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Moment of inertia of a disk by integration 
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#1
Oct2709, 05:07 PM

P: 226

1. The problem statement, all variables and given/known data
Show that the moment of inertia of a disk is [itex] 0.5 mr^2 [/itex]. 3. The attempt at a solution [tex] I = \int R^2 dm [/tex] Using [itex] dm = \lambda dr [/itex] such that [itex] m = \lambda r [/itex]: [tex] = \int_{r}^{r} R^2 \lambda dr [/tex] [tex] = \frac { \lambda } {3} ( 2r^3 ) [/tex] [tex] = \frac {2} {3} (\lambda r ) (r^2) [/tex] [tex] = \frac {2} {3} M R^2 [/tex] which should be the moment of inertia for a ring. Integrating this from 0 to 2pii relative to the angle gives me [itex] \frac {4} {9} m r^3 [/tex], which is wrong. How can you calculate the moment of inertia for a disk? 


#2
Oct2709, 05:33 PM

P: 383




#3
Oct2709, 05:36 PM

P: 383

The infinitesimal change in mass is given by
[tex] dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr [/tex] If you use that in your integral and integrate from [tex]0[/tex] to [tex]R[/tex], you should get the desired result. 


#4
Oct2709, 05:53 PM

P: 226

Moment of inertia of a disk by integration
[tex] I = \int R^2 dm [/tex] [tex] = \int R^2 \frac { 2m } {R^2} r dr [/tex] [tex] = \int_{0}^{r} 2mr dr [/tex] [tex] = [ m r^2 ]^{r}_{0} [/tex] [tex] = mr^2 [/tex] The result shoud be [tex] I = .5 mr^2 [/tex] 


#5
Oct2709, 05:55 PM

P: 226

I am not sure why I cannot do that. 


#6
Oct2709, 06:04 PM

P: 383




#7
Oct2709, 06:17 PM

P: 226

Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative [tex]r[/tex]. I feel that it is possible to integrate from [tex]r[/tex] to [tex]r[/tex] in a cartesian coordinate system. 


#8
Oct2709, 07:26 PM

P: 383

This problem is by far easier in polar coordinates: [tex] \begin{array}{ll}I&=\int_0^R r^2\frac{2m}{R^2}rdr \\ &=\frac{2m}{R^2}\int_0^Rr^3dr \\ &=\frac{2m}{R^2}\left(\frac{R^4}{4}0\right) \\ &=\frac{2m}{R^2}\cdpt\frac{R^4}{4} \\ &=\frac{1}{2}mR^2 [/tex] 


#9
Oct2709, 10:50 PM

P: 226

You do not get the x and y coordinates easily from the definition of the moment of inertia. You would get [tex] I = \int (x^2 + y^2) dm \\ &= \int (x^2 + y^2) m \frac { 2r } {R^2} dr [/tex] The calculations seem to get challenging, since we need to use Pythogoras such that [tex] r = \sqrt{ x^2 + y^2 } [/tex] which implies [tex] dr = \frac { 1 } { \sqrt {x^2 + y^2} } * 2x [/tex] We can get similarly the relation relative to [tex]y[/tex]. The next step is not fun at all: [tex] I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx [/tex], where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4 [/itex], since it is the maximum radius. This way the two 2s cancel out. I do not even know how to integrate this! Polar coordinate system really seems to be better in this case. 


#10
Oct2809, 07:12 AM

P: 383

[tex] I=\frac{m}{A}\int_{R}^R\int_{\sqrt{R^2x^2}}^{\sqrt{R^2x^2}}\left(x^2+y^2\right)dydx [/tex] [tex] I=\frac{m}{\pi R^2}\int_{R}^R \frac{2\sqrt{R^2x^2}\left(R^2+2x^2\right)}{3}dx [/tex] [tex] I=\frac{m}{\pi R^2}\cdot\frac{\pi R^4}{2} [/tex] [tex] I=\frac{mR^2}{2} [/tex] 


#11
Oct2809, 05:01 PM

P: 226



#12
Oct2809, 05:13 PM

P: 383




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