# Does every object rotate around its center of gravity?

It is said that rocket,plane rotate about center of gravity ,why this is is not case for boats?
Boat pivot point is not in center of gravity.

jbriggs444
Homework Helper
Also, boats have keels, daggerboards and centerboards.

anorlunda
Staff Emeritus
The picture shows the AC75 racing boat. The whole purpose of that foil in to move the center of rotation outside the hull. The boat lifts out of the water as it rotates around that point at the end of the foil. You must consider all forces, not just gravity.

Also, boats have keels, daggerboards and centerboards.
What do you want to say by this?
Plane also has rudder,wings,flaps,slats,horizontal stablizer etc etc...

jbriggs444
Homework Helper
What do you want to say by this?
Plane also has rudder,wings,flaps,slats,horizontal stablizer etc etc...
For this reason, planes can be regarded as pitching about a horizontal axis through the wings. And sailboats can be regarded as yawing about a vertical axis through the centerboard.

A relevant point is that the center of buoyancy/lift/pressure/whatever has little to do with the center of mass and much to do with the geometry of the shell.

James Demers
For this reason, planes can be regarded as pitching about a horizontal axis through the wings. And sailboats can be regarded as yawing about a vertical axis through the centerboard.

Does rocket, plane rotate around CG?
Only difference in air and water is fluid density,why then boat dont rotate around CG too?

jbriggs444
Homework Helper
Does rocket, plane rotate around CG?
Only difference in air and water is fluid density,why then boat dont rotate around CG too?
I do not understand your confusion.

Much of the question is psychological -- if we describe the motion of an object, we can describe it as a translation and a rotation. We pick a point on the body and ask how that point translates. Then we ask how the rest of the body rotates about that point.

There are many choices about which point to use. Any of them will work. All of them will yield a correct description of the motion of the body. Which should we choose?

Water is key to the successful operation of a boat.

If a boat is purely rotating and not moving then the choice is easy. Pick the point that is stationary in the water. This will be where the centerboard is located. The boat spins in place about its centerboard.

If the boat is purely translating and not rotating then the choice is irrelevant. Any point will do. There is no rotation to worry about.

If the boat is turning while moving then we have a choice. But we want to pick a point that is not moving sideways in the water due to the rotation. That means a point at the centerboard. Any point fore or aft of that would be moving sideways in the water due to the turn. Any point port or starboard of the centerboard could be used -- it does not matter much.

vanhees71, russ_watters, PeroK and 1 other person
A.T.
Does rocket, plane rotate around CG?
It's up to you. A plane flying a looping can be described in different ways:
1) Plane's CG moves around the center of the looping & plane rotates around its CG
2) Plane rotates around the center of the looping

FactChecker, vanhees71 and jbriggs444
It's up to you. A plane flying a looping can be described in different ways:
1) Plane's CG moves around the center of the looping & plane rotates around its CG
2) Plane rotates around the center of the looping
Why then center of pressure must be behind CG to have stable rocket?
This is exmple that rocket pivot point is at CG?

jbriggs444
Homework Helper
Why then center of pressure must be behind CG to have stable rocket?
The point you pick for center of rotation has nothing to do with the question you now ask. The rotation rate of the rocket does not depend on one's choice of reference point.

Accordingly, one is free to pick a reference point that makes for an easy explanation.

Since the rocket as a whole may be accelerating, it is convenient to place the reference axis at the center of mass so that any acceleration of the body as a whole has no effect on angular momentum assessed about that axis. Then one can simply ask: "as the object rotates clockwise, does the clockwise torque increase or decrease as a result".

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it is convenient to place the reference axis at the center of mass

If you put CG behind center of pressure,rocket will be unstable and crash..So it seems CG is real pivot point for rocket,not just math agreement.
Isnt it?

A.T.
If you put CG behind center of pressure,rocket will be unstable and crash..So it seems CG is real pivot point for rocket,not just math agreement.
The torque around CG determines whether the rocket starts rotating. But that rotation doesn't have a specific "pivot", it's just a change in orientation.

The torque around CG determines whether the rocket starts rotating. But that rotation doesn't have a specific "pivot", it's just a change in orientation.

I find here some answers,where people has same confusion as I:

https://physics.stackexchange.com/q...t-pushed-by-multiple-forces?noredirect=1&lq=1

https://physics.stackexchange.com/q...rotate-around-the-torque-vector-or-its-center

https://physics.stackexchange.com/questions/147870/will-an-object-rotate-when-we-apply-a-force-to-it

The rules of motion lead us the following equivalent statments that are valid for both 2D and 3D bodies:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

• Every body has a Centre of Mass, whatever its form. If an object has a regular shape and uniform, homogeneous distribution of mass its CoM coincides with its centre.
• If an object is fixed to a pivot, a fulcrum the axis of rotation will be at the pivot
• If an object is free, not fixed to an artificial axis of rotation any action outside the CoM will make it rotate around it

• Suppose now we have an object B (a board, for example, or a door, like in your other question. Its CoM lays at the middle: if we exert a force, an impulse, an impact bang on the CoM, the whole board will move in the same direction.
• If you apply a force on any point except the Com, let's say at one edge, you must specify if the force is rotating with the body. The body will rotate anyway, but if the force always act in the same direction, after a short time it will lose contact with the board.
• Lastly, if a stone, a point mass hits the edge of the board it will move forward and rotate at the same time. Supposing that the projectile has mass 1 and v 20 (p = 20, L = 10, E = 200) and the board has m = 9 and that the collision is elastic, the ball will bounce back at roughly v = -10 m/s , the board will translate at 3.33 m/s and the board will rotate with a frequency ν=2.6rps

hutchphd
Homework Helper
A rigid object set rotating in space far from other influences will rotate about its center of mass. If other "forces" are present it may not.

berkeman, etotheipi and russ_watters
A rigid object set rotating in space far from other influences will rotate about its center of mass. If other "forces" are present it may not.
When gust of side wind hit rocket , will rocket start rotation about CG?

jbriggs444
Homework Helper
A rigid object set rotating in space far from other influences will rotate about its center of mass. If other "forces" are present it may not.
The instantaneous center of rotation of a rigid object can be pretty much anywhere. Pick a frame. Any frame.

