introduction
we have here a lemniscate[1] with the equation in polar form as,
r2=a2cos2θ(1)
the graph of which at a=1 looks like:
pretty.
setting up coordinate and integration limits
i will consider the polar coordinates. therefore let’s clear out some basic transformations.
x=rcosθ
y=rsinθ
da=rdrdθ
moving on i shall set the limit of integration. pretty obviously r shall go from 0 to acos2θ√. for θ however we need to define a domain where the area will converge. setting r=0, we shall get from equation 1, θ=π4+nπ2 where n∈z. now from the graph we can see that the function is even, and the area converges between θ from −π/4 to π/4 to get covering one side. therefore we can integrate θ from −π/4 to π/4 to get one half and then multiply by 2.
calculations for the area
we need the area to calculate the density. therefore we have,
a=2∫π/4θ=−π/4∫acos2θ√r=0rdrdθ
=∫π/4−π/4a2cos2θdθ
=a2sin2θ2∣∣∣π/4−π/4
a=a2
calculation for ixx
since i don’t know which axis you specifically want, i will calculate over all the axis as well as the production of inertia and then get its inertia tensor.
i have taking density as σ,
ixx=2σ∫π/4θ=−π/4∫acos2θ√r=0r2sin2θrdrdθ
=σ2∫π/4−π/4a4cos22θsin2θdθ
=σ∫π/40a4cos22θsin2θdθ
=σ2∫π/40a4cos22θ(1−cos2θ)dθ
=σ2{∫π/40a4cos22θdθ−∫π/40a4cos32θdθ}
putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,
=σa44{∫π/20cos2ϕdϕ−∫π/20cos3ϕdϕ}
now using beta function[2] we have here, p=1/2 and q=3/2 for the first integral and p=1/2 and q=2 for the second integral. therefore we get using gamma-beta relation,
=σa44{π4−23}
expanding σ,
ixx=ma248(3π−8)
calculation for iyy
iyy=2σ∫π/4θ=−π/4∫acos2θ√r=0r2cos2θrdrdθ
=σ2∫π/4−π/4a4cos22θcos2θdθ
=σ∫π/40a4cos22θcos2θdθ
=σ2∫π/40a4cos22θ(cos2θ+1)dθ
=σ2{∫π/40a4cos32θdθ+∫π/40a4cos22θdθ}
putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,
=σa44{∫π/20cos3ϕdϕ+∫π/20cos2ϕdϕ}
now using beta function we have here, p=1/2 and q=2 for the first integral and p=1/2 and q=3/2 for the second integral. therefore we get using gamma-beta relation,
=σa44{23+π4}
expanding σ,
iyy=ma248(8+3π)
calculation for izz
since the object considered is 2d we can use the perpendicular axis theorem to get izz. we have,
izz=ixx+iyy
izz=mπa28
calculation for ixy and iyx
moving on i shall calculate products of inertia. we have here,
ixy=iyx=2σ∫π/4θ=−π/4∫acos2θ√r=0r2sinθcosθrdrdθ
=σ∫π/4−π/4∫acos2θ√0r2sin2θrdrdθ
=σ4∫π/4−π/4a4cos22θsin2θdθ
=σ2∫π/40a4cos22θsin2θdθ
putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,
=σa44∫π/20cos2ϕsinϕdϕ
now using beta function we have here, p=1 and q=3/2 for the first integral and p=1/2 and q=3/2 for the second integral. therefore we get using gamma-beta relation,
=σa412
expanding density,
ixy=iyx=ma212
calculation for ixz and izx
since the object is 2d, z=0, hence,
ixz=izx=0
calculation for iyz and izy
again since the object is 2d, z=0, hence,
iyz=izy=0
inertia tensor of a lemniscate
finally as promised, here is your inertia tensor:
i↔=⎡⎣⎢⎢⎢ma248(3π−8)ma2120ma212ma248(3π+8)000mπa28⎤⎦⎥⎥⎥
i was thinking of calculating the ellipsoid of inertia as well, however it’s quite late night and i am kind of sleepy, so maybe later. also since these calculations were pesky if you do find any mistake, comment it down or suggest an edit.
step-by-step explanation:
