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?
Find moment of inertia with respect to x axis.
x = -1, x = 1, y = 0, y = 1
Ix = 2pi(k) (int) (1)y^3 dy
boundaries 1 and 0
= (2pi(k))/4 = Ix
m = 2pi(k) (int) xy dy
= 2pi(k) (int) (1)y dy
= 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
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.
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
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
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