| which part of the question are you having trouble with?
are you having trouble setting up the integration itself, or
how to evaluate the integration once you set it up ?
in general the center of mass for a 2D section, has x,y coordinates
x= moment of mass around the x-axis
----------------------------------
mass of the section
y= moment of mass around the y-axis
----------------------------------
mass of the section
to find moment of mass around an axis, say x-axis, you just integrate
over the area multiplied by y.
so moment of mass around x-axis =
Int Int ( y * dA )
where dA = dx dy, and the limits of the integrations go over the
boundary, i.e. x=from..to, and y=from..to
if the density of the material is not unit, and is a function of x,y
also, just multiply that with dA above inside the integral.
again, i was not sure what part you having trouble with....
\nasser
|
| Nassar,
Thanks for your fast response, I am going to tackle this
problem over the weekend. It's the setting up of the integration thats
the problem. I've a few notes in the house that I am going to browse
over but if you could get me started that would be superb..
...........Alan
|
| .0> show how the results can be checked by elementary methods
Centroids 'add' and 'subtract'. Let s1 be a section with mass m1 and
centroid at c1=(x1,y1), and similarly for s2. The 'sum' of these sections
has mass m1+m2 and centroid at m1*c1+m2*c2 = (m1*x1+m2*x2,m1*y1+m2*y2).
Consider subtraction to be the addition of a negative mass.
The centroid for a semicircle was already given. To find the centroid
for a quarter circle (demicircle?), add together two semicircular regions
(with common center and radius) with one is rotated 90 degrees. One
quarter is 'two-ply', and two quarters are 'one-ply'. Subtract the two
'one-ply' quarters (their centroid is in the center, by symmetry). This
leaves the two-ply quarter circle, and so you have its centroid.
If you want the centroid for the section that's filled with x's below,
simply do one more subtraction.
y
|. . .
|xxxxxx'
|xxxxxxxx' Supposed to be circular
a|xxxxxxxxxx'
|xxxxx' ' 'xx'
|xxx' 'x'
|x' 'x'
|' 'x'
|_______|________'______ X
1/2 a 1/2a
|
| > y
> |. . .
> | '
> | ' Supposed to be circular
> a| '
> | ' ' ' '
> | ' ' '
> | ' ' '
> |' ' '
> |_______|________'______ X
1/2 a 1/2a
From the semi-circle info, it's easy to get the centre of mass of
a quadrant, just by folding the semi-circle over. Here, the quadrant has
centre of mass at (4a/3#,4a/3#), with mass #a�/4. The smaller semi-circle
has centre of mass at (a/2,2a/3#), with mass #a�/8. The difference therefore
has centre of mass at (8a/3#-a/2,2a/#), mass #a�/8. (Assuming unit mass per
unit area, and writing pi as #.)
Let I be the moment of inertia of a mass, M, about its centre of mass,
P. The moment of inertia, J, about some axis at distance r from P is given by
J = I + Mr�.
The moment of inertia of a circle of radius r, about its centre is
Mr�/2. So the moment of inertia of the quadrant is #a^4/8. The moment of
inertia of the smaller semi-circle about the origin is #a^4/64 + (#a�/8)*(a/2)�
= #3a^4/64. (There is a subtlety in the last step: the two summands are
(I+My�) and Mx�.) Anyway, the moment of inertia of the difference about the
origin is #5a^4/64.
Andrew.
|