- #1
KastorPhys
- 3
- 0
Hi everyone, I am trying to find out the Moment of Inertia of a sphere which is all known to be
2/5(m)(r^2)
I calculate this in 2 ways.
One by triple integration and one by disk method.
From the textbooks, moment of inertia should be in the form,
dI = (r^2) dm,
However, the textbook, University Physics, propose the method of disk by suming up,
dI = 1/2 (r^2) dm, which is the I of a extremely thin disk. It sounds logical.
However, I just think in a way that the equation,
dI = (r^2) dm,
must be suitable for any situations.
Thus, in the 2 ways, I just start by this and found that the triple integration way gives the right answer. But the disk method way just provide a doubled result.
Can anyone try me why this is happened?
Why, in the disk method, we should think in the way that suming up the dI of all thin disks, but not just start from the original equation, then
dm = (ro) dV = (ro) (pi) (r^2) dz
Treating the dV as the volume of a thin disk and add them up?
Thanks!
2/5(m)(r^2)
I calculate this in 2 ways.
One by triple integration and one by disk method.
From the textbooks, moment of inertia should be in the form,
dI = (r^2) dm,
However, the textbook, University Physics, propose the method of disk by suming up,
dI = 1/2 (r^2) dm, which is the I of a extremely thin disk. It sounds logical.
However, I just think in a way that the equation,
dI = (r^2) dm,
must be suitable for any situations.
Thus, in the 2 ways, I just start by this and found that the triple integration way gives the right answer. But the disk method way just provide a doubled result.
Can anyone try me why this is happened?
Why, in the disk method, we should think in the way that suming up the dI of all thin disks, but not just start from the original equation, then
dm = (ro) dV = (ro) (pi) (r^2) dz
Treating the dV as the volume of a thin disk and add them up?
Thanks!