Re: Moment of Inertia of a thin circular disc
- From: PD <TheDraperFamily@xxxxxxxxx>
- Date: Wed, 20 Jun 2007 10:12:26 -0000
On Jun 19, 5:27 pm, qwerty <qwe...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
The disc has uniform density. The mass of a slice (like a pizza)
of the disc is:
dm = (M/2pi) * dtheta
theta: the angle of the slice in radians
M: the mass of the disc
So to find the moment of inertia:
I = (Integral from 0 to 2pi)R^2 dm = M*R^2
That's incorrect, as the moment of inertia of a circular disc is
M*R^2/2.
So what am I doing wrong?
The integral you want to do is I = Int[r^2*dm] where in the integrand
the little mass element all has to sit at a particular distance from
the axis. In your thin little pie slice, not all of the mass sits at a
distance R from the axis. In fact, only the very outer part of the pie
slice sits at R. In fact, in your little mass element, you have mass
strung out all the way from 0 to R. This should indicate to you that
you've chosen the wrong mass element.
What shape mass element can you use such that all the mass in that
element has the essentially the *same* distance from the axis?
PD
.
- References:
- Moment of Inertia of a thin circular disc
- From: qwerty
- Moment of Inertia of a thin circular disc
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