# Thread: Basic moment of inertia question

1. ## Basic moment of inertia question

26-5-7

Find moment of inertia with respect to x axis.
x = -1, x = 1, y = 0, y = 1

Attempt:
Ix = 2pi(k) (int) (1)y^3 dy
boundaries 1 and 0
= 2pi(k)1/4(y^4)
= (2pi(k))/4 = Ix

m = 2pi(k) (int) xy dy
= 2pi(k) (int) (1)y dy
= 2pi(k)1/2(y^2)
= pi(k)(y^2) = pi(k)

R^2 (subscript x) = [(2pi(k))/4]/[(pi(k)]
K = 1 (from appearances of book example questions)
R^2 (subscript x) = 0.5
R (subscript x) = sqrt(0.5)

the book answer is 2/3k though. What did I do wrong? thanks

2. ## Re: Basic moment of inertia question

Find the moment of inertial of what with respect to the x-axis? The square bounded by x= -1, x= 1, y= 0, and y= 1? And what is "k"? The density?

3. ## Re: Basic moment of inertia question

Density is K pretty sure. K or k. So it has been used in the examples as such, and the example questions calculate K as 1. However the answer includes K for this question, so apparently they didn't just plug 1 into it. You could assume easily enough that this question considers the area of x = -1, x = 1, y = 0 and y= 1 as the area of interest.

4. ## Re: Basic moment of inertia question

So your problem looks like this?

Its not a square but anyway,
Consider a tiny strip of material with width dy a distance y away from the x axis

Inertia is given by $I= mr^2$

For this strip of material the mass is dm and the distance from the axis 'r' is y. Let the inertia for a tiny strip be I' while the inertia for the whole body is I
$I'= y^2dm$

As the length of the strip is 2 (from -1 to 1) the area of this strip dA is given by $dA= 2dy$

The mass per unit area, the density k is equal to $\frac{dm}{dA}$
So $dm= k(dA)$

$dm= k(2dy)$

Use the expression of dm in the formula for the inertia of the piece

$I'= y^2k(2dy)$

Sum up the inertia of all the tiny strips along the material from y=0 to y=1

$\int I'= \int_0^1 y^2k(2dy)$