Hi, So as not to lead anybody astray, I have decided to post a correction to my ill-fated attempt in this thread: https://www.physicsforums.com/showthread.php?t=158832 to derive the moment of inertia of a solid sphere of uniform density and radius R. [tex] dI = r_{\perp}^{2} dm = r_{\perp}^{2} \rho dV [/tex] where [itex] r_{\perp} [/itex] is the perpendicular distance to a point at r from the axis of rotation. Therefore, [itex] r_{\perp} = |\mathbf{r}|\sin{\theta} [/itex]. Integrating over the volume, [tex] I = \int \! \! \! \int \! \! \! \int_V \rho r_{\perp}^2 \,dV = \rho \int_0^{2\pi} \int_0^{\pi} \int_0^R (r \sin{\theta})^2 r^2 \sin{\theta} \,dr\,d \theta\, d\phi[/tex] [tex] = 2\pi \rho \int_0^{\pi} \sin^3{\theta}\, d\theta\, \int_0^R r^4\,dr [/tex] Make the substitution u = cos[itex]\theta[/itex]. [tex]= 2\pi \rho \frac{R^5}{5} \int_{-1}^{1} (1 - u^2)\,du = 2\pi \rho \frac{R^5}{5} \left(1 - (-1) - \left[\frac{u^3}{3}\right]_{-1}^1 \right) [/tex] [tex] = 2\pi\rho \frac{R^5}{5} \left(2 - \frac{2}{3}\right) = \underbrace{\frac{4}{3} \pi R^3 \rho}_{M} \left(\frac{2}{5}R^2\right) [/tex] [tex] = \frac{2}{5} MR^2 [/tex] where M is the total mass of the sphere. Please let me know if there is anything wrong with this derivation. Thanks, Cepheid
Interesting derivation, I never thought about doing it this way. I use cylindrical coordinates; it eliminates the [tex]sin(\theta)[/tex] factor, but introduces a [tex]\sqrt{R^2-r^2}[/tex] in one of the limits of integration for the z variable. Anyway, it seems to look fine to me.
Cotufa is doing homework on "moment of inertia" of uniform solid sphere and a uniform solid cylinder. And needs to solve in both spherical & cylindrical coordinate system. It won't help cotufa learn anything by looking at arunma's derivation. I recommend not to post that.
It is a very nice derivation but there is a small error in the limits of integration after substitution. The limits should be 1 to -1 (not -1 to 1). The answer will be the same (but would in fact be different if you used the other limits).
Actually there isn't any error in the limits of integration. If you pay attencion you will notice that: [tex]\int \sin^3(\theta)d\theta = -\int^{-1}_1 (1-u^2)du = \int^{1}_{-1} (1-u^2)du.[/tex] It's all right.
Exactly. When I did the substitution u = cos(theta), I got du = -sin(theta)d(theta), and I absorbed that negative sign by flipping the limits of integration.
hi, erm can you explain how you get your integral terms in the triple integral step? i understand why you use (rsinθ )^2, which is your r^2(perpendicular) term, but why is there a r^2sinθ after that?
A volume element in spherical coordinates is given by: [tex] dV = r^2 \sin{\theta} \,dr\,d \theta\, d\phi [/tex] You can think of the volume element as an infinitesimal cube. One dimension of the cube, the length in the radial direction, is given by dr. The length in the azimuthal direction (the arc swept out by [itex] d\phi[/itex]) is given by [itex] r \sin \theta d\phi [/itex]. The third dimension of the cube (the arc swept out by [itex] d\theta[/itex]) is given by [itex] r d\theta [/itex]. The following diagram, which I found just by Googling, will aid greatly in understanding *why* this is so. In particular, it explains where the sine factor comes from (it involves a projection onto the xy-plane). Be mindful, however, that whoever made this diagram uses [itex] \rho [/itex] instead of r, and his definitions of [itex] \theta [/itex] and [itex] \phi [/itex] are *switched* compared to mine. http://www.spsu.edu/math/Dillon/VolumeElementSphericalCoordinates.htm
oh now it all makes sense.. no wonder my lecturer said we had to take it for granted that for the projection on a xy plane of a circle, we had to change the dA to rdrd(θ) . i was wondering where the r came from. so when integrating for the area of a circle, i am actually integrating 1 dA = integrate 1 rdrd(θ) which is actually the radius times the arc length (rdθ term) where 0<θ<2pi so its like chopping the circle into tiny rectangular pieces and integrate over the entire circle so the integrand we are using is actually the radius(length) X arc length (width) is it?
