# Moment of inertia of this shape about the x-axis

#### jisbon

Homework Statement
Calculate moment of inertia of a 2d plane rotating about x axis. Mass per unit area of plate = $1.4g/cm^2$ , total mass = 25.2g
Homework Equations
$I=\int y^2 dm$

I've attempted this question, but the answer seems to be incorrect. Here's my workings:
$I=\int y^2 dm$ - standard equation
$dM = \mu * dy * x$ - take small slice and find mass of it
$x = 4y-16$ - convert equation in terms of x to sub in later
$dM = \mu * dy * 4y-16$
$I=\int y^2 \mu * dy * 4y-16$
$I=\mu\int y^2(4y-16) dy$
$I=\mu\int 4y^3-16y^2 dy$
$I=\mu \int_{0}^{3} 4y^3-16y^2 dy$ = 12.6, which is not the answer.
Any clues why?

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#### Delta2

Homework Helper
Gold Member
Something wrong in your final calculation of the integral, I found it to be
$\int_0^3 (4y^3-16y^2 )dy=-63$ and if $\mu=1.4$ then the final value should be -88.2. The minus sign can be explained (because you took $x=4y-16$ while i believe it is $x=16-4y$).

#### jisbon

Something wrong in your final calculation of the integral, I found it to be
$\int_0^3 (4y^3-16y^2 )dy=-63$ and if $\mu=1.4$ then the final value should be -88.2. The minus sign can be explained (because you took $x=4y-16$ while i believe it is $x=16-4y$).
The answer seems to be 37.8g/cm^2 though :/ Any ideas on this?

#### Delta2

Homework Helper
Gold Member
At the moment I cant spot any other mistake. Are we sure that the equation of y(x) as well as the boundaries of integration (from y=0 to y=3 or from x=4 to x=16) are correct?

#### jisbon

At the moment I cant spot any other mistake. Are we sure that the equation of y(x) as well as the boundaries of integration (from y=0 to y=3 or from x=4 to x=16) are correct?
Equation of y is y= -1/4x+4 , so it should be correct :/

#### Delta2

Homework Helper
Gold Member
I think I spotted a mistake, the mass element should be $dM=\mu(x-4)dy$ so the final integral should be $\mu\int_0^3 (12y^2-4y^3)dy$

Last edited:

"Moment of inertia of this shape about the x-axis"

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