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This question already has an answer here:

I found after searching that This question has been asked before . But all the answers were not convincing.

Suppose I have a body which is free, not constrained always rotate about its center of mass. Why is that so? A convincing answer that I found was that in most cases moment of inertial about center of mass is the least and that's why the body rotates about the center of mass.

But I ask it again with hope of the question not getting closed and getting a better succint answer.

Edit

I was thinking that motion about the COM is the most stable one and the rotation about other points degenerates. I don't think it's right . Is it ?

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marked as duplicate by ja72, Jon Custer, M. Enns, David Hammen, Kyle Kanos Aug 1 '17 at 11:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Because it happens to rotate about a point - and that point is named Centre of mass. I guess your actual question is: why is there such a point at all? $\endgroup$ – Steeven Feb 12 '17 at 16:01
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    $\begingroup$ Related: physics.stackexchange.com/q/53465/2451 , physics.stackexchange.com/q/81029/2451 and links therein. $\endgroup$ – Qmechanic Feb 12 '17 at 16:01
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    $\begingroup$ What were the answers which you found "unconvincing"? Why were they "unconvincing"? You need to explain, and to provide links. $\endgroup$ – sammy gerbil Feb 12 '17 at 19:42
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    $\begingroup$ @Steeven : The question Why does the body rotate about some point? is the same as asking Why does the body rotate at all? $\endgroup$ – sammy gerbil Feb 12 '17 at 19:46
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    $\begingroup$ If you cannot state objectively what you mean by convincing/unconvincing then how can anyone know that their answer will satisfy you? $\endgroup$ – sammy gerbil Feb 12 '17 at 19:50

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You presumably already know that in the absence of external forces, the center of mass of any collection of particles moves at a constant velocity. This is true whether they are stuck together in a single body or are just a bunch of separate bodies with or without interactions between them. We now move to a frame of reference moving at that velocity. In that frame the CofM is stationary.

Now suppose that the particles are indeed stuck together to form a rigid body. We see that the body is moving so that: 1) the CofM remains fixed, 2) all the distances between the particles are fixed. (This second condition is what is meant by a $rigid$ body after all).

A motion with these two properties, (1) and (2), is precisely what is meant by the phrase ``a rotation about the CofM''

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    $\begingroup$ Great explanation. So the body rotates about the COM because it has to stay rigid. Isn't it ? Had I analysed the motion in frame in which the COM was not at rest or say was in rotation ? What would be the case then ? Probably the rest frame in which COM is in rotation would give the trouble in this case. Isn't it ? And if I say that the moment of inertial about the com is least ( checked it for many discrete cases) that's why its the com about which the body rotates . Would it be correct ? $\endgroup$ – Shashaank Feb 12 '17 at 16:38
  • $\begingroup$ Im not sure about the role of the minimal moment of inertia. The rotation is just geometry. The moments of I. come into the dynamics: If the three pricipal moments differ, the angular velocity vector can be a complicated function of time. $\endgroup$ – mike stone Feb 12 '17 at 16:50
  • $\begingroup$ @mikestone Ok. No doubt your answer is a great explanation and I will accept it .But could you also provide an argument when we observe the rotation from the ground frame . $\endgroup$ – Shashaank Feb 12 '17 at 17:38
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    $\begingroup$ I'm sorry, I don't see how this explains that a centre-of-mass exists. Why couldn't all particles still keep the same distance to each other (remaining rigid) if they rotated around an edge-point instead? $\endgroup$ – Steeven Feb 12 '17 at 17:46
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    $\begingroup$ @Steeven, because then the COM would be rotating (accelerating) around some other axis, which violates the rule that a COM moves at constant velocity when no external forces are acting on it. $\endgroup$ – user1717828 Feb 12 '17 at 18:31
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Here is one more way to look at this:

You can consider an object with any shape as a single point where all the mass of the object is concentrated. This point is called the center of mass. From Newton's second law, as no force is acting on the object, the center of mass must either move in a straight line or be stationary. If the body rotates, the only way the center of mass can obey that law is if the rotation is around the center of mass.

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The reason that a body under free rotation rotates about its center of mass is that the moment of inertia tensor at the center of mass is a minimum. When you rotate about any point that is not the center of mass, you have to apply the parallel axis theorem.

$$I' = I_\mathrm{CM} + m\vec{r}_\mathrm{CM}^2$$

The minimum of this equation is when the radius from the center of mass to the axis of rotation is zero. Therefore, the center of mass is the point of rotation that provides the least resistance to rotation.

