Moment of inertia

by allergic
Tags: inertia, moment
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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 $$\frac{M}{12}(a^2+b^2)$$.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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 $$\frac{2}{5}\pi R^2$$.