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Let's say you have a body rotating about its centre of mass in a rotating frame. But the centre of mass of the body does not coincide with the rotation axis of the rotating frame.

We can often see such a set-up in real life when we go to amusement parks.enter image description here

This spinning tea cup ride (also called "Mad Tea Party") consists of many individual cups spinning about their own axes of rotation, which are all then placed on a large circular spinning plate.

Let the frame of the large spinning plate be $S_1$, spinning at angular velocity $\omega_1$ with respect to the lab frame. Let the frame of the tea cups with $S_2$, spinning at angular velocity $\omega_2$ with respect to their individual axes of rotation (labelled in the diagram as $\mathbf{z_1}$, $\mathbf{z_2}$...).

What will happen if one of the tea cup is tilted in a fixed direction wrt to the bottom spinning plate, such that it's angular momentum is not pointing straight upwards (but tilted to the side) How would you add up the 2 angular momentum in this case? Since the angular momentum changes direction, there must also be a torque. About which point is the torque applied?

In other words, in 3D space, where will be the "origin" of the angular momentum vector (for example of a pendulum, angular momentum vector will originate from the pivot point of the pendulum)

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3 Answers 3

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You need to add the spin angular momentum to the orbital angular momentum.

The spin angular momentum is the angular momentum of the body about its centre of mass and the orbital angular momentum is the angular momentum of the centre of mass of the body about the axis defined in the lab frame, shown as a red blob in your diagram.

So for the top left cup, $L_{\rm spin} = I_{\rm \text{CoM cup}}\,\omega_1$ and $L_{\rm orbital} = I_{\text {red blob}}\,\color{red}{\omega_1}$

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    $\begingroup$ Hi! Thanks for your answer. Can you write out the mathematical expression for that? How about the total angular velocity of the individual tea cups with respect to the lab frame? $\endgroup$ Jan 2, 2022 at 11:47
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    $\begingroup$ I was doing that as you were writing your request. $\endgroup$
    – Farcher
    Jan 2, 2022 at 11:50
  • $\begingroup$ How would the vector diagram look like (for adding up the 2 angular momentum vectors)? Thank you $\endgroup$ Jan 2, 2022 at 12:24
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    $\begingroup$ The reason I only stated what the magnitudes were is because you did not define the direction of the rotation of the turntable. If it is clockwise looking from above then you add the two magnitudes and the direction will be $-\hat z_1$. $\endgroup$
    – Farcher
    Jan 2, 2022 at 12:34
  • $\begingroup$ What will happen if one of the tea cup is tilted in a fixed direction wrt to the bottom spinning plate, such that it's angular momentum is not pointing straight upwards (but tilted to the side) How would you add up the 2 angular momentum in this case? Also like vectors? Since the angular momentum changes direction, there must also be a torque. About which point is the torque applied? Thank you (maybe I should edit my question and add these inside as well) $\endgroup$ Jan 2, 2022 at 13:24
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The answer is, that there are forces/torques you are overlooking.

Suppose, looking directly down from above, we focus on a teacup to the "south" (bottom). Quarter of a rotation of the entire ride later, its now to the "west" (left) instead. So the cups axis of rotation itself, has shifted.

That took place because of forces at the point it's mounted to the ride. Your summary overlooks those.

Suppose the original teacup, at the "south", had an axis tilted from bottom left to top right (rising west to east). When the ride turns quarter of a turn, that teacup now has an axis rising from bottom-south to top-north, and different rotation, as well as the teacup itself having a northerly velocity vs a westerly one.

All of those changes occurred because the ride can force the mount-point in the given direction, and the mountpoint can transmit that change to the teacup, with more than enough force that the teacup changes its movement and axis, if you want an informal explanation. Because reactions have an equal and opposite reaction, there will be an equal and opposite set of forces and torques on the base of the ride - but the opposite position teacups probably balance each other out, leaving little net force/torque (ride overall isn't changing position etc).

If you want a simple but not exactly the same parallel example, when you put a block on a rough level surface and push it very lightly, it doesn't move. The reason is that friction and reaction forces occurs at the junction, and unless you take that into account, it won't make sense when you analyse it.

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In other words, in 3D space, where will be the "origin" of the angular momentum vector (for example of a pendulum, angular momentum vector will originate from the pivot point of the pendulum)

Angular momentum is not in fact a vector; it can be put in bijection with a vector, but as this question shows, there are differences between them and actual vectors.

Each axis has its own angular momentum. The angular momentum has magnitude, and the axis has direction, so this combination of magnitude and direction looks like a vector, but it isn't really, and it doesn't have an origin in physical space. It makes sense to depict it as having its origin on the corresponding axis, and often the axis being considered goes through the center of mass, making the center of mass a natural choice for the origin, but the choice to depict its origin there doesn't really correspond to anything physical.

Even angular momentum "vectors" with the same "origin" can't be added like normal vectors if they have different axes (you get things like precession that wouldn't appear if they were truly vectors), and it's even more complicated to combine angular momenta from different axes.

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  • $\begingroup$ You mentioned that it is even more complicated to combine angular momenta from different axes. That is the case for the spinning tea cups right? Could you elaborate a bit more on how to do it? Thank you $\endgroup$ Jan 3, 2022 at 1:18

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