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dhernandez081
31.07.2019 •
Physics
Find the moments of inertia ix, iy, i0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the density at any point is proportional to the square of its distance from the origin. (assume that the coefficient of proportionality is k.)
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Ответ:
I(x) = 1444×k ×![{\pi}](/tpl/images/1301/7501/bd3e5.png)
I(y) = 1444×k ×![{\pi}](/tpl/images/1301/7501/bd3e5.png)
I(o) = 3888×k ×
Explanation:
Given data
function = x^2 + y^2 ≤ 36
function = x^2 + y^2 ≤ 6^2
to find out
the moments of inertia Ix, Iy, Io
solution
first we consider the polar coordinate (a,θ)
and polar is directly proportional to a²
so p = k × a²
so that
x = a cosθ
y = a sinθ
dA = adθda
so
I(x) = ∫y²pdA
take limit 0 to 6 for a and o to
for θ
I(x) =
y²p dA
I(x) =
(a sinθ)²(k × a²) adθda
I(x) = k
da ×
(sin²θ)dθ
I(x) = k
da ×
(1-cos2θ)/2 dθ
I(x) = k
× ![{θ/2 - sin2θ/4}^{\pi /2}_0](/tpl/images/1301/7501/1b24b.png)
I(x) = k ×
× ( ![{\pi /4} - sin\pi /4)](/tpl/images/1301/7501/84fb5.png)
I(x) = k ×
× ![{\pi /4}](/tpl/images/1301/7501/59c6b.png)
I(x) = 1444×k ×
.....................1
and we can say I(x) = I(y) by the symmetry rule
and here I(o) will be I(x) + I(y) i.e
I(o) = 2 × 1444×k ×![{\pi}](/tpl/images/1301/7501/bd3e5.png)
I(o) = 3888×k ×
......................2
Ответ:
The answer is D the rising of warm air pushing down cool air.