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Trouble Determining Moment of Inertia Formula

by bob1182006
Tags: determining, formula, inertia, moment, trouble
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bob1182006
#1
Nov2-07, 09:21 PM
P: 492
1. The problem statement, all variables and given/known data
Find the Moment of Inertia for a uniformly cut wedge of Length L and mass M.

Rotated about an axis perpendicular to the tip of the wedge. so it increases in height and mass as you go -> L from 0.

2. Relevant equations
[tex]dI=r^2 dm=\rho r^2 dV=\rho r^2 A dr[/tex]
[tex]\rho = \frac{M}{V}[/tex]

3. The attempt at a solution
I just can't figure out how to setup the intergral. or any of the Moment of Inertial Integrals for that matter..

If I cut the wedge into vertical slices I get 1 small triangle at the tip, and then you'd get rectangles with some small triangle at the top of them.

Do I need to do this using 3/2 integrals? Or is it possible to do it with just 1? Because so far I only have knowledge of up to 1 integral :/.
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saket
#2
Nov2-07, 10:05 PM
P: 169
You can neglect those small triangles, isnt it? 1 integral should suffice. (Well, what is the width of the base of the wedge? And, is it a right-angled wedge?)
bob1182006
#3
Nov2-07, 11:12 PM
P: 492
O I see, so I can ignore those triangles because making an infinite amount of those rectangles will account for their area.

The base is length L. and I'm assuming it's right angled from how the teacher drew it.

but when doing the integral I'd need x to vary from 0 to L. and some linear function f(x) that gives the height of the rectangles that increases from 0 to ?. since no height was specified and I think it doesn't matter how high it goes.

saket
#4
Nov2-07, 11:14 PM
P: 169
Trouble Determining Moment of Inertia Formula

Assume some height, say h, and start. I think, h will be required. (Or, was it an isosceles wedge?)
bob1182006
#5
Nov2-07, 11:26 PM
P: 492
no he said like a rod and you draw the diagonal connecting 2 corners, and remove one of the triangles. the other one will be the wedge, we kept the bottom and drew the diagonal from bottom left to top right.

but height will be changing at some rate, from 0 to h. so then I'd need to do a double integral?

cut the wedge into pieces, arbitrary piece with length dx, distance x from the axis of rotation.

[tex]dI=x^2 dm[/tex]

Not sure if this will be right.

[tex]\frac{dm}{dx}=\frac{\rho}{L}[/tex]
[tex]\rho = \frac{M}{V}=\frac{M}{\frac{1}{2}Lh}=\frac{2M}{Lh}[/tex]
[tex]dm=\frac{2M}{L^2h}dx[/tex] ?
saket
#6
Nov2-07, 11:53 PM
P: 169
Ok, I am assuming your wedge to be as shown in the attached file. Is it okay?
The wedge is being rotated about rightmost vertex, A.
Attached Thumbnails
Image(022).jpg  
bob1182006
#7
Nov3-07, 12:04 AM
P: 492
can't see the attached file.
but it resembles an inclined plane and you start at 0 with height 0 and end at L with height h.
saket
#8
Nov3-07, 12:11 AM
P: 169
yep, is it okay?
(Attachment is still in pending for approval. :( )
saket
#9
Nov3-07, 12:16 AM
P: 169
If that is the case, then don't you think, your following formula doesnt hold here: (Why?)


Quote Quote by bob1182006 View Post
[tex]dI=x^2 dm[/tex]
Simply because, it is a thin rod (of thickness dx) and length -- that can be obtained using similar triangles (or trigonometry). So, you should use moment of inertia for a thin rod (about its centre) and then use parallel axis theorem to get about the required vertex.
bob1182006
#10
Nov3-07, 12:17 AM
P: 492
it is if the rightmost vertex A is the tip of the wedge.

so you'd have
|          .B
|          |
.__________| h
A          C
|
where the | above/below A represent the axis of rotation.
with a line connecting AB and L=AC.

