# Moment of inertia of a disk by integration

by soopo
Tags: calculus, disk, moment of inertia
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 P: 226 1. The problem statement, all variables and given/known data Show that the moment of inertia of a disk is $0.5 mr^2$. 3. The attempt at a solution $$I = \int R^2 dm$$ Using $dm = \lambda dr$ such that $m = \lambda r$: $$= \int_{-r}^{r} R^2 \lambda dr$$ $$= \frac { \lambda } {3} ( 2r^3 )$$ $$= \frac {2} {3} (\lambda r ) (r^2)$$ $$= \frac {2} {3} M R^2$$ which should be the moment of inertia for a ring. Integrating this from 0 to 2pii relative to the angle gives me $\frac {4} {9} m r^3 [/tex], which is wrong. How can you calculate the moment of inertia for a disk? P: 383  Quote by soopo 1. The problem statement, all variables and given/known data Show that the moment of inertia of a disk is [itex] 0.5 mr^2$. 3. The attempt at a solution $$I = \int R^2 dm$$ Using $dm = \lambda dr$ such that $m = \lambda r$: $$= \int_{-r}^{r} R^2 \lambda dr$$ $$= \frac { \lambda } {3} ( 2r^3 )$$ $$= \frac {2} {3} (\lambda r ) (r^2)$$ $$= \frac {2} {3} M R^2$$ which should be the moment of inertia for a ring. Integrating this from 0 to 2pii relative to the angle gives me $\frac {4} {9} m r^3 [/tex], which is wrong. How can you calculate the moment of inertia for a disk? If you have a disc of radius $$r$$, how can you integrate from $$-r$$ to $$r$$?  P: 383 The infinitesimal change in mass is given by $$dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr$$ If you use that in your integral and integrate from $$0$$ to $$R$$, you should get the desired result. P: 226 Moment of inertia of a disk by integration  Quote by jdwood983 The infinitesimal change in mass is given by $$dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr$$ If you use that in your integral and integrate from $$0$$ to $$R$$, you should get the desired result. Your result gives me a wrong result: $$I = \int R^2 dm$$ $$= \int R^2 \frac { 2m } {R^2} r dr$$ $$= \int_{0}^{r} 2mr dr$$ $$= [ m r^2 ]^{r}_{0}$$ $$= mr^2$$ The result shoud be $$I = .5 mr^2$$ P: 226  Quote by jdwood983 If you have a disc of radius $$r$$, how can you integrate from $$-r$$ to $$r$$? I set the null point to the center of the circle such that I am integrating from -r to r. I am not sure why I cannot do that. P: 383  Quote by soopo I set the null point to the center of the circle such that I am integrating from -r to r. I am not sure why I cannot do that. If you set the origin to the center of the circle (which you should always try to do), the smallest value that $$r$$ can be is 0. So it is physically impossible to integrate from $$-r$$ to $$r$$, that is why you can't do it. P: 226  Quote by jdwood983 If you set the origin to the center of the circle (which you should always try to do), the smallest value that $$r$$ can be is 0. So it is physically impossible to integrate from $$-r$$ to $$r$$, that is why you can't do it. I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards $$r$$, while going to left means towards $$-r$$. Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative $$-r$$. I feel that it is possible to integrate from $$-r$$ to $$r$$ in a cartesian coordinate system. P: 383  Quote by soopo I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards $$r$$, while going to left means towards $$-r$$. Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative $$-r$$. I feel that it is possible to integrate from $$-r$$ to $$r$$ in a cartesian coordinate system. If you want Cartesian coordinates, then you'll need two integrals: one over $$x$$ and one over $$y$$. While technically you have two integrals in polar, $$r\, \mathrm{and}\, \theta$$, one is already done for you and reduces the integration to just one term: $$r$$. This problem is by far easier in polar coordinates: $$\begin{array}{ll}I&=\int_0^R r^2\frac{2m}{R^2}rdr \\ &=\frac{2m}{R^2}\int_0^Rr^3dr \\ &=\frac{2m}{R^2}\left(\frac{R^4}{4}-0\right) \\ &=\frac{2m}{R^2}\cdpt\frac{R^4}{4} \\ &=\frac{1}{2}mR^2$$ P: 226  Quote by jdwood983 If you want Cartesian coordinates, then you'll need two integrals: one over $$x$$ and one over $$y$$. While technically you have two integrals in polar, $$r\, \mathrm{and}\, \theta$$, one is already done for you and reduces the integration to just one term: $$r$$. This problem is by far easier in polar coordinates: If you use symmetry, it is enough to consider only the first quadrant that is where x > 0 and y > 0 such that four of these quadrants form the area of the disk. You do not get the x- and y -coordinates easily from the definition of the moment of inertia. You would get $$I = \int (x^2 + y^2) dm \\ &= \int (x^2 + y^2) m \frac { 2r } {R^2} dr$$ The calculations seem to get challenging, since we need to use Pythogoras such that $$r = \sqrt{ x^2 + y^2 }$$ which implies $$dr = \frac { 1 } { \sqrt {x^2 + y^2} } * 2x$$ We can get similarly the relation relative to $$y$$. The next step is not fun at all: $$I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx$$, where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4$, 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.
P: 383
 Quote by soopo The next step is not fun at all: $$I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx$$, where I assume that $R^2 = (x + y)^2 = (1 + 1)^2 = 4$, since it is the maximum radius.
Not quite. The integral you need is given by

$$I=\frac{m}{A}\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\left(x^2+y^2\right)dydx$$

$$I=\frac{m}{\pi R^2}\int_{-R}^R \frac{2\sqrt{R^2-x^2}\left(R^2+2x^2\right)}{3}dx$$

$$I=\frac{m}{\pi R^2}\cdot\frac{\pi R^4}{2}$$

$$I=\frac{mR^2}{2}$$

 Quote by soopo Polar coordinate system really seems to be better in this case.
For most moment of inertia problems, spherical or cylindrical coordinates are the best.
P: 226
 Quote by jdwood983 $$\int_{-R}^R \frac{2\sqrt{R^2-x^2}\left(R^2+2x^2\right)}{3}dx$$ $$= \frac{\pi R^4}{2}$$
How did you solve this part?

It has taken my some effort in trying to solve it by hand.
P: 383
 Quote by soopo How did you solve this part? It has taken my some effort in trying to solve it by hand.
There's a few extra steps between the two lines, like making a change of variables. But to be honest I used Mathematica and just wrote the lines because I forget what changes needed to be made. While it may be good to know the form of the equation, I'm not sure you would need the solution since it is far easier to do it in polar coordinates than in Cartesian coordinates.

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