| Topic: |
Science > Physics |
| User: |
"quat" |
| Date: |
01 Mar 2006 10:48:26 PM |
| Object: |
moment of intertia |
A sheet of metal has the shape of the surface z = x^2 + y^2, 0 <= x^2 + y^2
<= 2. The density at (x,y,z) is kz. Find the moment of intertia about the
z-axis.
My answer did not match the book's answer. Did I make a mistake?
z = R^2 ==> R = sqrt(z)
So this is a parabaloid aligned with the z-axis. Take an arbitrary ring on
it and the density is uniform. The mass of each ring is dm(z) =
kz*2*pi*R*dz = kz*2*pi*sqrt(z)*dz.
The moment of intertia is defined: I = integral( R^2*dm )
= integral( R^2 * kz*2*pi*sqrt(z) wrt z from 0 to 2)
= 2*pi*k*integral( z^(5/2) wrt z from 0 to 2)
= 6.46497*pi*k
But the book's answer is 7.0047*pi*k.
.
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| User: "Greg Neill" |
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| Title: Re: moment of intertia |
02 Mar 2006 08:09:42 AM |
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"quat" <spam@void.com> wrote in message news:6NuNf.1397$kp3.570@fed1read03...
A sheet of metal has the shape of the surface z = x^2 + y^2, 0 <= x^2 + y^2
<= 2. The density at (x,y,z) is kz. Find the moment of intertia about the
z-axis.
My answer did not match the book's answer. Did I make a mistake?
z = R^2 ==> R = sqrt(z)
So this is a parabaloid aligned with the z-axis. Take an arbitrary ring on
it and the density is uniform. The mass of each ring is dm(z) =
kz*2*pi*R*dz = kz*2*pi*sqrt(z)*dz.
The moment of intertia is defined: I = integral( R^2*dm )
= integral( R^2 * kz*2*pi*sqrt(z) wrt z from 0 to 2)
= 2*pi*k*integral( z^(5/2) wrt z from 0 to 2)
= 6.46497*pi*k
But the book's answer is 7.0047*pi*k.
The book's right. I get:
I = (1471/210)pi*k
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| User: "PD" |
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| Title: Re: moment of intertia |
02 Mar 2006 10:21:28 AM |
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quat wrote:
A sheet of metal has the shape of the surface z = x^2 + y^2, 0 <= x^2 + y^2
<= 2. The density at (x,y,z) is kz. Find the moment of intertia about the
z-axis.
My answer did not match the book's answer. Did I make a mistake?
z = R^2 ==> R = sqrt(z)
So this is a parabaloid aligned with the z-axis. Take an arbitrary ring on
it and the density is uniform. The mass of each ring is dm(z) =
kz*2*pi*R*dz = kz*2*pi*sqrt(z)*dz.
The moment of intertia is defined: I = integral( R^2*dm )
= integral( R^2 * kz*2*pi*sqrt(z) wrt z from 0 to 2)
= 2*pi*k*integral( z^(5/2) wrt z from 0 to 2)
= 6.46497*pi*k
But the book's answer is 7.0047*pi*k.
Here's a trick:
The location of the mass element along the z axis is immaterial. It is
only the distance *away* from the z-axis that counts. So collapse
(project) the whole paraboloid down onto the x-y plane.
However, in doing this, note two things:
1. The projection of the surface area onto the x-y plane is a steeper
and steeper projection, the further away from the axis you go. This is
simply because in going from R to dR, the z-height of that annulus
increases. That height dz is 2RdR.
2. On *top* of that projection effect, the density also increases with
R as k*sqrt(R).
PD
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| User: "arvee" |
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| Title: Re: moment of intertia |
01 Mar 2006 11:36:45 PM |
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quat wrote:
A sheet of metal has the shape of the surface z = x^2 + y^2, 0 <= x^2 + y^2
<= 2. The density at (x,y,z) is kz. Find the moment of intertia about the
z-axis.
My answer did not match the book's answer. Did I make a mistake?
z = R^2 ==> R = sqrt(z)
So this is a parabaloid aligned with the z-axis. Take an arbitrary ring on
it and the density is uniform. The mass of each ring is dm(z) =
kz*2*pi*R*dz = kz*2*pi*sqrt(z)*dz.
Not quite: the surface of the ring is slanted, so its area is not just
2*pi*R*dz. You need to figure out the angle, w, between the z-axis and
the normal to the surface at (x,y,z). Then, the ring's area is
2*pi*R*dz/sin(w).
R.G. Vickson
The moment of intertia is defined: I = integral( R^2*dm )
= integral( R^2 * kz*2*pi*sqrt(z) wrt z from 0 to 2)
= 2*pi*k*integral( z^(5/2) wrt z from 0 to 2)
= 6.46497*pi*k
But the book's answer is 7.0047*pi*k.
.
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