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Moment of inertia and integration

  • Thread starter nns91
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  • #1
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Homework Statement



A uniform rectangular plate has mass m and sides a and b.

a) Show by integration that the moment of inertia of the plate about an axis is perpendicular to the plate and passes through one corner is (1/3)*m*(a^2+b^2)

Homework Equations



I= integral( r^2, dm)

The Attempt at a Solution



do I account this problem as 3 dimensions or 2 dimensions ??

What should be r in this case since I have a and b
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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You could approach it in one of two ways.

1)find the moment for a rectangular plate a x b about the center and then use the parallel axis theorem to rotate it about the corner on a line through the center.

2)set up an integral along the diagonal from the opposite corner and relate Δ mass as little arcs that integrate over that distance.

Seems like 1) would be an easier approach.

Since it is uniform in thickness, then just treat it as a 2 dimensional object.
 
  • #3
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I can use the 1st approach for part 2 which they ask for moment of inertia about the axis going through the center o of mass.

If I use the 2nd approach, I will integrate (r^2,dm) from 0 to b (let say I choose b as the limit).

how can I represent r^2 as ? Is that a^2 ?
 
  • #4
LowlyPion
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Then do the second part first then.

To find the moment of a rectangle isn't it simply something like:

I = ∑m*r² = ρ*∫∫r²*dm = ρ*∫∫(x² +y²)*dx*dy

where ρ = m/(h*w) and the limits of your double integration are |½h-½h and |½w-½w

Then apply the || axis theorem along the diagonal at 1/2 the length of the diagonal away?
 
  • #5
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Thanks. A dumb question. How do I solve double integration ??
 
  • #6
LowlyPion
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Choose one variable - either one - to integrate keeping the other as a constant. Then you are left with a single integral that you can integrate treating the first variable as a constant.
 
  • #7
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Consider any plane shape and scale it up by two, keeping the mass
per unit area the same. Each "element" has quadrupled in mass and
doubled its distance from the C of M. It follows that the moment of
inertia about an axis through the C of M normal to the plane has gone up
by a factor of 16.

For the a x b rectangle, let I be the moment of inertia about the C of M
and J that about a corner.

The || axis theorem gives J = I + m(a^2 + b^2)/4

Now consider the double sized rectangle formed from placing 4 of the
originals together. We deduce that

16I = 4J = 4I + m(a^2 + b^2)

David
 
  • #8
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Somehow I cannot get the answer of (1/3)m(a^ 2+b^2).

I use the double integration technique and get

Icom=(1/12)*m*a^2*b^2

Then when I substitute in I= Icom+ m*h^2

I use h=a/2

I cannot get the answer. Did I do wrong somewhere ??
 
  • #9
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h is the distance from centre to corner, not a/2
 
  • #10
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Can I assume either it can be (1/2)a since stack the small dm along side a ??

Is not the center of mass is 1/2 from each corner ?
 
  • #11
LowlyPion
Homework Helper
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Somehow I cannot get the answer of (1/3)m(a^ 2+b^2).

I use the double integration technique and get

Icom=(1/12)*m*a^2*b^2

Then when I substitute in I= Icom+ m*h^2

I use h=a/2

I cannot get the answer. Did I do wrong somewhere ??
Are you sure you didn't get 1/12*(a2 + b2) ?

Because armed with that then and the || axis you can say the moment from the corner is Imiddle + m*(a2 + b2)/4

The hypotenuse2 being (a/2)2 + (b/2)2 from the center.
 
  • #12
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I am sure I did not get (1/12)*m*(x^2+y^2).

I got x^2*y^2 .

So I got out from the integration as (x^3/3)*(y^3/3)* m/(x*y).

So plug in the limit I got (x^3/12)*(y^3/12)*m/(x*y)

Then I cancel x and y and got x^2*y^2.

How did you get the "+" ??
 
  • #13
LowlyPion
Homework Helper
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For :
I = ∑m*r² = ρ*∫∫r²*dm = ρ*∫∫(x² +y²)*dx*dy

I'd recheck your calculation, being sure to recognize ρ*h*w as being the mass of the rectangular object.

There is also the approach suggested by the elegant thinking in this post by davieddy:
https://www.physicsforums.com/showpost.php?p=2057709&postcount=7
...that gets the same result.
 
  • #14
181
0
For :
I = ∑m*r² = ρ*∫∫r²*dm = ρ*∫∫(x² +y²)*dx*dy

I'd recheck your calculation, being sure to recognize ρ*h*w as being the mass of the rectangular object.

There is also the approach suggested by the elegant thinking in this post by davieddy:
https://www.physicsforums.com/showpost.php?p=2057709&postcount=7
...that gets the same result.
Thank you:)
I felt justified in presenting that solution because the question specifically
asked for integration.

By the sound of it, the original poster could use the practice with integrating.
Maybe Pythagoras needs some brushing up too :)

I = ∑m*r² = ∫r²dm = m/(ab)*∫(∫(x² +y²)dx)dy

integral limits 0 to a and 0 to b

David
 
Last edited:
  • #15
LowlyPion
Homework Helper
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Thank you:)
I felt justified in presenting that solution because the question specifically asked for integration.
I enjoyed it anyway because of the less traditional approach of deducing it from the || axis.
 

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