's calculations are not detailed enough to identify the source of their confusion.

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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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