Is the kinetic energy of a rotating system just the sum of the instantaneous kinetic energies of the individual particles?

Question

Take a look at this construction which shows linearly moving particles giving the illusion of circular motion. (Follow the instructions on the screen).

Now the question is when I calculate the total kinetic energy of the system of particles, do I calculate the sum of the instantaneous kinetic energies due to the oscillating motion of the particles in straight lines or the angular kinetic energy of the circular path of the centre of mass, together with the angular kinetic energy due to the rotation of the particles around their common centre of mass? Is linear kinetic energy just an alternative interpretation of angular kinetic energy and vice versa like linear momentum can be interpreted as angular momentum?

Answer 1

There's no "linear KE" nor "angular KE", there's only KE.

KE of a particle. The KE of a particle is defined as

$$K = \frac{1}{2} m |\mathbf{v}|^2 \ .$$

You just need to take the square of the absolute value of the velocity, so any information about direction is lost.

KE of a system of particles. Being (defined as) an additive quantity, the KE of a system of particles reads

$$K = \sum_i \frac{1}{2} m_i |\mathbf{v}_i|^2 \ .$$

KE of a continuum system. In a continuum, integral replaces summation

$$K = \int_V \frac{1}{2} \rho |\mathbf{v}|^2 \ ,$$

being $\rho(\mathbf{r},t)$ the density in the body with volume $V$.

KE of a rigid body. For a rigid body, the relation $\mathbf{v}_P - \mathbf{v}_Q = \boldsymbol{\omega} \times (\mathbf{r}_P - \mathbf{r}_Q)$ holds between the velocity and the position of two material points, being $\boldsymbol{\omega}$ the angular velocity of the rigid body. If you choose a reference point $Q$ and use this relation into the summation or the integral above to relate the velocity of the material points to $\mathbf{v}_Q$,

$$\mathbf{v}_i = \mathbf{v}_Q + \boldsymbol{\omega} \times (\mathbf{r}_i - \mathbf{r}_Q) \ ,$$

you can get a "simpler" expression of the KE as a sum of contributions containing the mass and inertia moments of the body w.r.t. the chosen reference point. As an example, choosing the center of mass $G$ as the reference point, the KE of a rigid body can be written as the sum of two terms, one due to the linear velocity of a point through its mass, the other due to the angular velocity of the body through the moment of inertia,

$$K = \frac{1}{2}m|\mathbf{v}_G|^2 + \frac{1}{2} \boldsymbol{\omega} \cdot \mathbb{I}_G \cdot \boldsymbol{\omega} \ .$$

You can find the whole derivation of different expressions of kinetic energy for rigid bodies in these hand-written notes:

• discrete systems: https://basics.altervista.org/test/Physics/Me/dynamical_quantities_rigid_discrete.html
• continuous systems: https://basics.altervista.org/test/Physics/Me/dynamical_quantities_rigid_continuous.html

Answer 2

There is no rotation of masses, you just have movement with v=sin(wt) of several masses in different direction. The video shows, that the projection of circular motion is a sin function.

Answer 3

What I am trying to establish is if kinetic energy calculated treating the system as a rotating system is just an alternative point of view from treating the kinetic energy as a sum of linear motions, which the other answers have not directly answered.

Consider the following system:

There are 4 particles each with the same speed v tangential to the origin. Note that these particles are freely moving and not connected to anything. For an instant this situation is identical to a rotating system where the particles are at a radius of 2 units from the centre. The kinetic energy of a single particle is $1/2mv^2 $ and the total kinetic energy of the system is $2mv^2$. Treating the system as rotational , the angular kinetic energy of a single particle is $1/2 I \omega^2$. Since the moment of inertia for a single particle is $mr^2$ and the angular velocity is $\omega = v/r$, the angular kinetic energy is $1/2 I \omega^2 = 1/2 (mr^2) (v/r)^2 = 1/2 mv^2$ and when multiplied by 4 we get the same total kinetic energy as when we treated the system as set of linearly moving particles.

The motion is not exactly circular. The radius expands over time. Consider the system at later time after a time t has elapsed. The system would look like the diagram below:

Now the radius of each particle from the centre is the length AP which is the hypotenuse of the triangle AUP, where one side is the original radius and side UP is equal to vt, so:

The new radius is: $r_2 = \sqrt{r^2 + (vt)^2}$.
The new angular velocity is: $\omega_2 = v/r_2 = v/\sqrt{r^2 + (vt)^2}$
The new moment of inertia is: $I_2 = mr^2 = m \left(\sqrt{r^2 +(vt)^2}\right)^2 = mr^2 +mv^2t^2.$

The new angular kinetic energy of a single particle is now $$1/2 \ I_2 \ \omega_2^2 = 1/2 (mr^2 +mv^2t^2) \left(\frac {v}{\sqrt{r^2 + (vt)^2}}\right)^2 = 1/2 m v^2,$$

so when multiplied by 4, the total kinetic energy is the same as calculate earlier, which is as it should be because no external forces are acting on the system and energy is conserved. This demonstrates that at least in this example, the kinetic energy is the same whether we treat the system as a collection of linearly moving particles or a system of particles with rotational motion. The app in the question states the circular motion is an illusion, but the reality is the system can really be treated as a rotating system. Despite the fact the radius is constantly expanding in this example, the angular kinetic energy remains constant because the angular velocity is continuously diminishing by just the right amount

Now let's consider an elastic collision and see if everything obeys the rules of physics. Consider the following set up:

Initially the 4 blue particles each of mass m are stationary and the red particle has velocity v and a mass of 4m. The initial linear momentum is $4mv$ and the initial linear kinetic energy is $4 \times 1/2 mv^2$. After the elastic coliision, the large red particle comes to a stop and the 4 blue particles are rotating around their common centre of mass as illustrated below:

Linear and angular momentum are separately conserved. The red particle is now stationary so all the linear momentum is in the blue particles. Each has a velocity of v and total linear momentum of the blue particles is now $ 4 \times mv$ which is the same as that of the red particle before the collision. The angular momentum of the blue particles is now $L = 4 \times 1/2 I \omega = 4 \times (mr^2)(v/r) = 4mvr$. Since angular momentum is individually conserved, the angular momentum must have been the same when only the red particle was moving. It turns out we can treat the red particle as having a radius of r relative to the centre of mass of the system and its angular velocity was $ L = I\omega = (4mr^2)(v/r) = 4mvr$ which is the same as the angular momentum of the blue particles after the collision.

Now lets consider angular kinetic energy before and after the collision. The angular kinetic energy of the red particle was $KE= 1/2 \ I \omega^2 = 1/2 (4mr^2)(v/r^2) = 2mv^2$. The angular kinetic energy of the blue particles after the collision is $KE= 1/2 \ I \omega^2 = 1/2 \times 4 \times (mr^2)(v/r^2) = 2mv^2$, which is the same as the red particle initially had. Again this example supports the argument that linear and angular kinetic energy are just two different ways of looking at the same thing. Any rotating system can be considered to be a system of instantaneously linearly moving particles and vice versa, from an energy point of view. However, we should be careful not to double count and not to add the the kinetic energy calculated assuming linear motion to the energy calculated assuming rotational motion.