In fact, the instantaneous center of rotation doesn't instantaneously shift to be at the center of mass of the object once the external forces stop acting on the object. Imagine you have a bowl and you drop a ball into it so that it's initial point of contact is close to the rim. The ball will tend towards the bottom of the bowl as that's the location with the lowest gravitational potential. However, before it gets there, it oscillates a bit before coming to rest. The bottom of the bowl is a stable point.

This is analogous to our rotation. The point about which the object rotates is initially offset from the center of mass. However, as time progresses, it tends towards the center of mass as it tries to find the path of least resistance. Rotation about the center of mass provides this least resistance.

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  • $\begingroup$ Um, the latter part of your answer cannot possibly be correct. You don't believe that bodies continue to accelerate for a while even after external forces stop acting on them, do you? Now, given that a body rotating around any axis that doesn't pass through its CoM must experience a non-zero net acceleration, I'm sure you can see the contradiction. $\endgroup$ – Ilmari Karonen Feb 12 '17 at 19:14
  • $\begingroup$ Perhaps you are right. But my intuition tells me that the center of rotation can't shift infinitely fast as that would entail an infinite angular acceleration about the axis of rotation. The inertia tensor at the principal axes of the object acts as a stable point, and rotation approaches that stable point during free motion. To put it succinctly, I see the angular momentum vector as shifting continuously toward the stable point once the external forces cease to act on the body. $\endgroup$ – gdbb89 Feb 13 '17 at 1:18
  • $\begingroup$ Easy to test: spin a large ring, such as one of those frisbee-like rings, on a stick . Clearly it's not spinning about its CM. Now release the ring (remove the stick). What does the ring start spinning around? --- $\endgroup$ – Carl Witthoft Feb 13 '17 at 16:39
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Imagine two stones tied together with a massless rod and let one stone to rotate around the second one being fixed.

In that case there must be a force that accelerate the first stone perpendicular to its velocity and causes it to rotate around the second one. The whole setup is free, so there is no counter force to equalize and this setup violates Newton's laws.

If we want to rotate this stone-rod-stone body with respect of Newton's laws we must add and arbitrary point it will be rotating around. In this case both stones are revolving around this point, radial force is applied to both of them and they have opposite direction. The forces must cancel out completely and they cancel out only if the arbitrary point is placed exactly in the centre of mass.

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I'm not a physicist, but I'll take a stab at it.

A simplified example of your spinning sphere that may help you with this concept would be a disk made all of one density of material. An example would be a child's top or a gyroscope that you can spin on a flat surface. Every part of the disk has a matching balancing part on the opposite side of the disk. Each balancing pair of parts of the disk have the same mass as each other, have opposite motions to each other when rotating and create opposite balancing centripetal forces that keep the disc's rotation balanced around the center of mass (which is also the disk's geometric center).

If you add more mass to the disc anywhere but at the center, the center of mass of the disc shifts away from the geometric center of the disk and toward the mass you just added. The object will now rotate around this new center of mass. This is because all the mass on the side away from the new added mass must create a balancing opposite force to the now heavier side of the disk. The mass of the disc between the geometric center of the disc and the new (shifted) center of mass shifts to becomes the opposing balancing force to the added mass.

The picture below may help you visualize this:

Spinning Disc

The green dot on the right is the original center of mass and center of the disc. The blue circle is an added mass. The green dot on the left is the new center of mass. The area between the two red lines is mass on the disk that balances the added mass when rotating. Adding more (blue) mass will shift the center of mass further from original center and move the left red line (and center of mass) further toward the added mass (left). If the original disk was very massive relative to the added mass, the center of mass won't shift as far (i.e. less area between the red lines needed to balance the new mass, and less shifting of center of mass to balance the added mass).

So to conclude, every time you add to (or subtract from) the mass of a rotating object, the object changes the location of it's center of rotation so that the forces caused by rotation remain in balance. The point of rotation is the center of all the mass of that object.

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I was thinking that motion about the COM is the most stable one and the rotation about other points degenerates. I don't think it's right . Is it ?

Let's go with this for a second. I'm not sure the term 'degenerates' it completely right here but I think you are on the right track. Consider a perfectly balanced wheel on an automobile. Its rotation is not free but rather fixed in its center which is also its center of mass (because it's balanced.) When it spins, there is no force on the axle.

