Moment of inertia and integration

In summary, the conversation discusses how to find the moment of inertia of a uniform rectangular plate about an axis that is perpendicular to the plate and passes through one corner. The two approaches suggested are to either use integration or the parallel axis theorem. The correct answer is (1/3)*m*(a^2+b^2) and the conversation includes a step-by-step solution using integration.
  • #1
nns91
301
1

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
 
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  • #2
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
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
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
Thanks. A dumb question. How do I solve double integration ??
 
  • #6
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
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
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
h is the distance from centre to corner, not a/2
 
  • #10
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
nns91 said:
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
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
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
LowlyPion said:
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
davieddy said:
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.
 

Related to Moment of inertia and integration

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is calculated by integrating the mass of each individual particle in the object with respect to its distance from the axis of rotation.

2. How is moment of inertia related to an object's shape?

The moment of inertia of an object depends on its shape and mass distribution. Objects with more mass concentrated towards the axis of rotation have a smaller moment of inertia compared to those with mass distributed farther from the axis.

3. What is the difference between moment of inertia and mass?

Moment of inertia is a measure of an object's resistance to rotational motion, while mass is a measure of an object's resistance to linear motion. They are not interchangeable and have different units of measurement.

4. How is integration used to calculate moment of inertia?

Integration is used to calculate moment of inertia by breaking an object into infinitesimal particles and summing their individual moments of inertia. This is represented by the integral of the mass of each particle multiplied by its distance squared from the axis of rotation.

5. How can moment of inertia be used in real-world applications?

Moment of inertia is important in many engineering and physics applications, such as designing rotating machinery, analyzing the stability of structures, and predicting the motion of objects in rotational motion. It is also used in sports science to understand the movements and balance of athletes.

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