# Homework Help: Moment Of Inertia Help

1. Sep 15, 2007

### metalInferno

Moment Of Inertia Help!!

people ,
i'm a student , and this is a complicated problem i came across ! please noe that i know da simplest of integration and differentiation and the like .

it might sound stupid , but can anyone tell why , during the calculation of moment of inertia of a ring the limits during integration are taken as R and 0 , instead of R and -R as in other cases like sphere and rod etc.

i noe a bit of calculus just to do about these things only .

please help !!!

got a hard exam !!

2. Sep 15, 2007

### arildno

Eeh?
What is R meant to be?
A radius?
An angle?
The length of a straight line segment?

And:
What type of ring are you talking about?
A 2-D ring with finite or infinitesemal thickness?
Or something else entirely?

3. Sep 15, 2007

### metalInferno

R is supposed to be the radius , and yes da figure is 2d . the answer is supposed 2 be 0.5 M(R^2) .

4. Sep 15, 2007

### arildno

Well, is it a RING or a DISK you are trying to find the moment of inertia to?

The disk has no hole in the middle, but a ring does.
Furthermore, if it IS a ring, is the thickness of it negligible, or is it an annulus?

5. Sep 15, 2007

### metalInferno

Man , it is a ring , and the thickness is negligible . Now please help!!

6. Sep 15, 2007

### arildno

So:
There is a big hole in the middle?

In that case, the given answer is just wrong.

Are you sure it is not a compact disc?

7. Sep 15, 2007

### Mindscrape

I'm guessing that your book exploited symmetry, and calculated half the disk, then multiplied by two.

8. Sep 15, 2007

### metalInferno

can u show the integration as u say , because it doesnt say anything like that

9. Sep 15, 2007

### arildno

Again:

Is it a compact 2-D disk, or is it a ring with a hole in the middle?

10. Sep 15, 2007

### metalInferno

sorry , sorry , my bad

1. it is a compact disc
2. thickness is negligible

now , i beg for your help , got an exam tomorrow !!

11. Sep 15, 2007

### metalInferno

please explain the integration part

12. Sep 15, 2007

### metalInferno

R = Radius of the disc

13. Sep 15, 2007

### arildno

Okay!

Now, for your question:

You can consider the disk to be composed of concentric infinitely thin rings at varying radii (hence, they have infinitesemal mass dm), and the disk's moment of inertia is the sum 8i.e, integral) of their inertias:

$$I=\int_{0}^{R}r^{2}dm=\int_{0}^{R}r^{2}(\frac{M}{\pi{R}^{2}})2\pi{r}dr$$
where:
$$\frac{M}{\pi{R}^{2}}$$ is the area density of the disk (M its mass!), whereas the "area" of a ring equals its circumference ($2\pi{r}$) times its width "dr".

At all r's between 0 and R, we must put such a ring; hence, the integration limits are 0 and R
Simplifying this expression, we get:
$$I=\frac{2M}{R^{2}}\int_{0}^{R}r^{3}dr=\frac{2M}{R^{2}}\frac{R^{4}}{4}=\frac{1}{2}MR^{2}$$

14. Sep 15, 2007

### metalInferno

aahhhhhhhh!!

but what about the moment of inertia of a sphere , where limits are +R and-R (R = radius) ??

15. Sep 15, 2007

### arildno

Okay!

First off, the variable lying between -R and R can't possibly be a radius itself, since it is meaningless to assign a negative number to a radius!

If we WERE to use a radius as our variable, then it would most naturally go from 0 to R, and we would regard the compact ball as composed of infinitely many spherical SHELLS.

That decomposition of the ball (or compact sphere if you like) is perfectly valid of course, but not the one implied with integration limits -R and R.

Rather, our variable in this case runs along a diameter of our ball, having its origin coincide with the ball's center. Let us call that variable z; we are essentially decomposing our ball as a set of stacked on disks/cylinders with infinitesemal heights, and with varying radii as we move along our diameter.

Get the picture?

