Moment of Inertia of a non-uniform density paraboloid

In summary: 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}
  • #1
emgram769
7
0

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.
 
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  • #2
Is it a three dimensional or a two-dimensional object?
Explain the notations. What is f(x,y)? What are r and theta?

ehild
 
  • #3
ehild said:
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 [itex] \sqrt{h} [/itex] with the appropriate density for that radius.
 
  • #4
gneill said:
It appears to be a standard notation.

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

ehild
 
  • #5
ehild said:
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 :blushing:
 
  • #6
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
 
  • #7
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
 
  • #8
Idisk = [itex]
\int_0^{2\pi}\int_0^r r^2\rho\,r\,\mathrm{d}r\,\mathrm{d}\theta
[/itex]

M = [itex]\int_0^V\rho\,\mathrm{d}v[/itex]

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

M = [itex]2\pi\int_0^r\rho\,r^3\,\mathrm{d}r[/itex]

which I think holds true with the relationship z=r2
 
  • #9
emgram769 said:
Idisk = [itex]
\int_0^{2\pi}\int_0^r r^2\rho\,r\,\mathrm{d}r\,\mathrm{d}\theta
[/itex]

M = [itex]\int_0^V\rho\,\mathrm{d}v[/itex]

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

M = [itex]2\pi\int_0^r\rho\,r^3\,\mathrm{d}r[/itex]

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
[tex]M=\int_0 ^{r_{max}}\int_{z_1(r)}^{z_2(r)} \int_0 ^{2 \pi}{\rho r d\theta dz dr }[/tex]

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

[tex]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}[/tex]

ehild
 

What is the moment of inertia of a non-uniform density paraboloid?

The moment of inertia of a non-uniform density paraboloid is a measure of its resistance to rotation. It is a physical property that depends on the mass distribution of the paraboloid.

How is the moment of inertia of a non-uniform density paraboloid calculated?

The moment of inertia of a non-uniform density paraboloid can be calculated using the formula I = ∫r^2 dm, where r is the distance from the axis of rotation and dm is the differential mass element of the paraboloid.

What factors affect the moment of inertia of a non-uniform density paraboloid?

The moment of inertia of a non-uniform density paraboloid is affected by the shape, size, and mass distribution of the paraboloid. A larger and more spread out mass distribution will result in a larger moment of inertia, while a more compact and concentrated mass distribution will result in a smaller moment of inertia.

How does the moment of inertia of a non-uniform density paraboloid compare to that of a uniform density paraboloid?

The moment of inertia of a non-uniform density paraboloid will generally be larger than that of a uniform density paraboloid with the same mass and shape. This is because the non-uniform density results in a more spread out mass distribution, leading to a larger moment of inertia.

What are some real-life applications of understanding the moment of inertia of a non-uniform density paraboloid?

Understanding the moment of inertia of a non-uniform density paraboloid is important in engineering and physics, particularly in the design and analysis of rotating systems. It can also be useful in fields such as sports, where understanding the moment of inertia can help in optimizing the performance of equipment, such as golf clubs or tennis rackets.

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