How could I do this using similar triangles? so you have the sum of the 2 wedges that make up the rod: [tex]I_r=I_{BW}+I_{TW}[/tex] where BW=bottom wedge, TW=top wedge. but they are both not equal so the bottom may be 80% of the total I_r but I have no knowledge of that so I can't go that way right?
saket
#11
Nov3-07, 12:32 AM
P: 169
Wait, I am a bit confused about the figure.
First thing, your wedge is a right-triangular prism, of some thickness, t.
The triangular face has sides L and h. (None of these are hypotenuse of the triangle.)
Now let us consider this triangular face on the plane of the paper (such that thickness, t, is along the perpendicular to the plane of the paper). Call it triangle ABC, right angled at C.
Your axis is passing through one of the vertices, A or B, of this triangle. Say, it is passing through A. Then, AC = L and BC = h.
Now, is your axis perpendicular to the plane of this triangle (i.e., parallel to thickness, t) or, is it in the plane of this triangle (and perpendicular to AC or BC) instead?
bob1182006
#12
Nov3-07, 12:39 AM
P: 492
does it need the thickness? all we were given was just that 2D figure.

say point A is (0,0) in the x-y plane then the axis of rotation is the y-plane.
so the axis is perpendicular to BC.

we even did a rod in class but in it's calculations no thickness was considered. I think since a thicker rod is the sum of thinner rods which will give the same Moment of Inertia?
The proffesor used that type of argument for a Disk -> Cylinder as the sum of disks
rl.bhat
#13
Nov3-07, 01:00 AM
HW Helper
P: 4,435
Assume thickness of the wedge t. Take a small section of wedge width dx and height h' at a distance x, perpendicular to AC. Now dm = D*h'*dx*t, where D is the density. In triangle h'/x = h/L. Hence h' = x*h/L;
dm = D*(h/L)*t*x*dx.
I = Int{x^2*dm} = D*(h/L)*t*Int{ x^3*dx} from 0 to L. In the answer put 1/2*D*L*h*t=M. Hence find I.
bob1182006
#14
Nov3-07, 01:19 AM
P: 492
wow I got the answer right away assuming some thickness....

But I'm curious, when doing a rod of uniform density why don't we assume some thickness? I'm trying to do it this way now and I'm getting an extra 2/(pi r)

Edit: nevermind doing it as a rectangular board instead of a cylinder getting it now!

Thanks alot!
rl.bhat
#15
Nov3-07, 01:28 AM
HW Helper
P: 4,435
In that the diameter of the rod itself is the thickness. and dm = D*pi*R^2*dx
bob1182006
#16
Nov3-07, 01:44 AM
P: 492
oo right and calculating D will give a M/L(pi*R^2) to cancel out that pi*R^2 so it has the same Moment of Inertia.

so when solving for dm is this process right?:
[tex]\frac{dm}{dV}=\frac{M}{V}[/tex]
where dV and V might be dA and A depending on the type of object, usually it's dV though and then dV=Adx or Adr right? where A is the area and then solve for dV and V and multiply both sides by dV?

and ran into some problem. calculating the moment of inertia of a uniform hoop. Radius R, height h, thickness t. rotated about the center of the hoop.
professor did:
[tex]I=\int R^2 dm=R^2\int_0^M dm= MR^2[/tex]
me doing it the longer way:
[tex]V=\pi R^2 t h[/tex]
[tex]dV=th*?[/tex]

Edit: got it now had a thickness that varied and canceled out after integration so left with the Moment of Inertia of a Hoop. Thanks Alot for the Help!
rl.bhat
#17
Nov3-07, 02:16 AM
HW Helper
P: 4,435
Your last experssion is wrong. What is the cross section of the hoop?
bob1182006
#18
Nov3-07, 01:39 PM
P: 492
the cross section would be a rectangle.
and the hoop can be cut into a rectangle in order to find the volume. the rectangle will have length 2piR, thickness w, height t

[tex]V=2\pi Rtw[/tex]

the cross section is also going to be a rectangle again with the same height/thickness but length going from 0 to 2pi R. Being rotated about the center of the hoop with radius R.

[tex]dV=wt dx[/tex]

so [tex]dm=\frac{M twdx}{2\pi Rtw}=\frac{M dx}{2\pi R}[/tex]

and the integral is [tex]I=\int_0^{2\pi R} R^2 \frac{M dx}{2\pi R}=\frac{R M}{2\pi}\int_0^{2\pi R} dx=MR^2[/tex]

right?


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