Moment of Inertia for Ellipsoid

In summary: The correct integral is ∫∫∫(r2)(ρ2sin∅)dρd∅dθ, with the same limits. Using the same approach as you did leads to ∫(ρ6/6)sin(2∅)dρ=0and ∫(ρ7/7)sin(2∅)dρ=0. Therefore the moment of inertia about the z axis is zero, as you suspected.In summary, the integration of the solid enclosed by the ellipsoid using the given transformation results in a moment of inertia of 0 about the z-axis due to the positive integrand for a second moment.
  • #1
ledphones
17
0

Homework Statement



a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw.

b)If the solid in the above has density k find the moment of inertia about the z-axis.

Homework Equations


∅=phi


The Attempt at a Solution


I got a correct 4/3∏abc

i thought b) would be 0 based on the following.

Jacobian is abc and the solid enclosed would be u^2+v^2+w^2≤1

x^2*y=(a^2*u^2)(bv)

V=∫∫∫(a^2bu^2v)(abc)dudvdw
=∫0→2∏ ∫0→∏ ∫0→1 (a^3b^2c)(ρ^2sin^2∅cos^2θ)(ρsin∅sinθ)ρ^2sin∅dρd∅dθ
=a^3b^2c∫∫∫(ρ^5sin^4∅cos^2θsinθ)dρd∅dθ
=a^3b^2c∫0→2∏ cos^2θsinθdθ∫0→∏ sin^4∅d∅ ∫0→1 ρ^5dρ
=0 since the 1st integral is 0
 
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  • #2
ledphones said:

Homework Statement



a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw.

b)If the solid in the above has density k find the moment of inertia about the z-axis.

Homework Equations


∅=phi


The Attempt at a Solution


I got a correct 4/3∏abc

i thought b) would be 0 based on the following.

Jacobian is abc and the solid enclosed would be u^2+v^2+w^2≤1

x^2*y=(a^2*u^2)(bv)

The integrand for a second moment is always positive. Your problem is in the last step above. The moment arm is the distance squared from the z axis: r2=x2+y2, not x2y.
 

1. What is Moment of Inertia for an Ellipsoid?

Moment of Inertia for an Ellipsoid is a measure of an object's resistance to changes in its rotational motion. It is a property of a rigid body that depends on its mass distribution and shape.

2. How is Moment of Inertia for an Ellipsoid calculated?

The formula for Moment of Inertia for an Ellipsoid is I = (2/5) * m * a^2 * b^2, where m is the mass of the object and a and b are the semi-major and semi-minor axes of the ellipsoid, respectively.

3. What is the significance of Moment of Inertia for an Ellipsoid?

Moment of Inertia for an Ellipsoid is important in understanding the rotational motion of objects and can be used to predict how an object will behave when subjected to external forces. It is also used in engineering and design to determine the stability and strength of structures.

4. How does the Moment of Inertia for an Ellipsoid compare to other shapes?

The Moment of Inertia for an Ellipsoid is typically larger than that of a sphere with the same mass and radius. This is due to the elongated shape of the ellipsoid, which distributes the mass further away from the axis of rotation, resulting in a larger moment of inertia.

5. Can the Moment of Inertia for an Ellipsoid change?

Yes, the Moment of Inertia for an Ellipsoid can change if the mass distribution or shape of the object changes. For example, if an ellipsoid is compressed or stretched, its moment of inertia will also change. Additionally, the moment of inertia can be affected by the object's orientation and the axis of rotation.

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