Consider a system of point particles , where the mass of particle $i$ is $μ_i$ and its position vector is $\vec{r}_i$. Let $\vec{r}_\text{cm}$ is the position vector of the center of mass of the system. Considering the system from a reference frame attached to the center of mass, the system may have a spin about the center of mass and it is given by the spin angular momentum $\vec{L}_{spin}$. It is given by the expression

$$\vec{L}_{spin} = \sum_i \mu_i \Bigl[(\vec{r}_i - \vec{r}_\text{cm}) \times (\dot{\vec{r}}_i - \dot{\vec{r}}_\text{cm}) \Bigr]$$

The rate of change of this spin angular momentum is the total torque acting on the system about the center of mass in the center of mass reference frame.

My question is, is there any (spin kinetic (may be)) energy associated with the spin angular momentum in the center of mass reference frame ? How is it defined ?

  • 1
    $\begingroup$ Your formula includes the "spin" energy inside. The decomposition into separate "spin" energy and translational energy is only useful when the relative motion is constrained, like for a rigid body. If you just have a collection of particles, there is no need to add a separate contribution. The word "spin" is not used for this type of thing, it's just called "the kinetic energy of the extended rotating object". The word spin is usually reserved for cases where the angular momentum is of a quantum point particle, or a particle which can be treated as a point, and the spin energy is in the mass. $\endgroup$
    – Ron Maimon
    Dec 24, 2011 at 13:07
  • $\begingroup$ @Ron Maimon : I need to separate the energy associated by the spin alone from the total kinetic energy. I have a system which is constrained but not rigid. Is there any way it is possible or it doesn't make sense ? $\endgroup$
    – Rajesh D
    Dec 24, 2011 at 13:27
  • $\begingroup$ It would help if you said what the constraints are. If it's nearly rigid, like rotating jello, you can perturb away from a rigid body. If it's a rotating double pendulum, you might want to keep the particle view. I don't think there's a unique answer for all systems. $\endgroup$
    – Ron Maimon
    Dec 24, 2011 at 14:16

2 Answers 2


Similar to the derivation of separation of angular momentum into $L_{CM}$ and $L_{internal}$, one can derive similar expression for Energy as $E = \frac{1}{2}M_{total}v_{CM}^{2} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$.

Proof: $$E = \frac{1}{2}\sum \mu_{i} v_{i}^{2}$$ $$v_{i} = v_{CM} + v_{i}^{'}$$ $$E = \frac{1}{2}\sum \mu_{i} v_{CM}^{2} + v_{CM}\sum \mu_{i} v_{i}^{'} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$$ Since in CoM frame $\sum \mu_{i} (r_{i}-r_\text{cm}) =0 \to \sum \mu_{i} v_{i}^{'}=0$. $$QED$$

L and E within Com frame can be related only if body is rigid. One can refer Klepner & Kolenkow Classical Mechanics.

  • $\begingroup$ the last statement in your answer is contradicting to the answer by Lubos. Hope someone help resolving this $\endgroup$
    – Rajesh D
    Dec 27, 2011 at 8:20
  • $\begingroup$ Dear Rajesh, I don't think there's a contradiction. This answer quantifies exactly what I meant, turns it into equations. The comment that "it" only works for rigid bodies is true in the sense that only rigid bodies have their own "constant" value of the moment of inertia. For non-rigid systems of particles, you may define the value of $I$ for which the energy splits in this way but it's useless because $I$ may change arbitrarily, too. $\endgroup$ Dec 27, 2011 at 9:12
  • $\begingroup$ @Luboš Motl : Dear Luboš Motl, The two equations in your answer cannot be interpreted as replacing the entire system with one body whose moment of inertia changes with time (that is waht you have tried to give an interpretation of instantaneous moment of inertia). Simply because the angular velocity of all the particles of the system is not same. Just try to answer what you mean by $\omega$ in your equation (in your answer) in terms of the individual velocities of the particles of the system, and the problem appears ! $\endgroup$
    – Rajesh D
    Jan 2, 2012 at 12:26

The energy associated with the internal rotations is the rotational energy


given by the formula

$$ K = \frac 12 I \omega^2 $$

where $\omega$ is the angular frequency and $I$ is the moment of inertia with respect to the axis of rotation

http://en.wikipedia.org/wiki/Moment_of_inertia $$ I = \int {\mathrm d}m \, \rho^2 $$

where $\rho$ is the distance of the infinitesimal mass ${\mathrm d}m$ from the axis of rotation. Also, $$ I = I_{ab} n_a n_b $$ in terms of the tensorial moment of inertia $I_{ab}$; $\vec n$ is the unit vector along the axis of rotation.

  • $\begingroup$ In this situation, the system is not a rigid body, which means the moment of inertia is not constant with time. Please let me know what is the formula in this situation ? $\endgroup$
    – Rajesh D
    Dec 24, 2011 at 12:55
  • 1
    $\begingroup$ It's still the same formula, just $I$ is time-dependent. It isn't as useful but the total energy may always be divided to the internal "rotational energy" above, and to the kinetic energy concentrated to the center of mass, $M_{total} V_{CM}^2/2$. That's an identity whether or not the pieces are rigid. $\endgroup$ Dec 24, 2011 at 14:17
  • $\begingroup$ Do you mean that $$K = \frac{\vec{L}_{spin}^2}{2I}$$ where $$I = \sum_i \mu_i \Bigl[(\vec{r}_i - \vec{r}_\text{cm})^2\Bigr] $$ . Please clarify me whether I am correct. $\endgroup$
    – Rajesh D
    Dec 25, 2011 at 3:01
  • $\begingroup$ Yes, that is correct... $\endgroup$
    – David Z
    Dec 28, 2011 at 3:44
  • $\begingroup$ @David : I think that moment of inertia of a system (about an axis through CoM) does make sense for only rigid bodies. When the system is not rigid, the angular velocities of particles about CoM vary from particle to particle, hence in the expression for angular momentum (summation of angular momenta of all particles) we cannot take out the angular velocity common out of summation, hence we cannot get $I\omega$ type of expression, hence there is no such concept of time varying moment of inertia for non rigid bodies. Please correct me if i am wrong. $\endgroup$
    – Rajesh D
    Jan 1, 2012 at 4:36

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