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Why it makes no sense is because centre of mass (COM) is a place where the mass of the whole body is present. Why can't we think like this: "When the semicircular wire is stretched and made to a straight wire the COM will be at the centre, right?". But COM of semicircular wire is $2R/\pi$ from the centre.

If we think in real life the COM of semicircular wire is in air? What? Can anyone please explain it to me? Or is the way I am thinking wrong?

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    $\begingroup$ Where do you think the COM of a circular ring is, or should be? $\endgroup$
    – G. Smith
    Commented Sep 30, 2020 at 19:54
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    $\begingroup$ The COM does not have to be in the body of mass. Look at a horse shoe. $\endgroup$ Commented Sep 30, 2020 at 21:04
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    $\begingroup$ Can you hold the circular ring at the centre? No. As with the semicircular wire, there’s just air there. Are you concerned about the COM being in air for a ring? $\endgroup$
    – G. Smith
    Commented Sep 30, 2020 at 21:59
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    $\begingroup$ Yes, that’s right. $\endgroup$
    – G. Smith
    Commented Oct 1, 2020 at 5:25
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    $\begingroup$ The answers have many counter cases, but I find one particularly useful. Given any object whose COM is "in" the object, you can easily construct an object whose COM is outside of the object by simply removing a bit of material around the COM. Start with any object you please. $\endgroup$
    – Cort Ammon
    Commented Oct 1, 2020 at 9:04

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The center of mass is an "average" position of the mass of the system, so there's no reason to suggest there's actually mass there. In the same way, the average of outcome when you throw a $6$-sided die is $(1+2+3+4+5+6)/6=7/2$, which is not the value of any face on the die. This average value is certainly real.

To take a more extreme example, the center of mass of a ring with uniform mass distribution is the center of the ring, something intuitively clear. Moreover, from the symmetry of the ring, it does't make any sense to have the COM on the ring itself as all points of the ring are equivalent.

The location of the COM is certainly real, and the force does act as it all the mass was located there, even if there is no actual mass at that point.

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You comment that the location of the center of mass "doesn't make sense" because we can't see it with our naked eyes. But in fact we can see it by doing a very simple experiment. Suspend the wire from a single point with a post or something and draw the line vertically downward from the point of suspension.

enter image description here

Then suspend the wire from a different point and do the same thing. The point where the two lines intersect gives the position of the center of mass.

enter image description here

So we don't have to take it on faith that the prediction for the location of the center of mass is correct. We can do an experiment to check that our prediction is consistent with the reality we observe.

(Strictly speaking, this procedure gives us the position of the center of gravity instead of the center of mass, but the difference between these is not really important for answering your question.)

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  • $\begingroup$ Does it mean that the gravity plays a role in centre of mass ? $\endgroup$ Commented Sep 30, 2020 at 20:22
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    $\begingroup$ The way that we define and calculate the center of mass has nothing to do with gravity. But the way that gravity acts on an extended object does depend on the center of mass / center of gravity of the object. $\endgroup$
    – d_b
    Commented Sep 30, 2020 at 20:26
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    $\begingroup$ @Angelinevarghese You can also imagine pushing on it in other ways. Push on one part of the ring so that it moves but does not rotate. Now push on another part of the ring and find another force that makes it move but not rotate. The directions of these two forces defines two lines, and the intersection of the lines is the center of mass. $\endgroup$ Commented Sep 30, 2020 at 21:26
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    $\begingroup$ @wizzwizz4 Technically yes, but for a wire with an everyday length scale near the surface of the earth, the difference is negligible. Try calculating what the field gradient will be over a wire with diameter 1 meter and then figure out what affect this has on position of the center of gravity relative to a uniform field. It's a really small effect. $\endgroup$
    – d_b
    Commented Oct 1, 2020 at 21:16
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    $\begingroup$ You're right. I drew it freehand with a pen, so it didn't come out exactly right, and I didn't want to redraw it. :) But the exact location of the c.o.m. is basically irrelevant to addressing the question. The point is that it's at some point outside the extent of the wire itself and we can see that point pretty easily with a simple experiment. $\endgroup$
    – d_b
    Commented Oct 1, 2020 at 22:19
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Why does it make no sense because centre of mass(COM) is a place where the mass of the whole body is present.

This is not true in general, and really is only true for a point mass, as by definition all of its mass is located at a single point. For an extended body or a system of particles the center of mass is not where all of the mass is present. For example, not all of your mass is located at a single point; it is located all throughout your body.

Why can't we think like this 'When the semicircular is streched and made to a straight wire the COM will be at the centre right? 'But COM of semicircular wire is 2R/3.14 ,from the centre. If we think in real life the COM of semicircular wire is in air? What?? can anyone please explain it to me ?

Yes, at first it is odd to think about the center of mass not being located where any mass is, but it follows directly from the definition of center of mass.

The center of mass of a system is just a weighted average of the position of the masses in the system, where the weights are the masses of each part of the system. For a discrete set of $N$ point masses, we have

$$\mathbf r_\text{COM}=\frac{\sum_{i=1}^Nm_i\mathbf r_i}{\sum_{i=1}^Nm_i}$$

and for a continuous mass distribution with density function $\rho(\mathbf r)$ we have

$$\mathbf r_\text{COM}=\frac{\int \mathbf r\,\text dm}{\int\,\text dm}=\frac{\int \mathbf r\rho(\mathbf r)\,\text dV}{\int\rho(\mathbf r)\,\text dV}$$

As an analogy, the reason the center of mass of a system can be outside of the body is the same reason why the average of a set of numbers does not need to be in the set of numbers. For example, the average of the set $\{1,2,2,4\}$ is $2.25$, which is not in the set. The average position of the mass of the system does not need to be the position where mass is actually located.