Now consider what happens if we attach a weight to the rim of the wheel and make it unbalanced. When wheel spins, it will now apply forces to the axle. If you've ever driven in a vehicle in such a situation, you will feel this as a vibration at most speeds as the wheel continually 'jumps'. Why does this happen? It's because the axle is forcing the wheel to rotate around a point that is not its center of mass. In other words, only rotation around the center of mass is neutral; in order for an object to rotate around another point, another force is required to keep it in place. By definition, a 'free' object is not subject to any such force.

One way to see this is to take a frisbee and spin it around a finger inside the rim. It will rotate around your finger (which is not at the center of it's mass.) Your muscles will need to constantly resist the motion in order to keep it in place. If you suddenly remove your finger, it will fly off in a straight line and continue to spin around its center of mass.

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I understand this to be a mathematical or psychological phenomenon rather than a physical one.

An object can rotate about any axis. However, in the case where the axis does not go through the center of gravity, we would typically decompose this motion into a movement of the object's center of gravity, combined with motion about the center of gravity. You can always do this. Just ask 'How did the center of gravity move?' and subtract that movement from the movement of each piece. By definition, the remaining movement is a rotation about the fixed center of gravity.

We don't have to breakdown movement in this way. It so happens that (in Newtonian mechanics) we know how to deal with momentum and angular momentum separately. Different decompositions would be possible. But they almost certainly would be more complex and less intuitive. For example, suppose that angular momentum always led to an additional 'linear force', which was directional, and depended on the relationship between the mass distribution and the axis. It would be much harder to understand what it actually consisted of. We are used to rotating things about their center of mass.

When you solve Newtonian problems of torque, you usually have to judiciously choose a point to resolve torques around. The solution would be the same, but the technique is much easier if you choose the right point, for which as many forces as possible cancel out. 'Center of gravity' is just the standard heuristic to the general case of this problem.

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Because Inertia Moment is mimimum when rotating around center of mass, so any force applied to the body will "pass" through the "path" of minimal resistance.

Basically it is the point for which the sum of all momentums is minimal.

Also water and electric current flows through paths of minimal resistance. I wanted to do a short answer on purpose, because I think the alternative answer is just a "show the calculations" which is not very intuitive.

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As you know from riding a bicycle is it possible to balance at a very slow speed without falling if you can keep the weight balanced equally on both sides. As with a seesaw. It turns out that you can always find a way to balance a rigid body regardless of how it touches the surface underneath by placing the mass exactly right so it is evenly distributed.

Man balancing laptop

You can see a video on how this can be done here: https://www.youtube.com/watch?v=OGRUf1PLJdY

If you now recognize that this is true regardless of how the rigid body is orientated in the first place, you just need to consider if all these "divide the mass in equal parts"-division lines have anything in common or not. It turns out they all go through a single common point (this can be mathematically proven) which you probably have already guessed has named the center of mass.

It can also be mathematically proven that a rigid body behaves the same as a single point with the full mass of the rigid body. This makes newtonian calculations (like the movement of the Earth around the Sun) easier.

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In fact, a solid body doesn't rotate about a point (COM), but around an axis on which the COM is situated.

For example in a solid, uniform sphere (every object with a non-uniform distribution of mass can continuously be transformed in a sphere with the same mass uniformly distributed), the only points that rotate about the COM lie in the equator plane perpendicular to the axis of rotation. All the other points rotate about another point on the axis of rotation.

If you let the sphere spin from zero angular velocity to an angular velocity x, without imparting a linear momentum on the sphere the linear momentum can only be conserved if all the moments of the dm's (infinitely small masses if we consider the sphere as a continuous mass) cancel, which is the case if the COM lies on the axis of rotation. Of course, if you consider different axes of rotation they have the COM point in common.

For two separate bodies bounded by an attractive force as gravity, it is possible to say that the bodies rotate about the COM of the two bodies. Like two masses connected by a rope, but with the rope removed. In this case, the rotation is also about the axis of rotation (perpendicular to the rotation plane), but as well about the COM point.

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To make the couple around the center of mass axis equally distributed so that the body can rotate. As rotation itself is defined as the axis of rotation and the center of mass should be in same line else it becomes revolving around the axis of rotation.
The other reason is that the rotation can be sustained.

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    $\begingroup$ Besides the last line , I couldn't get what you are saying ! $\endgroup$ – Shashaank Feb 12 '17 at 16:17

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