Calculating Moment of Inertia for Simple Function

In summary, a parabolic plate with a height of 1.96 m and a mass of 7.27 kg has a lower boundary defined by y=.35x^2. To find the moment of inertia (in kgm2) about the y-axis, the plate's area and density must be calculated using a double integral, with one integral already given. Once the area and density are found, the moment of inertia can be calculated using the formula I=\int((\rho)x^2dA).
  • #1
americanforest
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http://loncapa2.physics.sc.edu/res/msu/physicslib/msuphysicslib/20_Rot2_E_Trq_Accel/graphics/prob02_paraplate.gif

A uniform plate of height H = 1.96 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: [tex]y=.35x^2[/tex]. The plate has a mass of 7.27 kg. Find the moment of inertia of the plate (in kgm2) about the y-axis.

I have absolutely no clue where to start... I know [tex]I=\int(r^2)dm[/tex] but I'm not sure how I can relate this to this problem.
 
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  • #2
You would do well to start by finding the area of the plate. There is a double integral involved, and I think most calculus books do this problem two ways. Either way, one of the integrals is usually done for you by the way the problem is set up in the first place. Either you divide the plate into vertical strips of varying length L(x)(which is equivalent to doing the y integral in your head) and then integrating over x, or you divide the plate into horizontal strips of varying length L(y) (which is equivalent to doing the x integral in your head) and then integrating over y. For this problem, you will want to use the vertical strip approach because when it comes time to do the integral for the moment of inertia, vertical strips have only one r that applies to the whole strip.
 
  • #3
I started off by finding the value of x when y=H, which is 2.366 and -2.366. Then we know Area = [tex]\int(x^2dx)[/tex] between those values of x which is 3.09 m^2. I know [tex]\frac{dm}{dA}=\rho=\frac{M}{A}[/tex]. Therefore, [tex]dm=\frac{M}{A}dA[/tex]. We are given that [tex]\frac{M}{A}=2.353[/tex]. This leaves us with [tex]I=\int 2.353x^2dA=\int 2.353x^2dxdy[/tex]. I know how to do multiple integrals. Is that what is required here. I doubt it since this class is just a basic Physics class with Calc 1 and 2 prerequisites, no Multivariable stuff.

What do I do next?
 
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  • #4
americanforest said:
I started off by finding the value of x when y=H, which is 2.366 and -2.366. This looks good.
Then we know Area = [tex]\int(x^2dx)[/tex] This does not look good. You should be integrating dA = (H - y)dx here.
between those values of x which is 3.09 m^2. The area is more than half of 2*2.366*1.96, so I know this is not correct I know [tex]\frac{dm}{dA}=\rho=\frac{M}{A}[/tex]. Therefore, [tex]dm=\frac{M}{A}dA[/tex]. We are given that [tex]\frac{M}{A}=2.353[/tex]. This leaves us with [tex]I=\int 2.353x^2dA=\int 2.353x^2dxdy[/tex]. I know how to do multiple integrals. Is that what is required here. I doubt it since this class is just a basic Physics class with Calc 1 and 2 prerequisites, no Multivariable stuff.

What do I do next?
See annotations. Fix the area integral. Once you get that and find the density correctly you will need to do the same type of integral but with the integrand multiplied by x^2 as you have done.
 
  • #5
OK, I've got correct area and density (I think) [tex]A=6.184[/tex] [tex]\rho=1.1756[/tex]. Now I'm stuck on the next part. I know [tex] \int((\rho)x^2dA)[/tex] is the equation I want, right? But how do I go from here? Is this a double integral? I tried just [tex]\int((\rho)x^2dx)[/tex] but didn't get the right answer. Where do I include the y? I'm lost, help!:confused:
 
  • #6
americanforest said:
OK, I've got correct area and density (I think) [tex]A=6.184[/tex] [tex]\rho=1.1756[/tex]. Now I'm stuck on the next part. I know [tex] \int((\rho)x^2dA)[/tex] is the equation I want, right? But how do I go from here? Is this a double integral? I tried just [tex]\int((\rho)x^2dx)[/tex] but didn't get the right answer. Where do I include the y? I'm lost, help!:confused:
Assuming you used dA = [H - y(x)]dx for the area integral, you now do everything the same except this time you integrate x^2dA = x^2[H-y(x)]dx. ρ is a constant, so it is just a scale factor multiplying the integral.
 
  • #7
Thank you so much, this problem had me stumped. I'm pretty sure I got it now.
 

Related to Calculating Moment of Inertia for Simple Function

1. What is moment of inertia?

Moment of inertia, also known as rotational inertia, is a measure of an object's resistance to changes in its rotational motion. It is calculated by considering the distribution of mass and the distance of mass from the axis of rotation.

2. How is moment of inertia calculated for simple functions?

For simple functions such as point masses, rods, and disks, moment of inertia is calculated using the formula I = mr^2, where m is the mass of the object and r is the distance of the mass from the axis of rotation.

3. What units are used to measure moment of inertia?

Moment of inertia is typically measured in units of kilogram-meters squared (kg•m^2) in the International System of Units (SI).

4. How does the distribution of mass affect moment of inertia?

The distribution of mass has a significant impact on moment of inertia. Objects with the majority of their mass located further from the axis of rotation will have a larger moment of inertia compared to objects with the same mass but a more compact distribution of mass.

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

Moment of inertia is an important concept in engineering and physics as it helps determine the behavior of objects in rotational motion. It is used in designing machines and structures such as flywheels, gyroscopes, and bridges. It is also used in sports equipment design, such as in the construction of golf clubs and tennis rackets.

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