I always thought that whenever there's rotational kinetic energy, there's also translational (linear) kinetic energy. The opposite is not true. There can be translational kinetic energy without having any rotation. Is this true? I thought it was until I stumbled upon this question today:

A pulsar is a form of a neutron star, the core of collapsed star of between $1.4$ and $3$ solar masses, that rotates rapidly and gives off radio waves. Suppose a star, with radius $7 × 10^8$ m, before supernova rotates at an angular rate of $3 × 10^{-6}$ rad/s, and upon supernova, shrinks in radius to $15$ km. At what rate does it rotate?

They solve it as follows:

Conservation of kinetic energy:

$K_{rot_0} = K_{rot_1}$

$ \frac{1}{2} \times I_0 \times \omega_0^2 = \frac{1}{2} \times I_1 \times \omega_1^2$

with $ I_0 = \frac{2}{5} \times m \times R_0^2 , I_1 = \frac{2}{5} \times m \times R_n^2 $

$ \frac{1}{2} \times \frac{2}{5} \times m \times R_0^2 \times \omega_0^2 = \frac{1}{2} \times \frac{2}{5} \times m \times R_n^2 \times \omega_1^2$

$ \frac{1}{5} \times (7 \times 10^8)^2 \times (3 \times 10^{-6})^2 = \frac{1}{5} \times (15 \times 10^3)^2 \times \omega_1^2$

$\omega_1 = 0.14$ rad/s

Before looking the solution I also accounted for the transitional rotation energy which is why I got stuck and couldn't solve it. How can an object have only rotational kinetic energy? If it's rotating, then it's moving. Or am I missing something?

  • 1
    $\begingroup$ What do you mean by "transitional rotational energy"? I'm not sure if I'm missing your point, but recall that the expression for rotational KE is arrved at by applying $1/2mv^2$ to every element of the system. This accounts for the all of the movement of the system. $\endgroup$
    – garyp
    Mar 8, 2019 at 16:03
  • $\begingroup$ I think garyp has already accounted for the fallacy in your problem. Nothing called Rotational KE or Translational KE exists in physics.The fundamental expression for KE is $1/2mv^2$. The expression involving /omega is derived from the fundamental expression. $\endgroup$
    – user213933
    Mar 24, 2019 at 14:02

4 Answers 4


How can an object have only rotational kinetic energy? If it's rotating, then it's moving. Or am I missing something?

You can have pure rotation of a body without translation and therefore have rotational kinetic energy without translational kinetic energy.

This is because the translational velocity of a body, which results in translational kinetic energy, is the velocity of its center of mass along a straight line relative to some external (to the body) frame of reference. The center of mass of a body undergoing pure rotation does not move at all so there is no translational kinetic energy.


The above being said, the total kinetic energy of an extended object can be expressed as the sum of the translational kinetic energy of the center of mass ($\frac {mv^2}{2}$) and the rotational kinetic energy about the center of mass ($\frac {Iω^2}{2}$). The rotational inertia $I$ (moment of inertia) is analogous to translational inertia $m$ (mass) and the angular velocity $ω$ is analogous to the linear velocity $v$.

For an extended object undergoing pure rotation, the center of mass has no translational velocity, thus the translational kinetic energy component of the total kinetic energy is zero.

One very important final point. The terms “pure rotation” and “pure translation” only have meaning with respect to a defined frame of reference. I’m on a train going at constant velocity in a straight line. I have a globe of the earth resting on a table. I spin it. From my frame of reference on the train, the motion of the globe is pure rotation and it has rotational kinetic energy only. From the frame of reference of someone standing on the track watching the globe go by through the train window the globe has both rotational and translational kinetic energy.

Kinetic energy depends on the frame of reference.

Hope this helps.


It is instructive to look at the way that "rotational" kinetic energy arises in the theory of mechanics.1 If you start by considering the kinetic energy of a collection of point masses2 $m_i$ each with their own velocity $\mathbf{v}_i$, then the kinetic energy of the system is $$ K = \sum_i K_i = \sum_i \frac{1}{2} m_i v^2_i \;. $$ This is all "translational" kinetic energy.

But if one defines the center of mass $\mathbf{X}$, center of mass velocity $\mathbf{V}$, and the total mass $M$ of the system \begin{align} \mathbf{X} &\equiv \frac{\sum_i m_i \mathbf{x}_i}{\sum_i m_i} = \frac{\sum_i m_i \mathbf{x}_i}{M}\\ \mathbf{V} &\equiv \frac{\sum_i m_i \mathbf{v}_i}{\sum_i m_i} = \frac{\sum_i m_i \mathbf{v}_i}{M} \;, \end{align} and decomposes the position and velocity of each particle into a CoM part and a relative part \begin{align} \mathbf{x}_i &= \mathbf{X} + \mathbf{x_r}_i \\ \mathbf{v}_i &= \mathbf{V} + \mathbf{v_r}_i \; \end{align} you obtain a factorization of the overall kinetic energy into center-of-mass and internal parts3 \begin{align} K &= \sum_i \frac{1}{2} m_i \left( V^2_i + 2\mathbf{V} \cdot \mathbf{v_r}_i + v_{ri} \right) \\ &= \frac{1}{2}M V^2 + \sum_i \frac{1}{2} m_i \left( 0 + v_{ri}^2 \right) \\ &= \frac{1}{2}M V^2 + \sum_i \frac{1}{2} m_i v_{ri}^2 \\ & = K_\text{CoM} + K_\text{int} \;. \end{align}

In the case of a rigid (or rigid enough) body the only motion relative the center of mass allowed is rotation as a whole, and one can continue with the the work to express the internal kinetic energy in terms of moment of inertia and angular velocity.

But the philosophical point here is that the factorization of the kinetic energy into parts due to translation of the center-of-mass and due to rotations and deformations relative the center-of-mass is very much a convenience tool wrapped around the idea that notional point particles have only translational kinetic energy.

1 You can find this kind of thing done in detail in most upper-division and graduate mechanics texts.

2 You can, of course, treat the system as a continuum of infinitesimal mass elements $\mathrm{d}m$, in which case you replace all the sums herein with integrals. But the results are the same.

3 The origin of the $0$ on the second line is a nice puzzle: expand $\sum_i m_i \mathbf{x_r}_i$ in terms of the $\mathbf{x}_i$s and $\mathbf{X}$ to show that it is zero. But the $\sum_i m_i \mathbf{v_r}_i$ is the time-derivative of that so it is also zero.


$\let\om=\omega$ Mine isn't kind of an answer to your question, but you already had it. Rather I'd like to note that the solution you report to the collapse problem is plainly wrong.

Kinetic energy doesn't conserve during collapse, as gravitational potential energy strongly decreases - it's negative and varies like $1/R$. Instead angular momentum may be conserved, as there are no obvious ways for the collapsing part of the star to lose or gain it.

So, instead of having $$I_0\,\om_0^2 = I_1\,\om_1^2$$ you have $$I_0\,\om_0 = I_1\,\om_1$$ which is a much higher $\om_1/\om_0$ ratio.


There is no such thing like a body is performing "pure" translation or "pure" rotation about so and so point,etc.

Any motion can be expressed as a combined motion of translation and/or rotation wrt the center of mass of the body.

For example: the total KE of a body = $1/2 * M * v_{cm}^2 + 1/2 * I_{cm} * \omega ^2$ always holds true.


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