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A lever with loads and a fixed pivot is not a closed system, because it is connected to a pivot, which is itself connected to the floor/ground/Earth. This is why it shouldn't be surprising that when I exert a force $\vec{F}$ on one side, the total momentum does not obey \begin{align}\tag{1} \vec{F} = \frac{d\vec{p}}{dt} \end{align} (when we push down on one end, the center of mass doesn't accelerate downward into the ground thanks to the pivot).

However, it seems we do have \begin{align}\tag{2} \vec{\tau} = \frac{d\vec{L}}{dt} \qquad\text{and}\qquad \vec{F}\cdot\vec{v}_{\text{point of application}} = P \end{align} where $P$ is the power or in this case simply $d(KE)/dt$ of the entire lever setup. The first says that the torque applied equals the rate of change in angular momentum, and the second says that the force times velocity equals the rate of change in kinetic energy.

Why is it that we say equations in $(2)$ hold, but $(1)$ does not hold? After all, the fixed pivot connected to the Earth means that momentum can escape into the ground, so why don't angular momentum and energy escape as well?

Let me put it another, shorter, way: Suppose we throw a ball that elastically collides with one end of the lever. Conservation of angular momentum holds, conservation of energy holds, but conservation of momentum doesn't hold because this is not a closed system and the momentum transfers to the ground through the pivot. Why do the first two hold, but not the third? Why the exception?

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  • $\begingroup$ Is the pivot located at the centre of mass of the lever or at an arbitrary position on the lever? $\endgroup$
    – Farcher
    Jul 28, 2023 at 13:54
  • $\begingroup$ @Farcher Yes, it is located at the center of mass. (I realize this makes a difference.) $\endgroup$ Jul 28, 2023 at 14:08
  • $\begingroup$ Say you just throw your ball up into the air. Doesn't that already violate conservation of momentum? The answer is "No" because in both cases, you've left out something (Hint: it's very large and both you and the pivot are supported by it) $\endgroup$ Jul 29, 2023 at 11:29

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Why is it that we say equations in (2) hold, but (1) does not hold?

The pivot affects the change in linear momentum because of its reaction force. But an ideal (frictionless) pivot has no moment (torque) reaction. Thus it does not effect angular momentum.

Hope this helps.

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Neither of your equations 1 or 2 hold. The equations that hold are $$\sum_{i} \vec F_i = \frac{d\vec p}{dt}$$ $$\sum_{i} \vec \tau_i = \frac{d\vec L}{dt}$$ $$\sum_{i} P_i = \frac{dE}{dt}$$

Where the $i$ indicates the individual external interactions acting on the system of interest. In other words, you have to sum over all of the external interactions on the lever, and when you do so any change in the lever's momentum, angular momentum, or energy is due to the total external force, torque, or power respectively.

These equations assume mechanical forces only, but can be generalized to include fields and heat as well. But these equations apply equally to a lever or any other mechanical system.

Suppose we throw a ball that elastically collides with one end of the lever. Conservation of angular momentum holds, conservation of energy holds, but conservation of momentum doesn't hold because this is not a closed system and the momentum transfers to the ground through the pivot. Why do the first two hold, but not the third? Why the exception?

It is not an exception, it is simply applying the standard formulas for the calculation of each. There are two forces acting on the lever the force from the pivot, and the force from the ball.

Since the center of mass of the lever is not accelerating we know from Newton's 2nd law that $$\sum_i \vec F_i = m\vec a = 0$$ $$\frac{d\vec p}{dt}=0$$ so $\vec F_p = -\vec F_b$ where $\vec F_p $ is the force at the pivot and $\vec F_b$ is the force from the ball.

For torque $\vec \tau_i = \vec r_i \times \vec F_i$. Taking the torques about the pivot $$\frac{d\vec L}{dt} = \sum_i \tau_i =\tau_p + \tau_b = 0 \times \vec F_p + \vec r_b \times \vec F_b = \vec r_b \times \vec F_b \ne 0$$

For power $P_i = \vec F_i \cdot \vec v_i$. This gives $$\frac{dE}{dt}=\sum_i P_i = P_p + P_b = \vec F_p \cdot 0 + \vec F_b \cdot \vec v = \vec F_b \cdot \vec v \ne 0$$

So there are not execptions to the equations listed above. The difference is that for momentum the terms cancel each other, while for both torque and power one of the terms is 0 and does not cancel the other.

