Moment of inertiaFind the moment of inertia of the plate

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Homework Help Overview

The discussion revolves around calculating the moment of inertia of a uniform plate shaped like a parabolic section, defined by the equation y=1.75x^2, with a specified height and mass. Participants are attempting to derive the correct moment of inertia about the y-axis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the moment of inertia using integration but questions the validity of their result. Another participant suggests a different integral approach involving the area of the plate and its projection. There is a request for clarification on the area calculation and the shape of the plate.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the area calculation, but there is no consensus on the correct method or interpretation of the problem yet.

Contextual Notes

Participants are working under the constraints of the problem as posed, including the specific shape of the plate and the need to find the area for the moment of inertia calculation. There is some confusion regarding the setup and the calculations involved.

Punchlinegirl
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A uniform plate of height H=1.60 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y=1.75x^2. The plate has a mass of 4.94 kg. Find the moment of inertia of the plate (in kgm^2) about the y-axis.

I integrated 1.75x^2 to get .583x^3. Then put 1.60 in for x and multiplied by the mass 4.94. I got 4.02 kgm^2 as my answer, which isn't right. Can someone help?
 
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I have good reasons to believe that

[tex]I_{y}=\frac{M}{S}\int_{0}^{1.6} dy\int_{0}^{\sqrt{\frac{y}{1.75}}} x^{2} \ dx[/tex]

The area of the plate is

[tex]S=S_{rectangle}-\int_{0}^{l_{x}} \left(1.75x^{2}\right) \ dx[/tex]

,where [itex]l_{x}[/itex] is the "x" coordinate of the point obtained by projecting the point of coordinates
[tex]\left(\sqrt{\frac{1.6}{1.75}},1.6\right)[/tex]
onto the Ox axis...

I hope u see which rectangle I'm talking about.

Daniel.
 
Last edited:
I'm sorry but I have no idea what you've done in this problem...
 
U've got a plate whose contour is made up of 2 right lines and an arch of a parabola...U have to find the plates' area which will be needed to evaluate the moment of inertia.

Make a drawing and see how u can compute that area...

Daniel.
 

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