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

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?

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.

Re: Basic moment of inertia question

So your problem looks like this?

http://puu.sh/2j1FP

Its not a square but anyway,

Consider a tiny strip of material with width dy a distance y away from the x axis

http://puu.sh/2j1JU

Inertia is given by

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

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

The mass per unit area, the density k is equal to

So

From your expression of area

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

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