Consider two objects $O_1$ and $O_2$ of mass $m$ each. The objects are connected one to another with a rod of say length $2\cdot l$. At the middle of the rod, $M$, there is a rotational joint allowing the objects to rotate in a plane. The joint sits on a cart $C_1$ with the center of mass also at the point $M$.

enter image description here

Suppose the cart in the image above is restricted to move only up and down (that is on $y$ axis) and not allowed to rotate. Also suppose there are no other forces acting on the system but a torque applied to the bar, in the rotation joint such that the bar (along with the objects $O_1$ and $O_2$) are rotating with a constant angular velocity $\omega$ about the $z$ axis (coming out of screen).


Suppose that the object which happens to be in the upper half of the cart (that is having the angle with $x$ axis between $0$ and $\pi$ ) has an extra energy $E_x$ which when it approaches angle $\pi$ will be "instantaneously" transferred to the other object (which now enters the upper half). In simple words we are trying to make always the object in the upper plane have $E_x$ extra energy. Is the cart feeling a small positive acceleration on $y$ axis?

Answer (...)

I think the answer is yes if $E_x > 0$. For each of the objects during rotation the center of mass feels a centrifugal force $F = m\cdot l \cdot \omega^2$ in the direction of the object. However, because the cart is restricted to move only on $y$ axis the acceleration is given by the projection of this force on $y$ axis. Suppose $O_1$ is now on the upper half plane and let $\phi_1$ be the angle it forms with the $x$ axis. That means it has an extra energy, hence its inertial mass is $m_1 = m + \frac{E_x}{c^2}$. Therefore the force due to $O_1$ is $$ F_1 = \left( m + \frac{E_x}{c^2}\right)\cdot l\cdot \omega^2 \cdot \sin(\phi_1)$$ while the force due to $O_2$ (which is symmetric to $O_1$ about M) is $$ F_2 = - m\cdot l\cdot \omega^2 \cdot \sin(\phi_1) $$ Hence the total force felt by the cart is: $$F = F_1 + F_2 = \frac{E_x}{c^2} \cdot l\cdot \omega^2 \cdot \sin(\phi_1) > 0$$ for all $\phi_1 \in (0,\pi)$ Is this true?

PS: The linear momentum of the cart varies but no external force was applied to the center of mass ... ?! Therefore something might be wrong in my thinking, but I do not know what.


I want to add few words about that "instantaneous" energy transfer. Probably there are more methods (or none) but I was primarily thinking about energy transfer through radiation. In the simplest form say the object exiting the upper half of the plane is losing energy (it had some energy say stored in a capacitor) by illuminating the second object. The second object (which now enters the upper half plane) captures this some of this energy becoming therefore "heavier" ...

some analysis of variation of linear moments:

According to Newtonian mechanics the linear momentum of the object $O_1$ while in upper half plane is $$\vec{P}_1 = \left( m + \frac{E_x}{c^2}\right)\cdot \vec{v}_1 = \left(m + \frac{E_x}{c^2} \right) \cdot \begin{bmatrix} -l\cdot \omega \cdot \sin(\phi_1)\\ l\cdot \omega \cdot \cos(\phi_1) + v_{cart}\\ 0 \end{bmatrix}$$ while the linear momentum of the other object $O_2$ is
$$\vec{P}_2 = m \cdot \vec{v}_2 = m \cdot \begin{bmatrix} l\cdot \omega \cdot \sin(\phi_1)\\ -l\cdot \omega \cdot \cos(\phi_1) + v_{cart}\\ 0 \end{bmatrix}$$ However at the moment when $\phi = \pi$ the mass is being transferred hence the object $O_1$ undergoes a change in linear momentum: $$ \Delta P_1 = -\frac{E_x}{c^2} \begin{bmatrix} 0\\ -l\cdot \omega + v_{cart} \\ 0 \end{bmatrix}$$ while the object $O_2$ undergoes a change in linear momentum $$ \Delta P_2 = \frac{E_x}{c^2} \begin{bmatrix} 0\\ l\cdot \omega + v_{cart} \\ 0 \end{bmatrix}$$ hence $$\Delta P_1 + \Delta P_2 = \frac{E_x}{c^2}\cdot \begin{bmatrix} 0 \\ 2\cdot l\cdot \omega\\ 0 \end{bmatrix}$$ Is this correct?

