Moment of Inertia of a non-uniform density paraboloid

[emgram769]

Homework Statement

Find the moment of inertia of a paraboloid f(x,y)=x^2+y^2 whose density function is ρ(r)=cr=dm/dv. use mass M and height H to express your answer

2. The attempt at a solution

I took the double integral ∫∫r^2 ρ r dr dθ to find the I of a single disk as a function of r

Beyond that I need some help.

 

[ehild]

Is it a three dimensional or a two-dimensional object?
Explain the notations. What is f(x,y)? What are r and theta?

ehild

 

[gneill]

Is it a three dimensional or a two-dimensional object?
Explain the notations. What is f(x,y)? What are r and theta?

ehild

It appears to be a standard notation. If f(x,y) is plotted on the z-axis then the object is the surface of revolution of a parabola (a cross section of the object in any plane containing the z-axis is a parabola).

I'd suggest switching to cylindrical coordinates and expressing r in terms of z (or h since the problem wants the results in terms of Height and Mass). Integrate over h. A mass element is then a ring of radius \sqrt{h} with the appropriate density for that radius.

 

[ehild]

It appears to be a standard notation.

Gneill, I know it! I asked the OP and wanted him to find it out.

ehild

 

[gneill]

Gneill, I know it! I asked the OP and wanted him to find it out.

ehild

My apologies. Sorry for putting my muddy footprints all over your nice clean thread

 

[emgram769]

Yes, it is standard notation. Sorry if that wasn't clear. I integrated with z=r^2, dz=2rdr

I found c to be (5M)/(2πr^5). Final result was ()MH, can anyone verify this for me?

∫∫r^2 cr r dr dθ =
c∫∫r^4 dr dθ =
(c2πr^5)/5
∫(c2πr^5)/5 2rdr =
(c4πr^7)/35

M=∫ρ dv
V=∫πr^2 dz = ∫πr^2 2r dr = (1/2)πr^4
dv=2πr^3dr
M = ∫2πcr^4dr = (2πcr^5)/5
c = 5M/(2πr^5)

Plugging in c and setting r^2 = H
(2/7)MH

 

[ehild]

You get both I and M by volume integral, integrating with respect to θ, r and z between appropriate bounds. I do not see the bounds of integration. Make them clear, please. I do not think your result for M is correct.


ehild

 

[emgram769]

I_disk = <br /> \int_0^{2\pi}\int_0^r r^2\rho\,r\,\mathrm{d}r\,\mathrm{d}\theta<br />

M = \int_0^V\rho\,\mathrm{d}v

Change of variable to r gives different bound in the next part

M = 2\pi\int_0^r\rho\,r^3\,\mathrm{d}r

which I think holds true with the relationship z=r²

 

[ehild]

I_disk = <br /> \int_0^{2\pi}\int_0^r r^2\rho\,r\,\mathrm{d}r\,\mathrm{d}\theta<br />

M = \int_0^V\rho\,\mathrm{d}v

Change of variable to r gives different bound in the next part

M = 2\pi\int_0^r\rho\,r^3\,\mathrm{d}r

which I think holds true with the relationship z=r²

No, it is not true. z=r² is true only for one of the boundary surfaces of the paraboloid, not for the whole volume.

You need the mass and the moment of inertia of a paraboloid - a three dimensional object, bounded by surfaces f(x,y)=x^2+y^2 and g(x,y)=H.
If you integrate with respect to r , the bound of the definite integral can not be r. It has to be a specified value.
If you use cylindrical system of coordinates, the volume element is dV=rdrdθdz and
M=\int_0 ^{r_{max}}\int_{z_1(r)}^{z_2(r)} \int_0 ^{2 \pi}{\rho r d\theta dz dr }

Find the proper bounds z₁, z₂ and r_max. As the integrand depends only on r,

M=\int_0 ^{r_{max}}{\rho r \left( \int_{z_1(r)}^{z_2(r)}{dz}\right)dr} \int_0 ^{2 \pi}{d\theta }=2\pi \int_0 ^{r_{max}}{ \rho(r) r \left(z_1(r)-z_2(r)\right) dr}

ehild