Math question (calculus)
Hi.
I have a rectanglar area with one corner rwesting at the origin and the other corner resting at some point (A, B). The area weighs one unit of mass per one square unit of area. What is it's rotational inertia about some point (x,y), which may or may not lie within the rectangle?
The "insantaneous" rotational inertia of a point is proportional to the distance between the point in question and the chosen (x,y). I think this question involves this integral
the sum of (as u varies from 0 to A) the sum of (as v varies from 0 to B) sqrt((u-a)^2 + (v-b)^2) dv du
which I have no idea how to do
> The "insantaneous" rotational inertia of a point is
> proportional to the distance between the point in
> question and the chosen (x,y). I think this question
> involves this integral
>
> the sum of (as u varies from 0 to A) the sum of (as v
> varies from 0 to B) sqrt((u-a)^2 + (v-b)^2) dv du
>
> which I have no idea how to do
did you mean "sqrt((u-x)^2 + (v-y)^2) dv du"
respectively "and the chosen (a,b)" ?!
try
http://integrals.wolfram.com/
first solving the inner integral (x == v) then solving the outer integral using the solution from the first run using x == u.
the solution is not too complicated... see yourself!
> I have a rectanglar area with one corner rwesting at
> the origin and the other corner resting at some point
> (A, B). The area weighs one unit of mass per one
> square unit of area. What is it's rotational inertia
> about some point (x,y), which may or may not lie
> within the rectangle?
>
> The "insantaneous" rotational inertia of a point is
> proportional to the distance between the point in
> question and the chosen (x,y). I think this question
> involves this integral
>
> the sum of (as u varies from 0 to A) the sum of (as v
> varies from 0 to B) sqrt((u-a)^2 + (v-b)^2) dv du
I think you are wroung about the formula. The rotational inertia of a mass point is I = m * r^2. This will remove the square root in your integral. But you can do even better by first computing the rotational inertia with (x,y) at the center of the rectangle. You can also simplify your integral by dividing the rectangle into stripes and using Steiner's theorem. Then, for an arbitrary point (x,y) you can again use this theorem.
I don't know if you want to do this the long way, after all the rotational inertia of a rectangle can be looked up in tables. But I don't want to spoil your efforts, so I won't tell you (the quite simple) solution (for now :-).
> The rotational inertia of a mass point is I = m * r^2.
Are you sure? (.... google ...)
http://hyperphysics.phy-astr.gsu.edu/hbase/inecon.html
Holy cremona! So it is! Well, that makes it a whole 'nother ball game.
**** ...
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax
This completely stuffs up what I was hoping to do. I had thought that by taking a suitcase or a rock and wobbling it around several axes, one could map out the distibution of mass inside it. If it had been I=mr, then I think it would have worked because the moments of a distributed mass would not sum neatly. But the parallel axis theorem basically tells me that there's no way to distingish a disk from a hoop by wiggling it (if you don't know the density).
Oh well. There goes that idea.
> > The rotational inertia of a mass point is I = m *
> r^2.
Not quite. That would be for a point mass. You can restore
this formula if you interpret r as the radius of inertia
(the problem is equivalent to rotating a point mass if
placed at the right spot).
> Are you sure? (.... google ...)
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/inecon.html
I see you've found a reference, so there's no need for
me to show you the formula.
> Holy cremona! So it is! Well, that makes it a whole
> 'nother ball game.
>
> **** ...
> http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#p
> x
>
> This completely stuffs up what I was hoping to do. I
> had thought that by taking a suitcase or a rock and
> wobbling it around several axes, one could map out the
> distibution of mass inside it. If it had been I=mr,
> then I think it would have worked because the moments
> of a distributed mass would not sum neatly. But the
> parallel axis theorem basically tells me that there's
> no way to distingish a disk from a hoop by wiggling it
> (if you don't know the density).
You're right. Every solid has an inertial ellipsoid that
describes the rotational properties. There's no way to
distinguish among objects that have the same ellipsoid
except by it's interaction with other bodies--you could
map the gravitational field, but that's just about
impossible in practice because the field is so weak.
> ... if you interpret r as the radius of inertia
> (the problem is equivalent to rotating a point mass if
> placed at the right spot).
I now remember the correct term: the _radius of gyration_
is the effective radius. Usually it depends on the axis
about which the object rotates. The principal axes of the
inertial ellipsoid determine the ellipsoid itself. For
rotation about the centre of mass, the radius of gyration
is just the distance from the centroid to the ellipsoid.
The moments of inertial of a rectangular plate of with sides of length (b,c) lying in the yz-plane with its center at (0, 0) has these moments of inertia:
Ixx = m*(b*b + c*c)/12
Iyy = m*c*c/12
Izz = m*b*b/12
If you want to know its moment of inertia about an axis parallel to a principal axis at a distance d away from it, you'll need the parallel axis theorem:
I = I' + m*d*d
You can look this up in any statics or dynamics book. - MOD
"...The moments of inertial of a rectangular plate of with sides of length (b,c) lying in the yz-plane with its center at (0, 0) has these moments of inertia..."
Pretty poor English there. I must have been in too much of a hurry or had low blood sugar. I meant to write:
"...A thin rectangular plate with sides of length (b,c) lying in the yz-plane with its center at (0, 0, 0) has these moments of inertia..."
The formulas are correct. I got them out of Beer & Johnston's "Statics and Dynamics". - MOD
