Consider the lamina ( with uniform density .01m) having the shape of the region Q in the xy-plane: 1≤x^2+y^2≤4. The density at each point P varies directly with the distance from P to the origin, with density at the point P(2,0,0) being 10kg/m3.Find the polar moment of inertia Io
To find the polar moment of inertia Io for the given lamina, we need to integrate the density function over the entire region Q.
The given shape of the lamina is the region bounded by two circles centered at the origin with radii 1 and 2.
To set up the integral, we need to express the density as a function of x and y coordinates. The problem states that the density varies directly with the distance from P to the origin. The distance from any point (x, y) to the origin is given by r = sqrt(x^2 + y^2).
Let's express the density function as ρ(x, y). Since the problem states that the density at (2, 0, 0) is 10 kg/m^3, we can write:
ρ(x, y) = k * sqrt(x^2 + y^2)
where k is a constant of proportionality.
To solve for k, we can use the given density at (2, 0, 0):
10 = k * sqrt((2^2) + (0^2))
10 = k * sqrt(4)
10 = 2k
k = 5
Now we can express the density function as:
ρ(x, y) = 5 * sqrt(x^2 + y^2)
To find the polar moment of inertia Io, we need to integrate the density function squared over the region Q.
Io = ∬(Q) ρ^2(x, y) dA
Here, dA represents the differential area element in polar coordinates.
To set up the integral using polar coordinates, we substitute x = r * cos(θ) and y = r * sin(θ). The limits of integration for r will be from 1 to 2, and for θ, it will be from 0 to 2π.
Io = ∫(0 to 2π) ∫(1 to 2) (5 * sqrt(r^2))^2 * r dr dθ
Io = ∫(0 to 2π) ∫(1 to 2) 25r^3 dr dθ
Now we can integrate this expression to find the polar moment of inertia Io.