  • $\begingroup$ Obviously this question is very related to the other one where instead of a rotating mass I've considered a capacitor or something else ... and the energy is transferred using radiation . About the movement of the center of mass: in an ideal scenario the transfer occurs at $\phi_1 = \pi$ hence the center of mass cannot move sideways because its velocity on $x$ axis is zero due to constraints. $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 12:53
  • $\begingroup$ If I have answered satisfactorily to your objections, please consider up-voting the question back :) $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 13:11
  • $\begingroup$ ok I have removed my down-vote. I don't see any reason why you could not transfer mass instead of energy. The cart cannot move sideways, so a transfer of mass in the position shown will not alter the position of the CM. Isn't the movement of the CM what you are trying to avoid by transferring energy instead of mass? $\endgroup$ Commented Mar 5, 2018 at 14:01
  • $\begingroup$ The problem which I see is that, if the transfer of energy/inertia occurs when $O_1$ switches places with $O_2$ in the figure, then the cart will be back in its initial position. As $O_1$ moves up on the right in the +y direction the cart moves in the -y direction, and as $O_1$ moves down the left the cart moves back up in the +y direction. After each 'cycle' the cart is back where it started. $\endgroup$ Commented Mar 5, 2018 at 14:06
  • $\begingroup$ Sir, why are you saying that when $O_1$ is moving up the cart is moving down? I think the cart should also move up because of the greater centrifugal force $m\cdot l \cdot \frac{E_x}{c^2}$ acting on the object $O_1$ $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 16:06

2 Answers 2


just before the object is moving out of the upper plane, pointing-downward momentum changes by $\frac{E_x}{c^2} l \omega$, since you are reducing its mass somehow while moving, and you are increasing the mass of the opposite mass will moving in the opposite direction. So your hypothetical mass shifting is also shifting momentum around

  • $\begingroup$ I've made an edit with some rude analysis of variation of moments. Can you have a look at it? $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 16:36

Sometimes assumptions simplify problems, but sometimes they appear to allow the laws of physics to be violated. You want to transfer energy via radiation, rather than matter, to get an "instantaneous" transfer. But the fastest that radiation can travel is c. In this case you still have to direct the radiation slightly in the y-direction in order for the receiving object to receive it, because the receiving object will travel a finite distance during energy transfer.

However, even if energy (via radiation or matter) could be transferred instantaneously, you argue that at the moment of transfer all changes of momentum are in the x-direction. This cannot be true. It's easier to see why if you consider the case of energy transfer via matter. For the object giving up energy, at the moment of transfer, a small amount of it's matter has to be decelerated to zero in the y-direction and accelerated (to infinity) in the x-direction. So some matter makes an instantaneous 90-degree turn. Whatever force causes the deceleration in the y-direction must come from the cart, through the rod, so the rod must undergo some bending stress. The resulting forces from the rod will have x and y components. This will be true when transferring energy via radiation as well, as alluded to in the above paragraph. So the cart must lose some momentum in the y direction during energy transfer.

Does this make sense?

  • $\begingroup$ What are you saying, I think, is a sudden decrease in $P_1$ the linear momentum of the first object. But "in the same time" there is also a sudden increase in $P_2$, the linear momentum of $O_2$. $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 20:51
  • $\begingroup$ Yes, but both accelerations are in the positive y direction, because a deceleration in the negative y direction is an acceleration in the positive y direction. My explanation of what happens to the object losing energy is valid for the object gaining energy. But be careful of the plus and minus signs. So the total change in momentum for both objects is P1 + P2, (actually 2P1 because they are equal) in the positive y direction. The cart must provide the force for this, and hence gain momentum of 2P1 in the negative y direction during energy transfer. $\endgroup$
    – Tom B.
    Commented Mar 5, 2018 at 21:03
  • $\begingroup$ Have you looked at my analysis of linear moments? I arrive at the conclusion that $\Delta P_1 + \Delta P_2$ has zero component on $y$ axis ... $\endgroup$
    – C Marius
    Commented Mar 5, 2018 at 21:06
  • $\begingroup$ I have not checked your calculations, but I noticed you give a x-component of velocity as l w cos(phi), which is not zero at phi= zero, which I think is wrong. $\endgroup$
    – Tom B.
    Commented Mar 5, 2018 at 21:28
  • 1
    $\begingroup$ No. Y component of velocity is not zero at pi. Object is crossing X-axis at pi radians, so Y component of velocity is at a maximum. You have your sine and cosine reversed. $\endgroup$
    – Tom B.
    Commented Mar 5, 2018 at 21:48

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