I've tried so hard to understand the difference, yet no progress. There is a lot of answers here, on Quora, on Youtube,... but everyone give a different answer.

So can you please give a simple yet satisfactory answer?

Someone says that rotation is only about an axis that oass through the center of mass, other say that the axis can be anywhere inside the body but outside no because if it's outside it will be circular motion, but then if you search Wikipedia about Parallel Axis Theorem, they'll say : If the body "ROTATES" about an axis outside of it, you can use the Parallel Axis Theorem to...

So who's right?

And one more question : In circular motion, the kinetic energy formula for a body is $\frac{1}{2} MV^2$ or $\frac{1}{2} Iω^2$ (like in rotation)? I mean can we use the equation $x=\frac{1}{2} at^2 + Vt + X$ or $θ=\frac{1}{2}θ"t^2 + θ't + Θ$ (like in rotation)?

So many questions yet no one gives me a good answer, I hope that someone can here.

And what about this picture here, is it rotation? "The disc (D) can oscillate freely around a horizontal axis (A), perpendicular to its plane and passing through a point O of its periphery."




  • $\begingroup$ Thanks for whoever suggested the edit! $\endgroup$ Jul 17 at 16:04
  • $\begingroup$ What Wikipedia article are you quoting? $\endgroup$
    – Bob D
    Jul 17 at 16:36
  • $\begingroup$ Bob D - Paralell Axis Theorem $\endgroup$ Jul 17 at 16:44
  • 1
    $\begingroup$ Circular motion or rotation. They are the same for a point mass. Each of the descriptions can be converted into the other. But not $x_0 + v_0 t + \frac{1}{2} a t^2$, this is for "linear" acceleration motion. $\endgroup$
    – ytlu
    Jul 17 at 17:48

Take the planar case. Rotation of a rigid body is the change in angle between material line elements from one configuration to another.

For example:

enter image description here

The body in the diagram went from the configuration on the left to the configuration on the right. I painted a material line on the body from the nose to the left ear to get a sense of the rotation. You can see that the body has rotated about 30 degrees about an axis out of the page in this example.

Take a ferris wheel:

enter image description here

Buckets in a ferris wheel are not rotating. We say that these rigid bodies are in pure translation. To be precise, we call this case "curvilinear translation."

Sometimes problems are simple and you have rotation about a fixed point. This would be the case for a simple pendulum pinned at one end, where the pendulum is a bar.

For general motions in the plane, there is no fixed point about which you rotate. In my first example, you could either rotate about the nose 30 degrees and then translate the body, or you could translate the body and then rotate about the new configuration's nose. In a real process, the rotations and translations are likely to be happening simultaneously.

Rotation is not a concept that makes sense for a particle (material points). Particles only trace out paths. Often times though, there is an abuse of language. For example, in the case of a pendulum comprising a massless wire and particle. Or in the planetary model where all objects are point masses. You will often hear the term "rotation." Sorry that you are going through it. I had to go through it, too.

Edit: You may want to understand the "instantaneous center of rotation" for planar motions of rigid bodies. That may also help you classify things.

Edit: Fixed point rotation about $O$, where $O$ is not a material point belonging to the body:

enter image description here

If it helps you conceptually, you can consider a massless extension of material, so that $O$ is a fixed point that also belongs to a material point:

enter image description here

Edit 3:

I will be honest with you. You need to have a good working knowledge of vector calculus to understand your problem 100%. You are insisting on using scalars and it is hindering your understanding. With that said, I will try to put this into terms you will understand.

The kinetic energy of any rigid body moving in the plane is $K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$. In this formula, $m$ is the total mass of the body, $v$ is the speed of the center of mass, $I$ is the moment of inertia of the body about its center of mass about an axis normal to the page, and $\omega$ is the the angular speed of the body. In your problem, $\omega = \dot{\theta}$, and $\theta$ appears to be positive counter-clockwise. So if the body is rotating counter-clockwise, then $\omega$ will be positive. $\theta$ is the rotation angle of the body.

The formula $K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$ holds true for any planar motion of a rigid body, combining translation and rotation. You have a special case: rotation about the fixed point $O$. For this special case, $v = r | \omega |$. Substitute this into the formula: $K = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} I \omega^2$. Group terms: $K = \frac{1}{2} (I + m r^2) \omega^2$. We define a new quantity, $I_O$, as $I_O = I + mr^2$. $I_O$ is the moment of inertia of the disk about $O$ about an axis in and out of the page. That is, $I_O$ is about an axis parallel to the axis that $I$ was about, if you choose to think about the axes as passing through $G$ and $O$, which you can but you don't need to think of it that way.

The formula $K = \frac{1}{2} I_O \omega^2$ is only true in the case of rotation about a fixed point, which you have here. I will emphasize again that $O$ need not lie on the body. In your example, though, it does. $O$ is a fixed point in space that is also a material point belonging to the boundary of the disk.

  • $\begingroup$ so if the axis is outside the body, it still can be called rotation? $\endgroup$ Jul 17 at 19:56
  • $\begingroup$ Yep, see my example for rotation about a fixed point $O$, where $O$ lies off of the body. That is, $O$ is not a material point. $\endgroup$
    – Evan
    Jul 17 at 20:04
  • $\begingroup$ but in the first answer, Bob says that the axis can't be outside the body, it needs to be inside to call it rotation $\endgroup$ Jul 17 at 20:25
  • $\begingroup$ but nevertheless i understand the example, thanks for enlightening me, but i think that your explanation is more mathematically defined like in the chapter of Rotation, Translation, Dilation and stuff, but i don't think we can call the O example a rotation in physics, i don't know, hope you understand my point $\endgroup$ Jul 17 at 20:27
  • $\begingroup$ If you accept the concept of a "massless extension of material," then my answer is consistent with Bob's. $\endgroup$
    – Evan
    Jul 17 at 20:31

What is the difference between circular motion and rotation?

Basically, the difference is that in circular motion, an object just moves in a circle. In rotational motion, the object rotates about an axis passing through the object. That said, you can have both rotation and circular motion associated with a rigid body.

Consider some arbitrarily shaped rigid object. When that rigid object rotates about some axis, every part of it (every atom) moves in a circle (undergoes circular motion) about the axis of rotation, covering the same angle in the same amount of time (same angular velocity). The body as a whole is said to be in rotation yet the particles comprising it are said to be in circular motion.

Consider the earth rotating about the sun. Let's assume its orbit is circular (which of course it is not). The earth is then in circular motion about the sun. Meanwhile, the earth is rotating on an axis, so the earth is in rotation. All the atoms comprising the earth are in circular motion about the earths axis of rotation.

Someone says that rotation is only about an axis that oass through the center of mass,

Rotation does not have to occur about the center of mass of an object. It does if there are no net external forces acting upon the object. But if there are net external forces, the axis of rotation can be about some other point.

but then if you search Wikipedia about Parallel Axis Theorem, they'll say : If the body "ROTATES" about an axis outside of it, you can use the Parallel Axis Theorem to..

The parallel axis theorem has to do with calculating the moment of inertia. Not sure what that has to do with the difference between circular motion and rotation.

other say that the axis can be anywhere inside the body but outside no because if it's outside it will be circular motion,

If the axis is not passing through an object, then it cannot be said to be in rotation per the initial description. It can be said the object as a whole is in circular motion about an "external" axis. That would be the case if the earth were not spinning about its axis, but only in circular motion about the center of the sun.

And one more question : In circular motion, the kinetic energy formula for a body is $\frac{1}{2}MV^2$ or $\frac{1}{2}I\omega ^2$.

The first equation applies to circular motion where $V$ is the magnitude of the speed of the object in circular motion. The second applies to rotational kinetic energy where $\omega$ is the angular velocity.

And what about this picture here, is it rotation?

It's not clear, at least to me, what the picture involves. You refer to an axis of rotation A but it is not shown in the picture.

Hope this helps.

  • $\begingroup$ If the axis pass through the circumference of a disc for example, would it still be rotation?or the axis should be inside the disc only? $\endgroup$ Jul 17 at 17:49
  • $\begingroup$ this is the point of the picture above $\endgroup$ Jul 17 at 17:50
  • $\begingroup$ and for the parallel axis theorem, i added a picture taker from Wikipedia to show what i meant $\endgroup$ Jul 17 at 17:58
  • $\begingroup$ Sorry there is no edit button to make changes I don't know why, and for the circumference thing, i understand it thanks, so how to edit? $\endgroup$ Jul 17 at 18:03
  • $\begingroup$ I tried to use the edit feature to be able to copy something from your post but was unable to do so. It only shows that some other edit is awaiting approval. Sorry, but I don't what the problem is. $\endgroup$
    – Bob D
    Jul 17 at 18:06

This is semantics, not physics. Whether you call it “rotation,” “circular motion,” “circular rotation,” “circulation,” “circumambulation,” “rotary motion,” or any other permutation/synonym, the key is to clearly define what you mean. That way, everyone can be clear about the physics, regardless of their semantic opinions.

I recommend you focus on understanding the physics. You’ll get tripped up if you rely on memorizing definitions which are defined with vague terminology.

  • $\begingroup$ This should be included in all the answers. Focusing too much on labels is meaningless and counterproductive. It is obvious from the absence of a definite answer that it is not a significant question. $\endgroup$
    – nasu
    Jul 17 at 19:53

A rotary motion is one in which a body turns around its centre of mass. A purely circular motion is one in which a body follows a circular path but does not turn around its own centre of mass.

If you walk around a circle, you rotate. Although the centre of the circle is your centre of motion, your body also turns around its own centre of mass once every time it completes a lap of the circle.

If instead of walking normally around a circle, you sidle round so that your body always stays facing in one direction (north, say), then your motion is circular, as you do not turn about your own centre of mass as you complete each lap- you remain facing in one direction the entire time.


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