# Are translational KE and rotational KE exactly analogous?

Physics Asked on May 9, 2021

My textbook states that translational KE and rotational KE are completely analogous. The author states "They both are the energy of motion involved with the coordinated(non random) movement of mass relative to some reference frame. I can understand why this applies when comparing rotational velocity and translational velocity, but how can we justify that mass is equal to moment of inertia? It appears to me that moment of inertia isn’t exactly analogous to mass in that it involves both direction and a physical quantity.

Keep in mind what an analogy is. It's a comparison of forms, not of identical quantities. So if we look at the two forms, $$frac{1}{2}mv^2 text{ and } frac{1}{2}mathcal{I}omega^2,$$ we see that there are two types of quantities in the first form: a mass, which is independent of the motion of the object, and a square of a motion quantity. If we look at the 2nd form we see a square of motion quantity, $$omega^2$$, so what is that $$mathcal{I}$$ thing? Well, its exact form depends on the shape of the rigid object (if it's not a rigid object, then the analogy falls apart) and the mass of the object, but it's independent of the motion of the object. That's as far as the analogy can go, because of reference frame and rotational center point considerations.

To help begin thinking about how $$mathcal{I}$$ gets related to a motion-independent part of the form, consider a point mass, $$M$$, moving in a circular path of radius $$R$$, having some instantaneous speed $$v$$. The kinetic energy in the reference frame where the center of rotation is at rest (path of motion is a circle) is $$K = frac{1}{2}Mv^2.$$

But we can also describe the motion in terms of an angular speed, $$omega =v/R$$. if we do this we get $$K=frac{1}{2}Mleft(omega Rright)^2 = frac{1}{2}MR^2omega^2.$$

Here we see that $$M$$ and $$R$$ are speed independent, so we can lump them together and give them a single algebraic symbol and maybe even a name: $$MR^2 to mathcal{I}.$$

Correct answer by Bill N on May 9, 2021

Well this is how i understand this. The function that inertial mass has in translational motion is same as that of moment of inertia in rotational motion. Mass gives me an idea of translational inertia while the later gives me an idea of rotational inertia. So, in a sense they are analogous. Moreover, moment of inertia isn't a vector quantity. I rather perceive it as a statistical quantitiy, 'cause it actually is a sort of average value.

Answered by TanfeexUlhaqq on May 9, 2021

I think that a intuitive way to understand the rotation energy is to think of a real body, instead of an ideal rigid body.

A body can rotate if there are bonds between its atoms or molecules that prevents its dispersion in a dust of particles, each one following a straight line.

The result of that bonds is an elastic potential energy associated with the centripetal forces. There is a stress and an associated elastic strain.

Suppose a rod of steel rotating around an axis perpendicular to the it, passing by its center of gravity. The resulting centripetal tensile stress increases when the length of the rod increases. Theoretically, for a rod too long, (and/or too big angular velocity), the stress can be greater than the material strenght leading to material failure.

The elastic potential energy is $$w = frac{1}{2}sigma epsilon$$, and as for linear elasticity: $$sigma = E epsilon implies w = frac{1}{2}Eepsilon^2$$

A rigid body can be regarded as an elastic body when the modulus of elasticity (E) tends to infinity. So the potential energy corresponds to a very small elastic deformation $$epsilon$$.

Answered by Claudio Saspinski on May 9, 2021