bad lecturer! Very good! An even more precise way of looking at it, would be to use the difference between two sectors of the circle, both with obening [itex]d\theta[/itex], but radii r and (r+dr). Thus, we get: [tex]dA=\frac{1}{2}(r+dr)^{2}d\theta-\frac{1}{2}r^{2}d\theta=(rdr+\frac{dr^{2}}{2})d\theta[/tex] But, since the non-linear term in dr is vanishing relatively to the dr term as dr goes to 0, it follows that the appropriate area element is [itex]rdrd\theta[/itex] Due to the linearization argument above, yes.
If I understand you right, you are now talking about a 2D case in which you are using polar coordinates (r,θ) to represent a circle in the plane. Well now you know that, simply from definition of angular measure (in the radian system), rdθ is the arc swept out by angle dθ at radius r. Also you know that dA has to have units of area. r has units of length, and θ is dimensionless, so drdθ would not have had the right units. Yes, the area of a circle of radius R can be calculated using polar coordinates in exactly the way you have described: [tex] A = \int_A dA = \int_0^{2\pi} \int_0^R r\,drd\theta = 2\pi \int_0^R r \, dr = 2\pi \frac{1}{2}R^2 [/tex] Essentially, yes. Yes, where these are the length and width (respectively) of the infinitesimal area element dA.
thanks everyone! oh btw, just a side question, when we talk about the double integral of a rectangle in the xy-plane, say bounded by the x-axis, y-axis, x=2, y=1, then when we do a double integral of any variable say X, we take the integral limits of dx to be 0 to 2, and for dy from 0 to 1. but why sometimes we have to convert the limits to in terms of dx, for example, it becomes integrate from 0 to 1 dx, BUT example, integrate for dy from 0 to 5x-6 , where the y limits becomes expressed in x terms. so how do we know when to change the limits and when to just use the number? thanks. sry if its confusing
Well, I mean, if y is a well-defined function of x, then the boundaries of the region can be expressed entirely in terms of x, and the integral reduced to a single (one-dimensional integral).
How come when you use this we get: [tex] I = \int r^2dm = \rho\int^{2\pi}_{0}d\phi\int^{\pi}_{0}sin{\theta}d\theta\int^{R}_{0}r^4dr = \frac{3M}{4\pi R^3}*2\pi*2*\frac{R^5}{5} = \frac{3}{5}MR^2[/tex]
Hi Hells_Kitchen, You will notice that, in my original post, I wrote the moment of inertia as: [tex] I = \int_V r^2_{\perp}\, dm [/tex] That subscript on the r with the "perpendicular" symbol is very important. The distance [itex] r_{\perp} [/itex] is the perpendicular distance of the mass dm at [itex] (r,\theta,\phi) [/itex] from the axis of rotation, NOT its distance from the centre of the sphere, r. In other words, [itex] r_{\perp} [/itex] is the radius of the latitude circle being traced out by the point at [itex] (r,\theta,\phi) [/itex] as the sphere rotates. If you draw a picture, you'll see that [itex] r_{\perp} [/itex] is equal to [itex] r \sin \theta [/itex], giving you an extra sine squared factor in the theta integral. I showed all of this in my original post. That was the point...to post a full and correct derivation.
I understand! [tex] r_{\perp} = rsin{\theta} [/tex] Then you get [tex] \int^{\pi}_{0} sin^3{\theta} d\theta , [/tex] which is just [tex]\frac{4}{3}[/tex] using [tex]sin^3{\theta} = sin{\theta}(1-cos^2{\theta})[/tex] identity and it works out to be [tex] I = \frac{2}{5} MR^2[/tex]. Thanks!