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Good afternoon all,
I'm trying to model a physics problem and would greatly appreciate some help.

The situation that I'm looking at involves a person on a carousel, and the angular velocity required to send them flying off. From the research I've done so far, it seems that this can be modelled from both an inertial frame of reference (a stationary observer) and a rotating frame of reference (the person on the carousel), where the forces will be different depending on the frame used, however I believe everyone should be able to agree on the person's motion?

I've come across this set of notes (https://taleinav.github.io/Lectures/Ph%201a/Lecture%205%20-%202017-10-12.pdf) and I think I understand the example given on page 3, where the person is in uniform circular motion on the carousel, and the centripetal and centrifugal forces are of equal magnitude in their frame of reference. However, intuitively, I feel that if the carousel were to reach a high enough (constant) angular velocity, the person would go flying off.

This is the explanation I have so far:

Inertial Reference Frame
The person's weight is balanced by the normal force. The centripetal force is provided by friction between the person's feet and the carousel. For a high enough speed, the person's inertia (tendency to continue in a straight line) is too great and the centripetal force is no longer strong enough to keep them in circular motion. Therefore, they are sent flying off the carousel.

Rotating Reference Frame
The person's weight is balanced by the normal force. The centripetal force is provided by friction between the person's feet and the carousel. The person experiences a centrifugal force, which pushes them away from the centre of the carousel.

My questions are:

  1. Is there anything I've got so far that is wrong?
  2. In the inertial frame, is there a way of quantifying this inertia in order to calculate the angular velocity to send a person flying off the carousel?
  3. In the rotating frame, is the magnitude of the centrifugal force equal to $mr\omega^2$, or is it something else?

Thanks :)

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  • $\begingroup$ Minor nit pick: You said, "...and the centripetal force is no longer strong enough to keep them in circular motion." That would make sense if you change "centripetal force" to "static friction." Centripetal force, by definition, is the force that keeps the person on the circular path. Before the person slips, the static friction between their feet and the carousel is centripetal. It is exactly the force required to maintain their circular motion. After the person slips, there is no longer any centripetal force because they are no longer following that path. $\endgroup$
    – Solomon Slow
    Commented Mar 13, 2023 at 13:41
  • $\begingroup$ Ah I see, that makes sense. Thanks very much @SolomonSlow $\endgroup$
    – Thomas
    Commented Mar 13, 2023 at 14:46

1 Answer 1

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  1. I believe you're correct in your verbal analysis.

For 2, what you can do is quantify what acceleration an object going in a circle is experiencing ($mr\omega^2$).

From there, since you're in an inertial frame, you know that it must come from a force acting on the object (friction, in this case).

Then, you can see if the cause of the force in question (friction) can actually produce the required force. Eg $|F_{friction}| \le \mu mg$, so, the circular motion is possible if $ \mu mg \ge mr\omega^2$

  1. Correct.
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  • $\begingroup$ Thanks very much, Alex. Something else that I'm now wondering based on the third point, is what happens when friction is greater than the centrifugal force? I would have guessed that the person stays in uniform circular motion while not moving with respect to the carousel in their reference frame, but this doesn't seem right, as the forces on them are not balanced? @AlexK $\endgroup$
    – Thomas
    Commented Mar 13, 2023 at 16:28
  • $\begingroup$ @Thomas static friction is never greater than the forces it opposes -- that's why we write $\le$. $\endgroup$
    – Alex K
    Commented Mar 13, 2023 at 16:36
  • $\begingroup$ Ah I see, I think I slightly misread that line of your answer and didn't see the inequality, that makes perfect sense. Thanks very much, your answers have been really helpful. @AlexK $\endgroup$
    – Thomas
    Commented Mar 13, 2023 at 16:41

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