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

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
Relevant 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?

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##).

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?

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##

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