Boost Your Understanding of Moment of Inertia with These Helpful Tips!

[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 !

 

[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?

 

[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) .

 

[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?

 

[metalInferno]

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

 

[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?

 

[Mindscrape]

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

 

[metalInferno]

can u show the integration as u say , because it doesn't say anything like that

 

[arildno]

Again:

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

 

[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 !

 

[metalInferno]

please explain the integration part

 

[metalInferno]

R = Radius of the disc

 

[arildno]

sorry , sorry , my bad

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

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

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}

 

[metalInferno]

aahhhhhhhh!


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

 

[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?

 

[metalInferno]

soooooooo ,

aahhhhhh!

all right , it means that in a sphere , we don't 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 ? thanks a lot for all ur help !

 

[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!

 

[arildno]

soooooooo ,

aahhhhhh!

all right , it means that in a sphere , we don't take radius as the limit

We MAY, but not in this particular decomposition!

1. what are +R and -R then ?

In a ball with radius R, the diameter has length 2R.

2. or in other words , what are the limits !

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.

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 ? thanks a lot for all ur help !

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?

 

[metalInferno]

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

 

[metalInferno]

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

 

[arildno]

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

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

 

[arildno]

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

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<br /> dm=\frac{M}{\frac{4}{3}\pi{R}^{3}}\pi{r}^{2}dz=\frac{3}{4}\frac{M}{R^{3}}r^{2}dz<br /> <br /> Each of these cylinders have moments of inertia 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!).<br /> <br /> Now, the diameter variable z runs from -R to R in the ball, and at each value of z, such a cylinder lies.<br /> <br /> The moment of inertia I for the ball is the sum/integral of the moments of inertia of the cylinders, that is:<br /> 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<br /> <br /> Now, the cylinder radii r varies as to where they lie on the axis, hence, r is a function of z.<br /> <br /> By the Pythagorean theorem, we have the relationship:<br /> r^{2}=R^{2}-z^{2},<br /> and thus, we gain the expression for the moment of the inertia for the ball:<br /> I=\frac{3}{8}\frac{M}{R^{3}}\int_{-R}^{R}(R^{2}-z^{2})^{2}dz<br /> Calculating this yields the well known result:<br /> I=\frac{2}{5}MR^{2}

 

[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 ??

 

[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.

 

[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..

 

[metalInferno]

I GET IT ! thanks a bunch ! i think i still have a chance to do something in my test ! neways i am glad i signed up for this site ! thanks once again !