Why Is the Moment of Inertia of a Uniform Disc 1/2MR^2?

[CuriousJonathan]

I'm in AP physics because I was really bored in acc physics, but I'm not actually in calculus, so I may ask you to explain basic concepts further if I have not yet had the chance to figure them out for myself.
I was wondering about the equation for the moment of inertia of a uniform disc rotating about it's central axis. My book tells me that it's 1/2MR^2, but I attempted to arrive at that same equation myself and got MR^2, which is wrong-o. I've attempted several times trying different ways, but I think the problem isn't that I don't know how conceptually, it's that I haven't learned to think in calculus yet. If someone could explain to me how to do this that would be great . Thank you very much for your help.

Also, I would like to know if mathematical symbols are built into this forum anywhere.

 

[CuriousJonathan]

I figured out the mathematical symbols!

\int \ r^{2} \ dm

This is what my book tells me to start with as the generic moment of inertia equation and then to substitute for dm.

 

[dextercioby]

I figured out the mathematical symbols!

\int \ r^{2} \ dm

This is what my book tells me to start with as the generic moment of inertia equation and then to substitute for dm.

Chose a mass element "dm" at a distance "\rho" from the center of the circle.Due to the symmetry of the problem,it's wise to chose polar coordinates
The mass element is given wrt to its surface by
dm=\frac{M}{\pi R^{2}} dS(1)
The surface element in polar coordintes is
dS=\rho \ d\rho \ d\phi(2)

Therefore the integral
I=\int \rho^{2} dm (3)

becomes:
I=\int_{0}^{R}d\rho \int_{0}^{2\pi} d\phi \frac{M}{\pi R^{2}} \rho^{3}(4)

Integrate and find
I=\frac{M}{\pi R^{2}} 2\pi \frac{\rho^{4}}{4}|_{0}^{R} =\frac{1}{2}MR^{2}(5)


Daniel.

 

[dextercioby]

ALTERNATIVE METHOD
Think the disk as a reunion of concentric pieces in forms of coronas.Then:
The surface element of such a corona situated at a distance "r" from the center of the circle is
dS=2\pi r \ dr (1)
If the disk has homogenous surface density,then the mass element "dm" has the value:
dm=\rho_{Superficial} dS=\frac{M_{disk}}{S_{disk}} dS(2) (this assertion is valid for the prior post as well).

Then the mass element of such an infinitesimal corona is
dm=\frac{M}{\pi R^{2}} 2\pi r \ dr =\frac{2M}{R^{2}} r \ dr(3)

The integral giving the moment of inertia is
I=\int r^{2} dr =\int_{0}^{R} \frac{2M}{R^{2}} r^{3} dr (4)
,which is
I=\frac{2M}{R^{2}} \frac{r^{4}}{4}|_{0}^{R} =\frac{1}{2}MR^{2} (5)

Daniel.

 

[Galileo]

In general. If you know the density \rho(x,y,z) of the body, using dm=\rho dV will enable you to solve the problem.

I=\int \limits_E r^2\rho dV
where E denotes the region you integrate over.

In this particular case \rho is constant, so you can bring it outside the
integral.
Then, switching to polar coordinates, letting r run from 0 to R and \theta from 0 to 2\pi:

I=\rho\int_0^{2\pi}\int_0^Rr^2(r)drd\theta
You can eliminate \rho in favor of the mass to get the answer.

 

[CuriousJonathan]

Thanks for the help

thank you for all of the help. to tell the truth, I don't really understand any of it just reading it through, I'm going to have to print it out and look at it, but having the right answer will help me greatly in understanding why.

I have a question of the last post before this. I was wondering if the author of that post could supply me with the rest of the math, because I don't have any idea how one would proceed with integrating with respect to two different things. thank you again for the help

 

[CuriousJonathan]

Also, when I tried to accomplish this on my own, I attempted to think of it not as a union of concentric rings, but as a union of pie-piece-shaped wedges. This didn't work, but it did get me the moment of inertia equation for a ring. At least I got somewhere, even if it wasn't where I was intending to go.

 

[CuriousJonathan]

I was hoping someone could explain to me why my previous approach did not work.

 

[dextercioby]

I have a question of the last post before this. I was wondering if the author of that post could supply me with the rest of the math, because I don't have any idea how one would proceed with integrating with respect to two different things. thank you again for the help

It is called double integral and i doubt u'll see any of these in HS.Maybe if u go to a college/university in science domain,u'll learn about these "animals"...

Daniel.

 

[dextercioby]

Also, when I tried to accomplish this on my own, I attempted to think of it not as a union of concentric rings, but as a union of pie-piece-shaped wedges. This didn't work, but it did get me the moment of inertia equation for a ring. At least I got somewhere, even if it wasn't where I was intending to go.

Rings have no width.They don't have a surface.They have only one dimension,namely legth.You need to find the moment of inertia for a disk,which has both width and legth.The assumption of infinitesimally thin circular coronas yields the result,however,because these mathematical objects are similar to a disk,in respect that a finite reunion of such concentrical coronas cover the disk...

Daniel.

 

[Galileo]

Also, when I tried to accomplish this on my own, I attempted to think of it not as a union of concentric rings, but as a union of pie-piece-shaped wedges. This didn't work, but it did get me the moment of inertia equation for a ring. At least I got somewhere, even if it wasn't where I was intending to go.

Sorry about the double integral. Fortunately it can be done without it.

For a ring (circle) the moment of inertia is really simple. Since all the mass is at a distance R from the center, r in the integral is constant. So you can bring it outside of the integral and the integral over dm simply gives the mass.

I_{ring}=\int_{C} r^2 dm = R^2 \int_{C}dm=MR^2
where C is the circle you are integrating over.

If you consider the uniform disc as a collection of polar 'rectangles', you'll end up with a double integral.
I think the only way to get a single integral is to consider the disc as a collection of concentric rings with width dr. Then express dm in terms of r:
dm = (\frac{M}{\pi R^2})(2\pi rdr)=\frac{2M}{R^2}rdr[/itex]<br /> <br /> Now that you know dm in terms of dr, can you set up the integral and get the moment of inertia?<br /> <br /> BTW: This is identical to the method in dexter&#039;s second post and to a double integral after having integrated wrt theta.

 

[CuriousJonathan]

yes

yes, I can proceed with the ring approach, and actually some friends I know are going to be doing double integrals second semester, but since I'm not in calc I don't have to worry about them.

--My friend says "He's an idiot!. He's taking AP physics without taking calc"--

Anyways, thank you for the help!