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If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).

And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.

enter image description here

In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.

I guess that it is falling because of some torque but why is there no torque when it is upside down.

What is the reason behind this?

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  • $\begingroup$ In the second picture your position of finger is different. I guess it has to do with weight of air inside cup $\endgroup$
    – Danny LeBeau
    Commented Dec 12, 2020 at 7:26
  • $\begingroup$ "Today I noticed something that if I try to handle a tumbler or cup on my fingertip (as shown in fig) , it is quite hard to do so(that's why I used my all fingers to capture the photo)." Are you trying to convey that you used all your fingers to hold the cap? This text is a bit difficult to understand. $\endgroup$
    – Cantor Dust Drachen
    Commented Dec 12, 2020 at 8:23
  • $\begingroup$ @Buraian by saying I used my all fingers to capture the photo) , I meant that had I not placed my other fingers close to the cap I would have not been able to capture the picture in a stable mode. $\endgroup$
    – Ankit
    Commented Dec 12, 2020 at 9:13
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    $\begingroup$ This video could help you a little. $\endgroup$
    – Linkin
    Commented Dec 12, 2020 at 14:24
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    $\begingroup$ Does this answer your question? Balancing Utensils: Center of Mass $\endgroup$
    – Peter Duniho
    Commented Dec 13, 2020 at 3:02

7 Answers 7

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Take a look at this picture of a cup slightly out-of-balance :

enter image description here

In case (A), generated torque is directed out of your reference axis and in case (B) - towards your reference axis. So in case A), you need to compensate out of balance movement with your finger contra-movement. But in case B), torque assists you and makes balancing for yourself, so that you need minuscule additional efforts.

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    $\begingroup$ As a way to maybe intuitively showcase the difference: when cup A tilts, it will want to keep tilting further and further, awayt from the balance point. Cup B, however, when it tilts, it naturally wants to tilt back to its balanced position. This is a self-correcting system whereby any imbalance wants to correct itself automatically. The center of mass always wants to go down (due to gravity). With A, the COM goes down by tilting further. In B, the COM goes down by tilting less, which is why this cup "wants" to be balanced. $\endgroup$
    – Flater
    Commented Dec 12, 2020 at 17:03
  • $\begingroup$ @Flater Yes, true. In addition to that, system B is pretty much like a pendulum, aka damping oscillator. $\endgroup$
    – Agnius Vasiliauskas
    Commented Dec 12, 2020 at 18:13
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    $\begingroup$ The mathematical terms are stable and unstable equilibriums. Objects which balance on stable equilibriums are called "self-balancing". There's a neat application to some toys in this Numberphile episode $\endgroup$
    – BlueRaja - Danny Pflughoeft
    Commented Dec 13, 2020 at 10:34
  • $\begingroup$ @BlueRaja-DannyPflughoeft Harmonic oscillator can only be in a stable equilibrium, such as B case. $\endgroup$
    – Agnius Vasiliauskas
    Commented Dec 13, 2020 at 19:52
  • $\begingroup$ Your explanation employing the complex notion of torque hasn't been most likely understood by the person asking. $\endgroup$
    – Nick The Dick
    Commented Jan 7, 2021 at 21:09
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Maybe because when the cup is the right way up, it’s centre-of-mass is above the point on your finger meaning that as your finger tries to balance the cup any small motion will generate a torque about this c.o.m making it harder to balance.

When the cup is upside down, you have your finger on or going through the c.o.m and so any small motion by your finger will not generate torque about the c.o.m making it much easier to balance.

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  • $\begingroup$ but one more thing : does it make the cap more stable on the fingertip if the open part was covered with a circle of same radius ? $\endgroup$
    – Ankit
    Commented Dec 12, 2020 at 12:17
  • $\begingroup$ @Ankit more tendency to be unstable i suppose if the cap was in position of picture 1 $\endgroup$
    – user281869
    Commented Dec 12, 2020 at 13:05
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    $\begingroup$ Motion of your finger will cause a torque, though. The force of such motion is applied on the same place of the cup relative to the c.o.m, so, say, moving your finger to the side will give the same torque in both cases. It's just that gravity works either with or against any such torque, depending on which way the cup is rotated. $\endgroup$
    – Arthur
    Commented Dec 12, 2020 at 16:00
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enter image description here

Image courtesy google.

I could not manage a better picture than this one.

EXPLANATION

This is because when the cup you hold normally on one finger is in unstable equilibrium and mild deviations from the vertical line can cause it to fall. But there can be a case when you can manage it to be one one finger. Due to centre of mass of cup being above a decent height(even a mild deviation results in its falling.

That is why you need to use your other fingers to manage it to be in equilibrium.

You might have done the task of balancing book on one finger and spinning it or just balancing it.

Isn't that task easier?

You just need to find the equilibrium point and tada!

enter image description here

This is easier because the height of centre of mass above your point of contact is very small. Thus making it easier to keep it straight(sweet intution without using the name of torque).(and don't forget the friction there)

Another scenario

When the cup is upside down.

