Why is moment of inertia dependent on $r^2$ and not on $r$? (physical reason)

Moment of inertia is the mass equivalent in rotational dynamics. I know, by mathematical arguments, moment of inertia of a particle is $$I = \text{mass} \cdot r^2.$$ But what is the physical reason? When I want the other linear equivalents like velocity,acceleration, we only multiply the linear component times the distance from the centre. But, in moment of inertia, it is not distance but distance-square. Why is it so?

• Analogous to mass (the "linear" moment of inertia) being the ratio of momentum and velocity, the rotational moment of inertia is the ratio of angular momentum and angular velocity. The $\propto r^2$ relation follows from that. What exactly about that is unclear to you? – ACuriousMind Dec 1 '14 at 11:20
• @ACuriousMind: Angular velocity : linear velocity. r ; angular accln. : linear accln. . r . So, rotational inertia: mass . r . But this is not! – user36790 Dec 1 '14 at 11:26
• Have you looked at Wikipedia's derivation? – ACuriousMind Dec 1 '14 at 11:28
• @ACuriousMind: Even,sir, I do know the derivation. When I saw the other quantites, there was a single $r$ but in rotational inertia, it is $r^2$ . That surprised me! – user36790 Dec 1 '14 at 11:43
• There are often different functions and functional dependences in completely different laws and relationships, right? Why is it surprising? If you want to add something to 9 to get 15, you have to add 6. But if you want to get from 64 to 100, you need to add 36, more than that (actually a square). Is that surprising that the two differences are different? – Luboš Motl Dec 1 '14 at 11:48

This builds partially off of TZDZ's answer. In order to fully understand the moment of inertia, you need to fully understand the idea of torque. The only sensible way to fully understand torque that I've seen is to first understand it as the derivative of rotational work $W_R$ being done with respect to the angle $\theta$ through which the work is done. So: $\tau = \dfrac{dW_R}{d\theta}$. To explain further:

First, a proof that $Fr = \alpha I$, where $I$ is defined to be $\sum \limits_{i} m_ir_i^2$. I will not invoke the ideas of angular momentum or torque until the end where I mention their definitions, because I do not want to make a circular argument.

It is pretty well agreed upon that $\int_a^b F\cdot~ds = W{ab}$. In words, the integral of a force $F$ with respect to displacement from some point $a$ to some point $b$ is the amount of work that is done by that force from point $a$ to point $b$. When the force is parallel to the displacement, the dot product above simplifies down simply to $\int_a^b F~ds = W{ab}$.

In the context of rotational work, this simplifies down when the force is tangential to the arc of rotation to: $\int_a^b F~ds = W_R$. Here, $ds$ is the differential arc length that you get as you rotate the object about some axis. For any arc length $s$, $s = \theta r$. So $ds = d\theta r$. Substituting that in and fixing the bounds of integration $a$ and $b$ to become angles rather than positions:

$W_R = \int_{\theta_o}^{\theta_f} Frd\theta$ $If you note that$\frac{d\theta}{dt} = \omega$, rearrange to get$d\theta = \omega dt$, and substitute for$d\theta$in the above expression, making sure to change your bounds of integration yet again, you get:$W_R = \int_{t_o}^{t_f} Fr\omega dt$As TZDZ noted,$W_R = \Delta KE = \frac{1}{2}\omega^2 \sum \limits_i m_ir_i^2 = \frac{1}{2}\omega^2I$. Setting up the final equality:$\int_{t_o}^{t_f} Fr\omega dt = \frac{1}{2}\omega^2I $Differentiating both sides with respect to time gives:$F r \omega = \omega\alpha I$And finally canceling$\omega$on both sides leads to:$F r = \alpha IFr$is what we call torque. This is just the familiar equation:$\tau = I \alpha$The full statement:$\sum \limits_{i=1}^{net} \tau_i = \alpha I$can be shown through the same method. The big questions related to this: Why is Torque important and why is it so much different than everything learned earlier? Also, when should torque be used, as opposed to Newton's laws and the kinematic equations? The answer to the first question: In most introductory physics courses, inertia is introduced right after a chapter on center of mass, which follows several chapters on the motion of rigid bodies in which every particle moves with the same general motion. The kinematics equations can be applied to rigid objects in those earlier chapters because each particle in the object moves with roughly the same velocity as every other particle in the object. When you reach the chapters on center of mass, this is no longer the case, but because of the nature of the center of mass the kinematic equations can still be applied. Then, in chapters on rotation, the concepts of torque, inertia, and angular momentum become necessary because not every particle in the objects that you are dealing with moves with the same translational velocity. However, they do still move with a regular motion and thus can be modeled, but require different models. The answer to the second question: Torque should be used when the individual particles within the system of particles that you are modeling do not all move with roughly the same velocity. IOW, it should be used when not every particle in the system has the same motion. More specifically though, it should be used when the system does not have uniform translational motion but does have uniform angular motion. tl;dr Basically, the reason that inertia is proportional to$r^2$is because one of the$r$'s comes from the fact that$a_T = r \alpha$, where$a_T$is the tangential acceleration. The other one comes from the fact that$s = r \theta$. Since torque is the derivative of$W_R$with respect to$\theta$and not with respect to$s$, the$r$from$s = r \theta$remains in the equation. Inertia also becomes doubly useful to us because it is also present in the calculation of rotational kinetic energy, and so it is a full-fledged concept now with its own name and wikipedia page. • Sorry,why is torque the derivative of rotational work wrt angle? – user25849 Feb 27 '17 at 8:32 The moment of inertia is linked with the kinetic energy. $$E_c=\sum_i \tfrac12 m_i v_i^2 = \sum_i \tfrac12 m_i (\omega r_i)^2 = \tfrac12 \omega^2 \sum_i m_i r_i^2$$ (wikipedia) It seems quite logical to me. Other reason is dimensional. First, there is no reason why a physical quantity must be linear in r. That only happens for some variables (perhaps the ones you are familar with). In the case of the moment of inertia, it is actually a multiplication of two linear variables in r, that is why it end result is quadratic in r. By definition: the moment of inertia is$I=L/\omega$, so$L=I\omega$But L is also defined as$L=mrv$,and because$v=\omega r$you get$I=mr^2$• So the two effects captured in a mass moment of inertia value is a) linear momentum (or speed) increases linearly with distance$r$and b) the moment arm of momentum (like torque for linear momentum) increases linearly with distance$r$. Combined they produce an$r^2$dependency. – ja72 Dec 1 '14 at 15:00 There is a very simple, intuitive and physical (or at least geometric) reason why$r^2$appears in the Moment of Inertia. It is simply because the area of a circle is$\pi r^2$. What does the area of a circle have to do with Moment of Inertia? Let's go through it step by step: The area of a circle is: $$\pi r^2$$ The area of a sector of a circle, is the area of the circle, multiplied by the ratio of the angle and the total angle of the circle: $$A_{sec} = \pi r^2\,{\theta\over 2\pi}$$ This tells us that angle is proportional to sector area: $$A_{sec} = {r^2\over 2}\theta$$ It was common in certain periods of history to examine rotational physics problems in terms of sector areas instead of angles. See kepler's second law and Patrick d'Arcy's formulation of angular momentum for example. We can also define the period as a ratio between the circumference of the circle and the tangential velocity: $$T = {2\pi r\over v_t}$$ Now we can introduce the idea of Areal Velocity, the "sector area" version of angular velocity. Which can be derived by the ratio of the area of a circle to the period: $${dA\over dt} = {\pi r^2\over T} = {1\over 2} v_t r = {1\over 2}\omega r^2$$ Where the tangent velocity:$v_t = \omega r$and$\omega$is the angular velocity. Note that angular momentum (for simple motion in 2D) is: $$L = mv_t r = 2m {dA\over dt}$$ So angular momentum is proportional to areal velocity times mass! We can also define Areal Acceleration: $${d^2A\over dt^2} = {1\over 2} a_t r = {1\over 2}\alpha r^2$$ Where the tangent acceleration:$a_t = \alpha r$and$\alpha$is the angular acceleration. Giving us torque (for simple motion in 2D) in terms of areal acceleration: $$\tau = 2m {d^2A\over dt^2} = m\alpha r^2$$ Then we can see that$\tau = m\alpha r^2 = (mr^2)\alpha = I\alpha\$.

We can even put the canceled pi s back in and get an interesting geometric relationship: $$\tau = 2m {\pi r^2\over 2\pi} {d^2\theta\over dt^2}$$

Notice the ratio between area and circumference of a circle.

Remember how the magnitude of the cross product equals the area of the parallelogram formed by the two input vectors? It turns out that in more complicated and in 3 dimensional Rotational Mechanics you can also apply this "angles to areas" interpretation by simply using the area interpretation of the cross products that appear!

protected by Qmechanic♦Mar 26 '15 at 7:53

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