As a much simpler example, imagine two identical point masses separated by a distance $d$. The weighted average of the position of these masses will be found a distance $d/2$ from each mass on the geometric line joining them. Of course there is no mass here, but the weighted average of the position of the particles is located here.

From comments:

Classical Physics is about the things that we see in our daily life with our naked eyes. But this case it is something we have to believe and which don't make sense when we really see a semicircular wire in real life

You can definitely "see the center of mass" even if it is not located within the body. For example, take your rigid semicircle wire and throw it through the air so that it spins around while doing so. Even though various parts of the wire will be moving around, if you were to track the center of mass location of the wire you would find it follows a smooth parabolic shape, consistent with Newton's second law for extended rigid bodies.

This is getting somewhat philosophical, but one could make the same argument about velocity vectors. It's not like when we move around there are actual arrows pointing in the direction of our motion, yet we still talk about velocity vectors, and they are useful in explaining the world around us. In the same way, even though the center of mass is not an actual physical thing, it is a useful concept that can describe a lot of physical phenomena. If you think that this isn't sufficient to be considered as "real", then that's fine. The physics and the universe don't really care how to interpret their behaviors.

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  • $\begingroup$ Maybe it is good to note that for some calculations it is common to treat bodies as point masses, with the mass located at the center of mass. For simple linear dynamics, but also for planetary motion, this works quite well. $\endgroup$
    – Bernhard
    Commented Sep 30, 2020 at 19:29
  • $\begingroup$ Does is it mean that COM is not real? Is it just an assumption?But we see it around us eg:when we hold a pencil at the centre without touching other parts of the pencil $\endgroup$ Commented Sep 30, 2020 at 19:30
  • $\begingroup$ @Angelinevarghese It depends on what you mean by "real". $\endgroup$ Commented Sep 30, 2020 at 19:32
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    $\begingroup$ @Bernhard Yes, that is true, although I don't see what value it adds here. However, if you wanted to make an answer with that point then you should! I am sure the OP would welcome many different points of view. $\endgroup$ Commented Sep 30, 2020 at 19:38
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    $\begingroup$ @Angelinevarghese - I'm no physics expert, but for concepts like this (COM, potential/kinetic energy, etc.) I think of them like this: they're not "real" in the sense that there's nothing like them in the physical world. In the real world you just have quantum particles and the forces between them. But these "abstract" concepts are useful as mathematical shortcuts. They make it A LOT easier to think about some problems. You might find that they're not appropriate for ALL problems, but for many they work quite well. $\endgroup$
    – Vilx-
    Commented Oct 3, 2020 at 22:06
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"(COM) is a place where the mass of the whole body is present."

Well, that's where your misconception lies - you are taking this statement too literally.

When you want to simplify a physical situation, and pretend that the body is just a point mass, the COM is, conceptually, just a place (anywhere in space) where that point should be so that the physics and the associated math work out.
That's really all there is to it.

It's a simple form of modeling: you're representing the actual physical object as something else - in this case, a mathematical dimensionless point endowed with mass, and you're ignoring the actual shape of the object. This works as long as you're solving a problem that doesn't require taking the shape (or other ignored properties) into account (i.e., the problem is such that it suffices to abstract away the details of the object and think of it as being a point mass).

The COM turns out to be just a weighted average of the positions of all the individual bits that comprise the object; for details, see BioPhysicist's answer.

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Bend your semicircular wire to transform it into a ring. If the center of mass of the semicircle were at the center of the wire, then it should (by the same logic/intuition) remain there after you connect the ends of the wire. But that makes no sense: the ring has rotational symmetry, so the only sensible placement of the center of mass is at the center of the ring.

This also means that, while you were bending the wire to get the ring, the COM should have been moving from its initial position to the center. You can see how the distribution of mass influences the position of the COM.

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I noticed your comment on d_b's answer. Let me try to give a way to figure the COM that does not rely on gravity. Let's try no (low) gravity.

The COM is essentially the point of rotation when no forces are applied.

In this video, you see a gyroscope rotating in the International Space Station, after a gentle nudge. Let's focus in the first 20 seconds where the astronaut displays random motion and NOT gyroscopic effects. The whole gyroscope seems to be rotating around its centre point.

Now, imagine that instead of the gyroscope, you had just one ring. Kind of like this (mspaint skills): enter image description here

What do you think the ring would do in that case? Exactly the same thing. It would rotate around its centre point. Because that's where the COM is.

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If the wire shape is antisymmetric with respect to the central point that center point is COM.

Else in general COM will be away.

By coiling spiralling arrangement of a flexible wire COM can be adjusted to be on the wire.

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As pointed out, the center of mass definition does not require it to be located in the object. An interesting example is a belt balancing off of a finger. The center of mass of the belt is in the air directly below the pivot point so it's balanced (no net torque).

Question of Balance

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The center of mass is the point where if a force that passes directly through the COM is applied to any point on the mass it will cause translation of the mass without rotation. When the wire is straight its COM is in the center, as you thought, but when we bend the wire into a semi circle we are changing the position of much of the mass. This also changes the position of its COM to a point that is not necessarily inside the mass. Many objects may have a COM that is outside of the mass itself, most empty drinking glasses, rings, horseshoes, and other objects.

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The center of mass is in some sense the average position of the bodies constituents. Now the average value of some set of numbers can be defined variously depending on context, mean, median, mode, rms etc. In any context we can define what sort of average will be most helpful for our purpose.

For an insect standing at one end of the wire and wishing to visit other parts, the average length of its journey is defined in the way you would like. It is hard to think of many other situations where your average would be useful.

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