Last edited: Sep 15, 2007
16. Sep 15, 2007

### metalInferno

soooooooo ,

aahhhhhh!!

all right , it means that in a sphere , we dont take radius as the limit . so
1. what are +R and -R then ?
2. or in other words , what are the limits !

it would be really nice if u explain it like u did above .
and can give the integration part of the moment of inertia of a sphere ? thanx a lot for all ur help !

17. Sep 15, 2007

### metalInferno

also the same thing happens in a rod (thin , compact ,NO holes ) , where limits are l/2 and -l/2 where , it is given l=length (please excuse if i am wrong ) . please help!!

18. Sep 15, 2007

### arildno

We MAY, but not in this particular decomposition!
In a ball with radius R, the diameter has length 2R.
Explained above.
You have a line segment of length 2R; you choose its midpoint to be the origin; hence, its end points lie at -R and R, respectively.

Okay, but you need to do a bit yourself as well!

First off, if the ball's mass is M, its radius R, what is then its volume density?

19. Sep 16, 2007

### metalInferno

density should be M/(4/3 * pi * r^3)

20. Sep 16, 2007

### metalInferno

also can u tell me how is area = circumference * width ?

21. Sep 16, 2007

### arildno

Think of the arc as a bent rectangle of height equal circumference and width infinitely small.

22. Sep 16, 2007

### arildno

Correct. Note that I'll use R for the radius of the ball (mere convention).

Now, we decompose the ball as infinitely thin cylinders of height dz, and with variable radii r.

These cylinders have infinitesemal volume $dV=\pi{r}^{2}dz[/tex], and hence, mass $$dm=\frac{M}{\frac{4}{3}\pi{R}^{3}}\pi{r}^{2}dz=\frac{3}{4}\frac{M}{R^{3}}r^{2}dz$$ Each of these cylinders have moments of inertia [itex]dI=\frac{1}{2}dmr^{2}=\frac{3}{8}\frac{M}{R^{3}}r^{4}dz$, as we found in the last exercise as the moment of inertia for a disk (i.e, infinitely thin cylinder!).

Now, the diameter variable z runs from -R to R in the ball, and at each value of z, such a cylinder lies.

The moment of inertia I for the ball is the sum/integral of the moments of inertia of the cylinders, that is:
$$I=\int{d}I=\int_{-R}^{R}\frac{3}{8}\frac{M}{R^{3}}r^{4}dz=\frac{3}{8}\frac{M}{R^{3}}\int_{-R}^{R}r^{4}dz$$

Now, the cylinder radii r varies as to where they lie on the axis, hence, r is a function of z.

By the Pythagorean theorem, we have the relationship:
$$r^{2}=R^{2}-z^{2}$$,
and thus, we gain the expression for the moment of the inertia for the ball:
$$I=\frac{3}{8}\frac{M}{R^{3}}\int_{-R}^{R}(R^{2}-z^{2})^{2}dz$$
Calculating this yields the well known result:
$$I=\frac{2}{5}MR^{2}$$

23. Sep 16, 2007

### metalInferno

man , u seriously have one of the best brains i have ever seen . thanks a lot for all your help . i get the moment of inertia of both the bodies (disc and sphere ) . one last stupid question :

is area = circumference * width ( in the moment of inertia of disc derivation) true for width with finite values .

i think its no . what do u have to say ??

24. Sep 16, 2007

### arildno

Remember that a bent arc region of non-zero width has a SHORTER inner bounding arc than outer bounding arc.

Therefore, by unbending it, you get a trapezoid, not a rectangle.
That trapezoid has equal area as the bent figure.

In the limit of vanishing width, the trapezoid turns into a rectangle.

25. Sep 16, 2007

### arildno

I'll prove it to you like follows:

Suppose you have an annulus with inner radius r, greater radius R.
Its area A is the difference between the two disk's areas,
$$A=\pi(R^{2}-r^{2})=\pi(R+r)h, h=R-r$$

Now, consider a trapezoid with parallell sides equal in length to the circumferences, and height h. That trapezoid area's T equals:
$$T=\frac{2\pi{R}+2\pi{r}}{2}*h=\pi(R+r)h=A$$

Which was to be proven..

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook