Moment of Inertia of rectangular plate

[shizzle]

How do i find using integrals, I of a rectangular plate with sides a and b with respect to side b?

I know i have to use the equation I = integral of (a^2 dm) and i know
density = m/ab but I'm having trouble figuring out the mass element. Can someone tell me what the mass element is? and walk me through how to find I from that point?

 

[krab]

It's not integral of a^2 dm; it is integral of r^2 dm, where r goes from 0 to a. dm is the mass of the increment between r and r+dr.

 

[shizzle]

Thanks, you're right. Its r^2 dm. I'm still not sure what dm will be in this case though. I'm thinking...rho*a*db? or is it rho*a*da? I'm really confused.

 

[dextercioby]

Thanks, you're right. Its r^2 dm. I'm still not sure what dm will be in this case though. I'm thinking...rho*a*db? or is it rho*a*da? I'm really confused.

If u're calculating the moment of inertia wrt to side "b",then u should apply the definiton.By "rho" i assume u meant superficial mass density of the plate which is obviously [itex] \frac{m}{ab} [/tex].Find the mass of an element of surface of rectangular shape which has the width "dr" and the length "b".Then integrate the final result (after making all mutiplications) from 0 to "a".

Daniel.

EDIT:I maybe wrong,though.I haven't done such calculations in years.

 

[shizzle]

from following your suggestions, this is what i have so far:

I = integral (r^2) *dm
rho = m/ad
dm = rho*b*dr
therefore, I = r^2 * rho*b*dr

If you integrate this from 0 to a, you get
I = (m*a^2)/3b

Could someone please check this and let me know if this makes sense?

 

[dextercioby]

from following your suggestions, this is what i have so far:

I = integral (r^2) *dm
rho = m/ad
dm = rho*b*dr
therefore, I = r^2 * rho*b*dr

If you integrate this from 0 to a, you get
I = (m*a^2)/3b

Could someone please check this and let me know if this makes sense?

I underlined the wrong part of your work.

Daniel.

 

[shizzle]

The first underlined part was a typo and should read:
rho = m/ab. anyway, let's try this again:

I = integral (r^2) *dm
rho = m/ab
dm = rho*b*dr
therefore, dI = r^2 * rho*b*dr. plugging in rho we get
dI = (r^2 * m *b*dr )/ab

If you integrate this from 0 to a, you get
I = (m*a^2)/3

yay! i think i got it. Someone please confirm:)

 

[dextercioby]

The first underlined part was a typo and should read:
rho = m/ab. anyway, let's try this again:
I = integral (r^2) *dm
rho = m/ab
dm = rho*b*dr
therefore, dI = r^2 * rho*b*dr. plugging in rho we get
dI = (r^2 * m *b*dr )/ab
If you integrate this from 0 to a, you get
I = (m*a^2)/3
yay! i think i got it. Someone please confirm:)

Yap,it's okay.This time. It's quite curious it doesn not depend on both sides of the rectangle,right??But that's not weird,because you've chosed as rotation axis one of the sides of the rectangle.So that side (the length of it) does not enter the final result.If your rotation axis would have been different from any of the rectangle's sides,both sides would have made it through the final expression.U can convince of that by computing the MI for an axis which is perpendicular on the plane of the rectangle and passes thrugh any point of the rectangle.Both "a" and "b" will be in the final result.

Daniel.

 

[shizzle]

So for example, let's say that i wish to calculate the MI for an axis perpendicular to plane of rectangle and passes through center of mass, this is how i'd go about it:

Using perpendicular axis theorem, Iz = Ix + Iy so
Iz = [m (a^2 + b^2)]/3

is this right?

What would happen then if the axis was still perpendicular to plane but passing through a corner?

 

[dextercioby]

So for example, let's say that i wish to calculate the MI for an axis perpendicular to plane of rectangle and passes through center of mass, this is how i'd go about it:
Using perpendicular axis theorem, Iz = Ix + Iy so
Iz = [m (a^2 + b^2)]/3
is this right?
What would happen then if the axis was still perpendicular to plane but passing through a corner?

No,it's wrong.Try to apply the definiton of the moment of inertia.Imagine an infinitesimal rectangle of sides 'dx' and 'dy' at a square distance of 'x^{2}+y^{2}' from the center.The center of the rectangle is the meeting point of the axis system.

Daniel.

 

[shizzle]

Isn't the point of the parallel axis theorem and perpendicular axis theorem to make such calculations simpler? Do you know how we could apply the 2 theorems to solve this problem?

 

[dextercioby]

Isn't the point of the parallel axis theorem and perpendicular axis theorem to make such calculations simpler? Do you know how we could apply the 2 theorems to solve this problem?

I think you mean Steiner's theorem.In this problem the two axis of rotation are not coplanar.So you cannot apply the theorem.Take my advice and do the calculations with the definition.

Daniel.

 

[shizzle]

Okay, let's see...:)

I = r^2 dm --according to definition:)
rho = m/ab
dm = rho * da*db (since x and y are really a and b in this case--right?)
dI = r^2 * rho* da*db
dI = (r^2 * m * da* db )/ab

i will then have to do a double integration right? over limits 0 to a/2 and 0 to b/2 respectively?.. hmm..

 

[dextercioby]

Okay, let's see...:)
I = r^2 dm --according to definition:)
rho = m/ab
dm = rho * da*db (since x and y are really a and b in this case--right?)
dI = r^2 * rho* da*db
dI = (r^2 * m * da* db )/ab
i will then have to do a double integration right? over limits 0 to a/2 and 0 to b/2 respectively?.. hmm..

1.Put "x" and "y" back.It avoids confunsion.
2.The limits of integration are wrong.You're computing only for half of the rectangle.Think again.
3.Yes,double integration.

Daniel.

 

[dextercioby]

I = r^2 dm
rho = m/ab
dm = rho * dx*dy
dI = (x^2 + y^2) * rho* dx*dy
i would then integrate this twice over limits -a/2 to a/2 and -b/2 to b/2 (is this stupid? - it makes sense to melso, the reason u asked me to replace x and ys is because we integrate over limits involving a and b so x and ys get replaced anyway. right?

Yes it is correct to pick those integration limits.
Yes,of course they get replaced,since it is a definite integral with limits.

Daniel.

 

[shizzle]

assuming this is right, i have no clue how we'd find the MI when the axis is perpendicular to plane but passing through a corner instead of cm?

 

[dextercioby]

assuming this is right, i have no clue how we'd find the MI when the axis is perpendicular to plane but passing through a corner instead of cm?

It is right.Use Steiner's theorem,knowing in this case that the 2 axis of rotation are coplanar (are actually parallel to each other).The answer is easy to get.

Daniel.

 

[shizzle]

I = Icm + mr^2
I = (M(a^2 + a^2))/12 + m (a^2 +b^2) ---I'm b=not sure what r to use.
I = the sum once i know what r is

 

[dextercioby]

I = Icm + mr^2
I = (M(a^2 + a^2))/12 + m (a^2 +b^2) ---I'm b=not sure what r to use.
I = the sum once i know what r is

The distance between one corner of the rectangle and the center?

Daniel.

 

[shizzle]

The distance between one corner of the rectangle and the center is going to be
(a^2 + b^2 )/4 right? so if i plug this into the equation as my 2, that shd work.

I'm using pythagorean theorem to find this r.

 

[darsh_guy]

so it is by considering -just for instance -that we are working with 4 rods ...
each of which is at distance dr as r changes from 0 to a/2 in two cases and from 0 to b/2 in the other two cases...
by adding them we get :
I=$$\int$$r^2.dm
r²=x²+y²

therefore I=$$\int$$(x²+y²)dm

I_z=I_y+I_x

((all that as r changes from 0 to a/2 in two cases and from 0 to b/2 in the other two cases))
we shall take one case from each one and multiply the final result by 2:
so nw we get:
I_x=Mb²/12
I_y=Ma²/12
by adding them you get what you want

Mostafa Nabil