Moment of inertiaFind the moment of inertia of the plate

[Punchlinegirl]

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?

 

[dextercioby]

I have good reasons to believe that

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

The area of the plate is

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

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

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

Daniel.

 

[Punchlinegirl]

I'm sorry but I have no idea what you've done in this problem...

 

[dextercioby]

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.