The center of mass and a frame where the object is not translating is a useful choice because it means that the instantaneous center of rotation will not be gyrating along a spiral path.

etotheipi
russ_watters
Mentor
When gust of side wind hit rocket , will rocket start rotation about CG?
There is no single answer. It may or may not rotate, in either direction....and if it does, you get to pick the point it rotates about.

jbriggs444
jbriggs444
Homework Helper
There is no single answer. It may or may not rotate, in either direction....and if it does, you get to pick the point it rotates about.
Regardless of the choice, the change in orientation of the object is an invariant fact of the matter. It does not depend on which choice you make. [Which I know is what you just got done saying]

russ_watters
russ_watters
Mentor
Regardless of the choice, the change in orientation of the object is an invariant fact of the matter. It does not depend on which choice you make. [Which I know is what you just got done saying]
Yes, I didn't mean to imply otherwise. I was mostly trying to say that what happens depends on the specifics of the scenario. I can imagine scenarios where a rocket might rotate toward or away from a wind or not at all.

jbriggs444
The instantaneous center of rotation of a rigid object can be pretty much anywhere. Pick a frame. Any frame.
I really dont understand your "mathematical thinking".I am to stupid to understand this abstract thinking..

The instantaneous center of rotation of a rigid object can be pretty much anywhere. Pick a frame. Any frame.

The center of mass and a frame where the object is not translating is a useful choice because it means that the instantaneous center of rotation will not be gyrating along a spiral path.

2)earth
3)sun
4)galaxy
5)any point

and what answer has the most physics importnace for human that look at this weather vane?

hutchphd
Homework Helper
The center of mass. But of course this is not simple:

etotheipi
The point is that, under a carefully chosen time-dependent transformation involving a boost and rotation of your coordinates, the rigid body can be made to perform any possible motion.

Anyway, suppose you did settle on a particular frame. Now, select any point ##\mathcal{O}## at ##\mathbf{x}_{\mathcal{O}}## on the body, and let's consider how the position of another point ##\mathcal{P}## at ##\mathbf{x}_{\mathcal{P}}## on the rigid body changes between times ##t## and ##t + \delta t##. The point ##\mathcal{O}## undergoes a translation ##\mathcal{O}(t+\delta t) - \mathcal{O}(t) = \delta \mathbf{x}_{\mathcal{O}}##, and the change in the position of ##\mathcal{P}##, i.e. ##\mathcal{P}(t+\delta t) - \mathcal{P}(t) = \delta \mathbf{x}_{\mathcal{P}}##, can be described by compounding the translation ##\delta \mathbf{x}_{\mathcal{O}}## with a rotation by ##\delta \phi## about an axis ##\mathbf{n}## passing through the new position of ##\mathcal{O}##, i.e.$$\delta \mathbf{x}_{\mathcal{P}} = \delta \mathbf{x}_{\mathcal{O}} + \delta \phi \mathbf{n} \times (\mathbf{x}_{\mathcal{P}} - \mathbf{x}_{\mathcal{O}})$$Use the Physicist's trick of dividing by ##\delta t##,$$\frac{\delta \mathbf{x}_{\mathcal{P}}}{\delta t} = \frac{\delta \mathbf{x}_{\mathcal{O}}}{\delta t} + \frac{\delta \phi \mathbf{n}}{\delta t} \times (\mathbf{x}_{\mathcal{P}} - \mathbf{x}_{\mathcal{O}}) \implies \dot{\mathbf{x}}_{\mathcal{P}} = \dot{\mathbf{x}}_{\mathcal{O}} + \boldsymbol{\omega} \times (\mathbf{x}_{\mathcal{P}} - \mathbf{x}_{\mathcal{O}})$$As it turns out, the point ##\mathcal{O}## that we chose need not even be "on" the rigid body, the only requirement is that it is fixed with respect to the rigid body. Furthermore, you can show (try it!) that whichever such point ##\mathcal{O}## you choose, we get the same angular velocity vector ##\boldsymbol{\omega}##. To actually solve mechanics problems, you either take ##\mathcal{O}## to be the centre of mass [for general motion], or the point on the body fixed in the lab frame [for pure rotation].

You can also choose a particular point ##\tilde{\mathcal{O}}## such that the transformation between ##t## and ##t + \delta t## is carried about purely the rotation ##\delta \phi \mathbf{n} \times (\mathbf{x}_{\mathcal{P}} - \mathbf{x}_{\tilde{\mathcal{O}}})##, i.e. with ##\delta \mathbf{x}_{\tilde{\mathcal{O}}} = 0##; this is the instantaneous centre of rotation. [Although, for motion with no rotation, this will be at infinity...]

sysprog and hutchphd
jbriggs444
Homework Helper

2)earth
3)sun
4)galaxy
5)any point

and what answer has the most physics importnace for human that look at this weather vane?
The pivot point is convenient because that is the point that will remain stationary in the ground frame. In this case, the center of mass choice is inconvenient because it would be gyrating in such a frame.

Nonetheless, one can do physics in any frame of reference. Some choices simplify calculations. Some choices complicate them.