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  • $\begingroup$ I think you did not see that one does not account for dp/dt from Earth, since the mass of the earth is so large compared to the math of your lever system, that you can not measure $\frac{dp}{dt}=ME*v_E$ $\endgroup$
    – trula
    Jul 28, 2023 at 13:46
  • $\begingroup$ @trula the $\vec p$ is the momentum of the system of interest. Here that is the lever, not the earth. We are accounting for $d\vec p/dt$ of the system only, not the earth. $\endgroup$
    – Dale
    Jul 28, 2023 at 14:05
  • $\begingroup$ So then my question reduces to, which $P_{i}$'s, $\vec{F}_{i}$'s, and $\vec{\tau}_{i}$'s are non-zero and why? $\endgroup$ Jul 28, 2023 at 14:32
  • $\begingroup$ @MaximalIdeal those are all calculated according to their standard equations. Would you like me to add the standard equations to the answer? $\endgroup$
    – Dale
    Jul 28, 2023 at 15:12
  • $\begingroup$ @Dale Basically yes. I already see how to answer my own question, but I think your reply could be more complete if you explain why the pivot "transfers momentum" but not angular momentum and energy (actually it does transfer energy but it is infinitesimal and zero in the limit of infinite mass). $\endgroup$ Jul 28, 2023 at 15:38
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The torque is force times distance from the fulcrum. Since the fulcrum is the system's connection to the Earth, force between the the system and the Earth must be applied at the fulcrum, and so the distance from the fulcrum will be zero, and the torque transferred will be zero.

Let me put it another, shorter, way: Suppose we throw a ball that elastically collides with one end of the lever. Conservation of angular momentum holds, conservation of energy holds, but conservation of momentum doesn't hold because this is not a closed system and the momentum transfers to the ground through the pivot.

Huh? Is the ball hitting "one end of the lever, or is it hitting "through the pivot"? If a ball hits one end of the lever, then the momentum is not being transferred through the pivot, it's being transferred through one end of the lever, and so there will be transfer of torque. There will be another force applied through the pivot, but the initial force is not.

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Why is angular momentum conserved in general? You may know that conservation laws come from symmetries of the Lagrangian.

The Lagrangian is symmetric under space translation? The system will conserve momentum

Symmetric under time translation? Conservation of energy.

Symmetric under rotation around a certain axis? Will conserve angular momentum.

All fundamental forces that we know of possess all these three symmetries, and therefore all the other, derivative, forces will as well, when you consider a closed system.

Think about gravity. The attraction between two particles depends only on the masses and the distance. If we keep the distance and masses constant, the force won't change if the two particles are in Europe or in Australia (space translation), nor does it change if we measure the force today or tomorrow (time translation), nor if we rotate the two particle in space around any axis (rotation). So gravity conserves energy, momentum and angular momentum.

One example in which there is no conservation of energy: a system is inside an elevator that is going up and down. The potential energy of everything in the system depends explicitly on time. Doing experiments while the elevator is accelerating upwards is not the same as doing them when the elevator is going downwards. So energy is not conserved inside the elevator.

In your case the lever has a pivot, which is a rigid constraint that defines a special axis of the system. You still conserve angular momentum, because the system is symmetric under rotation around the pivot. You still conserve energy because the pivot is not changing in time, is just staying there. But linear momentum is not conserved because the pivot fixes a point in space. The system is not invariant under space translation because the pivot moves. So linear momentum is not conserved.

Let me know if this was clear enough. If you are familiar with Lagrangians, I can throw some maths in to make my statements more precise

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  • $\begingroup$ Huh, I was aware of the Noether correspondence you mentioned, but I never thought of applying it here. Thank you for the novel perspective. I really like it! $\endgroup$ Sep 4, 2023 at 19:50
  • $\begingroup$ @MaximalIdeal I like your questions! I'm working on Newton's Cradle one, it's very cool $\endgroup$
    – Prallax
    Sep 4, 2023 at 20:06

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