0
$\begingroup$

The moment of inertia of a point and ring are both $m R^2$. It is interesting that the formula for moment of inertia is exactly the same for both. Is there any physical reason why this is the case?

I think it might have something to do with the fact that when rotated about a center point, a point traces out a ring in space.

$\endgroup$
1
$\begingroup$

Moment of inertia depends on how far away mass is from the axis. In a ring of radius $R$, all the mass is $R$ from the axis. For a single particle $R$ away from the axis... well, all the mass is $R$ from the axis.

$\endgroup$
4
$\begingroup$

Moments of inertia are additive. Suppose you have particle $A$ with a moment of inertia $I_A$ and particle $B$ with a moment of inertia $I_B$. Then the total moment of inertia of both particles is just $I_A + I_B$.

You can imagine a ring as being made up from lots of point particles, all at a distance $R$ from the central axis. In that case the moment of inertia of the ring would be the sum of all the moments of inertia of all the point particles:

$$ I_{ring} = I_A + I_B + \, ... = m_A R^2 + m_B R^2 + \, ... $$

Since $R$ is a constant we can take $R^2$ outside the sum to get:

$$ I_{ring} = \left(m_A + m_B + \, ...\right)R^2 $$

and $m_A + m_B + \, ...$ just adds up to the total mass of all the particles $M$. so we end up with:

$$ I_{ring} = MR^2 $$

This is a special case of a general principle in calculating moments of inertia. For any body of any shape we can divide it up in to infinitesimal elements then add up their moments of inertia to get the total. In general this is done by integration, as Wet Savanna Animal explains in his answer.

$\endgroup$
2
$\begingroup$

Firstly, you must qualify moment of inertia by the axis it is taken about. If you translate this point, the inertia also changes as described by the parallel axis theorem.

So you question relates to the moment of inertia of a ring about the ring's axis of symmetry normal to the ring's plane, and to a point on this plane at the same distance from this central point.

In both cases, there is only one distance, namely the ring's radius, from the axis to every point mass that makes up the bodies. The fact that in one case these masses are spread over a ring and in the other case are concentrated at a point doesn't change the formula $\int r^2\,\mathrm{d}m$ for the scalar component of the mass moment of inertia.

$\endgroup$

protected by Qmechanic Jul 31 '15 at 12:23

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.