- #1

- 173

- 0

This may sound like a stupid question and probably is, but I'll take my chances anyway.

If the moment of inertia of an N particle distribution is defined as

[tex]I=\sum_{n=1}^N m_ir_i^2,[/tex]

where [itex]r_i[/itex] are the perpendicular distances repect to the axis of rotation, then if I consider a solid disk of radius R with its mass concentrated in two points in the circunference, then the moment of inertia is [itex]I=m_1 r_1^2+m_2 r_2^2=(m_1+m_2)R^2[/itex], right?

If I consider the center of mass of those two points, shouldn't the moment of inertia would be the same? If so,

[tex]r_{cm}=\frac{m_1r_1+m_2r_2}{m_1+m_2},[/tex]

so

[tex]I=\frac{\|m_1r_1+m_2r_2\|^2}{m_1+m_2}=\frac{(m_1^2+m_2^2)R^2+2m_1m_2 r_1\cdot r_2}{m_1+m_2}.[/tex]

So, what am I doing wrong?

What trows me out about the first calculation, is that no matter how the points are distributed along the circumference, the moment of inertia is the same. This can't be right, or can it?

Any help will be much appretiated.

If the moment of inertia of an N particle distribution is defined as

[tex]I=\sum_{n=1}^N m_ir_i^2,[/tex]

where [itex]r_i[/itex] are the perpendicular distances repect to the axis of rotation, then if I consider a solid disk of radius R with its mass concentrated in two points in the circunference, then the moment of inertia is [itex]I=m_1 r_1^2+m_2 r_2^2=(m_1+m_2)R^2[/itex], right?

If I consider the center of mass of those two points, shouldn't the moment of inertia would be the same? If so,

[tex]r_{cm}=\frac{m_1r_1+m_2r_2}{m_1+m_2},[/tex]

so

[tex]I=\frac{\|m_1r_1+m_2r_2\|^2}{m_1+m_2}=\frac{(m_1^2+m_2^2)R^2+2m_1m_2 r_1\cdot r_2}{m_1+m_2}.[/tex]

So, what am I doing wrong?

What trows me out about the first calculation, is that no matter how the points are distributed along the circumference, the moment of inertia is the same. This can't be right, or can it?

Any help will be much appretiated.

Last edited: