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Moment of Inertia of rectangular plate 
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#1
Dec1704, 02:57 PM

P: 21

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? 


#2
Dec1704, 03:09 PM

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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.



#3
Dec1704, 03:15 PM

P: 21

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.



#4
Dec1704, 03:25 PM

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Moment of Inertia of rectangular plate
Daniel. EDIT:I maybe wrong,though.I haven't done such calculations in years. 


#5
Dec1704, 04:07 PM

P: 21

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? 


#6
Dec1704, 04:15 PM

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Daniel. 


#7
Dec1704, 04:25 PM

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The first underlined part was a typo and should read:
rho = m/ab. anyway, lets 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:) 


#8
Dec1704, 04:32 PM

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Daniel. 


#9
Dec1704, 07:40 PM

P: 21

So for example, lets 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? 


#10
Dec1704, 08:01 PM

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Daniel. 


#11
Dec1704, 08:11 PM

P: 21

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?



#12
Dec1704, 08:17 PM

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Daniel. 


#13
Dec1704, 08:40 PM

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Okay, lets 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 caseright?) 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.. 


#14
Dec1704, 08:44 PM

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2.The limits of integration are wrong.You're computing only for half of the rectangle.Think again. 3.Yes,double integration. Daniel. 


#15
Dec1704, 09:10 PM

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Yes,of course they get replaced,since it is a definite integral with limits. Daniel. 


#16
Dec1704, 09:19 PM

P: 21

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?



#17
Dec1704, 09:27 PM

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Daniel. 


#18
Dec1704, 10:06 PM

P: 21

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 


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