Then small deviations instead of opposing you help to bring the cup straight.(thus making your task easier)

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  • $\begingroup$ I don't know of which doubt you talked about but I have a doubt : Due to centre of mass of cup being above a decent height. What is this decent height ? $\endgroup$
    – Ankit
    Commented Dec 12, 2020 at 13:25
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    $\begingroup$ @Ankit decent height here refers to a considerable height $\endgroup$
    – user281869
    Commented Dec 12, 2020 at 13:29
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    $\begingroup$ also a sphere on a fingertip is I guess in unstable equilibrium.. $\endgroup$
    – Ankit
    Commented Dec 12, 2020 at 13:31
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    $\begingroup$ @Ankit But your fingertip isn't a sphere; it deforms. It's better to model it as a ring (with a flat surface in the middle that doesn't really matter). So long as the centre of mass is within the ring (when viewed from the top), the book's centre of mass has to go up for the book to tilt, so it wants to remain on your finger. $\endgroup$
    – wizzwizz4
    Commented Dec 13, 2020 at 16:55
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    $\begingroup$ @Ankit: A cone balanced on your finger would be unstable with point side up or down, because it's a tall shape with the centre of mass relatively high up. (Such that tipping a small angle will move the CM outside the contact patch). It's stable on a flat surface because the contact patch is the whole base of the cone, not just a point in the middle. Unfortunately this answer doesn't fully explain that different between the image and your finger-balance example, but the image is a good example of stable vs. unstable equilibriums. $\endgroup$
    – Peter Cordes
    Commented Dec 14, 2020 at 6:20
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Dr jh has explained by taking torque about the center of mass but it could be explained better if we take it about the point of contact of your finger and the cup.

When the cup is upright gravity provides a torque by a force acting on its center of mass which is above your finger. Small perturbations will cause rotation about the point of contact and it rotates away from its original position. It's an unstable equilibrium and it's kind of like balancing a pencil on your fingertips.

When it's upside down it's a stable equilibrium and any perturbations would be restored back to its original position with the help of gravity. Kind of like a pendulum.

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Think about the position of the center of mass of the cup.

Assuming it's not one with an oddly heavy bottom or flimsy walls, the wallwill place that center of mass around the middle of the hollow interior of the cup.

Your finger acts now as a fulcrum point. Thus, the center of mass and your finger tip, in effect, form a very weird pendulum, with the center of mass being the bob, and your finger tip the anchor.

Now, is a pendulum more stable and easy to control when it's pointing down, or when it's pointing up? Which way is this "pendulum" pointing (draw an arrow from your fingertip to the center of mass)?

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Think of it like this: The center of gravity of the cup wants to be as low as possible, right? Consider your fingertip as a fulcrum point. So when the cup is above your fingertip, so is the center of mass, so the slightest tilt allows the cup to rotate so that the center of mass drops. When the cup is the other way, the center of mass is below your fingertip. Even a large tilt will go back to equilibrium without falling off, because it rotates back to normal.

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Apparently simple question which teaches a lot about constraints.

I think the simplest one-sentence summary is that 'different' initial configurations dictate the final kind of equilibrium that is reached with the passage of time.

As in, the first configuration involves your finger the cap being balanced on the tip of your finger, now unless you have wide fingers, it's gonna be very difficult to keep the force you apply right on the center. If there is some inaccuracy from the center, then the gravity will twist and turn the object to tumble onto the least energy state where it will be on the ground.

To successfully achieve a balance, you need to position your fingers in such a way that the total torque generated by each finger is zero. It's a similar problem to figuring out where we should we keep the 'legs' of a table so that the structure is sturdy.

In the second configuration, your finger is being depressed a little bit and it's creating an opposite force to counteract that deformation, this counteracts gravity. Think of putting a weight on a trampoline, once it depresses, it exerts an opposite force for the object to bounced outwards.

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  • $\begingroup$ I don't understand why you are saying all the fingers . May be my question is not clear. I tried the experiment with just one fingertip $\endgroup$
    – Ankit
    Commented Dec 12, 2020 at 9:23
  • $\begingroup$ you had written this " I meant that had I not placed my other fingers close to the cap" , which means you were using more than just one finger to balance it. If I misunderstood kindly correct me (This was in the comments of the original post btw) $\endgroup$
    – Cantor Dust Drachen
    Commented Dec 12, 2020 at 9:26
  • $\begingroup$ unless you have narrow fingers - actually, wider fingers would help you balance it upside down more easily, letting it move farther while still having the CM above the contact patch. (You do of course want the contact patch centred under the CM). And yes, in both cases the cap's weight deforms your fingertip into a pad, not a point. That's not exclusive to the unstable configuration. Balancing it inverted on a pin would only be possible in theory, not practice; even if you found the perfect spot, Brownian motion might be enough to start it tipping. i.e. even harder than a fingertip. $\endgroup$
    – Peter Cordes
    Commented Dec 14, 2020 at 6:22
  • $\begingroup$ @PeterCordes Thank you for the response! The reason I mentioned deformation only for the inverted way is that I think the deformation is much more noticeable in that case than the other one. I'll add that point in soon. Could you explain that brownian motion point?/ Lead me to where I could read more about it $\endgroup$
    – Cantor Dust Drachen
    Commented Dec 14, 2020 at 7:01
  • $\begingroup$ In a truly unstable equilibrium, literally any perturbation will lead to a self-amplifying offset from the balance point. (snowball effect). I'm guessing that balancing a bottle cap on the tip of a pin would have a small enough balance point that air molecules hitting it (brownian motion; google it) could create such an effect, if you hypothetically managed to balance it in the first place in air (not a vacuum). Or if not, then any vibration of the earth that affects the pin would be a problem, moving it enough to start the cap tipping. $\endgroup$
    – Peter Cordes
    Commented Dec 14, 2020 at 13:51

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