Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Where exactly does the equation $MR^2$ for moment of inertia come from? The quantity itself seems fairly arbitrary.

share|cite|improve this question
As a plausibility argument, we expect that it should have $M$ to the first power, because we expect it to add linearly when we combine two objects into one. We also expect $R$ to an even power, because in one dimension $R$ is similar to the $x$ coordinate, and mass placed at negative $x$ shouldn't make the object easier to rotate. Or if we expect a relation of the form $\tau=I\alpha$, then the exponents are fixed simply based on units. – Ben Crowell Jun 13 '13 at 13:58
up vote 11 down vote accepted

The equation you are referring to is the expression for the moment of inertia of a point particle of mass $m$ at a distance $R$ away from some axis. This is expression is really the definition of the moment of inertia for a point mass, so the question becomes "where does this definition come from, and why is it useful?"

Well for simplicity, suppose that such a point mass is rotating around the aforementioned axis on a string, then if a tangential force $F_t$ is applied to it, its resulting tangential acceleration $a_t$ will satisfy $$ F_t = ma_t $$ so the torque on it is $$ \tau = Rma_t $$ On the other hand, recall that the tangential acceleration $a_t$ and angular acceleration $\alpha$ are related by $a_t = R\alpha$, and plugging this into the right hand side of the expression for the torque gives $$ \tau = mR^2 \alpha $$ Notice that the quantity $mR^2$ has magically appeared. Well clearly it's not actually magic; the point really is that if we think of torque as a sort of rotational analog of force, and angular acceleration as the rotational analog of acceleration, then this shows that the quantity $mR^2$ is a sort of rotational analog of mass for a point mass. As a result, we give it a special name: moment of inertia.

It's important to point out that although I used the example of a point mass undergoing uniform circular motion to motivate the definition of moment of inertia, there are significantly more involved and general derivations that lead to a quantity called the inertia tensor which is the generalization of the moment of inertia for non-pointlike bodies undergoing arbitrary rotation.

See, for example, the following answer:

share|cite|improve this answer
Euler first used the term in his Theoria motus corporum solidorum seu rigidorum: Ex primis nostrae cognitionis principiis stabilita et ad omnes motus, qui in huiusmodi corpora cadere possunt, accommodata – Larry Harson Jun 13 '13 at 17:45

It looks arbitrary because the concept of Inertia is defined that way. If someone can digest $F=ma$ (linear), the same should be followed for $\tau=I\alpha$ (angular)

Moment of Inertia is just the rotational substituent for the inertia coefficient (mass) in linear motion. It's not some kind of arbitrary value. It's actually given by $I=Kmr^2$. The $mr^2$ normally seen everywhere is approximated only for a point mass because we always consider point particles to derive something. In case of rigid bodies, we go throught the derivation stuff, by choosing certain axis of rotation, make it spin and finally plug it into our formula $I=r^2\int dm$

So, it can come in various delicious flavors like - for $K=\frac{1}{3}$, we get the moment of inertia for a rod where $r$ acts as the distance from the axis of rotation (which is at one end) with respect to the mass $m$ of the rod. Different derivations "do" exist for different rigid body systems...

share|cite|improve this answer

The moment of inertia naturally arises when trying to account for the energy of motion of a rotating object. Break the object up into small parts with mass $m_i$. The kinetic energy of part $i$ is $$K_i = \frac{1}{2}m_i v_i^2 = \frac{1}{2} m_i r_i^2\omega^2,$$ where $r_i$ is the distance from mass $i$ to the axis of rotation and where $\omega$ is the angular velocity of the object. The rotational kinetic energy of the entire object is the sum of all the $K_i$s, $$K = \sum_i K_i = \frac{1}{2}\sum_i m_i r_i^2 \omega^2 = \frac{1}{2}I\omega^2,$$ where $$I = \sum_i m_i r_i^2$$ is the moment of inertia. In the limit in which the masses become infinitesimally small we have $$I = \int dm\, r^2 = \int dV\, \rho r^2.$$ The moment of inertia is not arbitrary, being the rotational analogue of mass. It tells us how hard it is to rotate an object.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.