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The textbook that I'm reading states that translational kinetic energy is:

(I'm doing high school physics so please go easy with the crazy calculus)

From what I understand, a system has translational kinetic energy if the system of mass is moving at a velocity but the textbook I'm reading provides this example in which the system as a whole isn't moving and yet still has kinetic energy. How does the system have kinetic energy if the centre of mass is stationary?

In the chapter on Newton's Laws, the book says that forces between elements INSIDE the system have no effect on the overall motion of the system so I thought kinetic energy worked the same where elements inside can have kinetic energy but the overall system doesn't because it's the centre of mass is stationary.

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    $\begingroup$ Consider a rotating wheel on a fixed axle - it certainly has kinetic energy. $\endgroup$
    – Jon Custer
    Jan 13, 2023 at 16:00
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    $\begingroup$ The word, "internal," in that description is important. The system-as-whole has no kinetic energy with respect to the coordinate system in which its COM is stationary, but it has internal kinetic energy. I'm not really a physicist, so I'm not going to say more about it myself, but with a little Googling... $\endgroup$ Jan 13, 2023 at 16:21
  • $\begingroup$ @SolomonSlow: That has the gist of what I was just going to type in as an answer. I believe that the question isn't about the detailed math of this; rather it's about the meaning of the phrase "how much energy does the system have?" Which you've answered, succinctly. $\endgroup$
    – TimWescott
    Jan 13, 2023 at 16:28
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    $\begingroup$ Technically nearly all energy is kinetic energy. Heat energy is mainly the kinetic energy of the individual atoms although the bulk object is not moving, etc. $\endgroup$
    – RC_23
    Jan 13, 2023 at 18:11
  • $\begingroup$ A system whose center-of-mass is stationary can't have momentum because the system's momentum is the vector sum of each component's momentum, so opposite directions tend to cancel. But the kinetic energy is the sum-of-squares, so opposite directions add up rather than cancel. $\endgroup$
    – eigengrau
    Jan 14, 2023 at 12:30

6 Answers 6

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It is often said that a quantity that has a direction and magnitude is a vector. For example, momentum is a vector. Two carts of equal mass headed toward each other at equal speed each have momentum. The two momentum vectors are equal and opposite, so the total momentum is $0$. This shows the overall momentum of the system is $0$. The center of the system doesn't move. Another way of saying it is that the two cart system as a whole is stationary.

But not every such quantity is a vector. Kinetic energy is an example. Each cart has kinetic energy. But the total kinetic energy of the system is the sum of the two energies. Kinetic energy adds like numbers, not like vectors. That is the key difference.

Energy and momentum measure different things. The two cart system has enough energy to dent the carts when they collide.

Two carts traveling in the same direction have the same energy. But they also have non-$0$ momentum. The system as a whole moves and can dent something else when the carts crash into it.

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I'll give an example. Let's say two blocks are going in opposite directions with equal speeds, so that the center of mass isn't moving. Now just add a spring in front of each of their paths. They will both eventually come to rest after compressing their respective springs.

Now you must agree that the final configuration of the system does have potential energy stored in the springs. This potential energy must have come from the kinetic energy of the blocks, because energy is conserved.

So the answer to your question is that kinetic energy is a scalar. Each of the components of the system carries kinetic energy, which adds like a scalar.

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In this case the system is composed of two carts, which we will approximate as point masses. So the kinetic energy of the system is $$KE = KE_1 + KE_2 = \frac{1}{2} m_1 \vec v_1^2 + \frac{1}{2}m_2 \vec v_2^2$$ Now, the center of mass has a velocity $$v_c=\frac{m_1 \vec v_1 + m_2 \vec v_2}{m}$$ where $m=m_1+m_2$. Then the so-called translational KE can be written $$KE_c=\frac{1}{2}m v_c^2$$

So, suppose that $m_1=m_2$ and $\vec v_1 = -\vec v_2$ then $v_c=0$ so $KE_c=0\ne KE$. In other words, the KE is not the same as the so-called translational KE. Sometimes all you care about is the translational KE, but it is important to realize that it does not represent the entire KE of the system.

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  • $\begingroup$ Missing 1/2 for KE2 in the first equation? $\endgroup$
    – Bob D
    Jan 16, 2023 at 16:57
  • $\begingroup$ @BobD oops, yes. Thanks for the catch. I have fixed it $\endgroup$
    – Dale
    Jan 16, 2023 at 20:16
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The acceleration of the center of mass (CM) of a system is determined by the total external force on the system; no external force no acceleration of the CM. The total kinetic energy (KE) of the system is the KE of the CM plus the total KE with respect to the CM. Physics mechanics textbooks prove these statements.

Consider the carts with the spring all as one system. There is no external force on the system therefore the center of mass (CM) of the system does not move even after the spring releases. However, the total KE of the system increases because the KE of the carts with respect to the CM increases.

Consider a mass of high explosive that is stationary. After the explosion the CM does not move but the fragments have KE and the total KE of the system increases.

(The total linear momentum of the system is the momentum of the CM of the system. With no external forces, the total momentum of the system does not change. KE is different from momentum.)

Additional information

For both the carts/spring and high explosive examples, internal to the system there is conversion of spring potential energy and chemical energy to kinetic energy, respectively.

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How can a system have kinetic energy if the centre of mass of the system is stationary?

The short answer is it can have rotational kinetic energy without kinetic energy of the center of mass (COM). That is, it can have

$$KE_{rotational}=\frac{1}{2}I\omega ^2$$

without

$$KE_{com}=\frac{1}{2}mv^2$$

because it is strictly rotating without translating.

In the equations, $I$ is the moment of inertia about the center of mass (COM), $\omega$ is the angular velocity, and $v$ is the velocity of the COM,

Likewise it can have translational KE without rotational KE if it is strictly translating without rotating, or it can have both if it is both translating and rotating.

Hope this helps.

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Kinetic energy is not a vector like velocity. Total or average velocity is still zero because a positive component of velocity ($+v$) is cancelled by a negative component ($-v$) of it present elsewhere on the body. But what if we square velocity of each small component and take the sum ? The body part having negative component of velocity which was acting earlier as a nullifier to positive component will not act like that now. Because we are ensuring that we are dealing with a value which is positive. Similarly, $K.E.$ is a value not dependent on $v$ but $v^2$ which is always a positive number.

A more simple example :

$$1 + (-1) = 0$$

But,

$$(1)^2 + (-1)^2 ≠ 0$$

Because the nullifier (say $-1$) being squared becomes positive and do not act like before.

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