Does every object rotate around its center of gravity?

  • #106
John Mcrain
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I think @jbriggs444 was thrown by the errors in the description and diagram. If only engine #2 is firing, which is not at/through the COM, it rotates.
I made mistake ,becuase I forgeot stick will accelerate in space becasue there is no any drag to counter force ,so question don't make sense at all..
 
  • #107
jbriggs444
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A.T. said it will..
I had incorrectly assumed you were firing the engine under the big mass, not the engine at the tip of the rod.
 
  • #108
John Mcrain
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If only rocket 2 is OFF, the stick will not travel with constant speed.

If engine 2 in ON,stick will accelerate again,because in space drag don't exist.
Stick will accelerate in any combination of engines,both ,only left or only right..

Isnt it?
 
  • #109
russ_watters
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I made mistake ,becuase I forgeot stick will accelerate in space becasue there is no any drag to counter force ,so question don't make sense at all..
You also appeared to locate engine 1 at the COM of the large object, not at the COM of the system.

But anyway, since we don't know where you are going with these questions, it is tough for us to error-check your overall logic. If you only want to know about rotation, it shouldn't matter that the system is also accelerating linearly -- we could simply ignore that. Maybe there's something else on your mind...
If engine 2 in ON,stick will accelerate again,because in space drag don't exist.
Stick will accelerate in any combination of engines,both ,only left or only right..

Isnt it?
Yes. Didn't you do this experiment with a pencil like I suggested?
 
  • #110
John Mcrain
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You also appeared to locate engine 1 at the COM of the large object, not at the COM of the system.

But anyway, since we don't know where you are going with these questions, it is tough for us to error-check your overall logic. If you only want to know about rotation, it shouldn't matter that the system is also accelerating linearly -- we could simply ignore that. Maybe there's something else on your mind...
No this is CoM from all system..Forget about it becuase it doesn't make sense at all,becuase system is accelerating,so mass show resistance to change speed again
 
  • #111
russ_watters
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No this is CoM from all system..Forget about it becuase it doesn't make sense at all,becuase system is accelerating,so mass show resistance to change speed again
K
 
  • #112
John Mcrain
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Yes. Didn't you do this experiment with a pencil like I suggested?
Yes I know for this test.

I try to prove something else,but now I see that this what I mean is impossible.
Now I realize that for constant speed you must allways have two opposite forces equal in magnitude,so net force again is zero.In space this second opposite force- drag force don't exsit.
 
  • #113
A.T.
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Now I realize that for constant speed you must allways have two opposite forces equal in magnitude,so net force again is zero.
This is true for constant velocity. For constant speed (magnitude of velocity) you can have a net force, that is perpendicular to velocity.
 
  • #114
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For rotation as change of orientation of the rigid body the reference point doesn't matter.
Yes, but the term for the rate of change of the orientation of a classical object is "spin", rather than "rotation" which is more general.
 
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  • #115
John Mcrain
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We pick one that is convenient.

For a rocket on a sub-orbital trajectory aimed at a target, we might use the rotating frame in which the surface of the Earth is stationary.

For a rocket in orbit around the Earth, we might use the inertial frame in which the center of the Earth is stationary.

For a rocket making a trip to the moon, we might switch frames in the middle of the trip. [It would be foolhardy for Neil Armstrong to be using the earth-centered inertial frame when descending the last 30 meters to land on the moon].

For a rocket making a slingshot maneuver, we can shift frames and observe how a trajectory which conserves spacecraft mechanical (kinetic + potential) energy in one frame increases it in another. This might confuse us until we realize that energy is conserved but is not invariant.
I am in rocket when engine is turn off.I get out of rocket and rocket turn ON engine and going away from me..

What is rocket speed in relation to me and how would I call this place in space where I am now if I want that place would be my origin of my reference frame?
 
  • #116
A.T.
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  • #117
Richard R Richard
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Will stick start rotating counterclock wise or it will keep going in straight line without rotation?

View attachment 276287
Suppose you are in an inertial frame of reference at rest or not with the ship.

cmss.png


Case 1 Both engines off, the ship continues at rest or in uniform rectilinear motion.

Case 2 You turn on motor 1, you apply force 1, on the MC, it accelerates in translation at a rate of ##a_1 = \dfrac{F_1}{M}##, but it does not rotate, the angular acceleration is zero ## \alpha = 0 ## since the lever stroke regarding the MC is null.

Case 3 Turn off engine 1 and turn on 2 The MC accelerates in translation at the rate of ## a_2 = \dfrac{F_2}{M} ## in the same direction as the force is applied and in turn rotates with angular acceleration ## \alpha = \dfrac MI = \dfrac{F_2L}{KML^2} ## Where K is a value that depends on the design of the ship, which is irrelevant, the important thing is that ## \alpha \neq 0 ## The speed of the MC in both cases results from ## v =\displaystyle \int_0^T \dfrac {F (t)}{M (t)} dt ## only for brief instants ## v \cong at ##, because both a The force and the mass can be variable in time, because we know that we can vary the thrust by increasing fuel consumption, the mass also varies because we consume the fuel, so these formulas are strictly not true, so it is more I must study what they propose you about the Tsiolkovski rocket.

Considering the same limits, in a few moments you can say that the rotational speed or angular velocity will be ## \omega = \displaystyle \int_0 ^ T \dfrac {F (t) L}{KM (t) L^2} dt ## then we can say that ## \omega \cong \alpha t ##

If you introduce resistive forces, these can be summarized as a single force with direction and direction applied at a point of thrust by modifying the modulus of the translational acceleration, It will also influence the acceleration of the turn, if the axis of the resultant force does not pass just through the MC, since it will have a non-zero lever arm, and will create torque.
 
  • #118
Shane Kennedy
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They all try to. Without any other forces, they will, but even a spacecraft will deviate when thrusters are fired,
 
  • #119
swampwiz
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It's because the mathematics are such that when dealing with a *rigid* body, the body force - i.e., gravity or centrifugal acceleration (if the radius of the spin is much larger than the volumetric extent of the body) - cancels out at the barycenter, and a net applied load (i.e. force & force couple moment) can be abstracted out to be a force & moment applied to the barycenter, with the inertia-acceleration equations being decouple.
 
  • #120
kinsler33
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Turn off the boat's engine and lift it out of the water. Now fling it into the air. It will then rotate around its center of mass.
 
  • #121
A.T.
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It will then rotate around its center of mass.
Or around any other reference point you choose to describe its motion.
 
  • #122
kinsler33
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But only the center of mass will precisely follow the trajectory of a thrown and spinning object. I think every physics text has a stock photograph of a thrown adjustable wrench.
 
  • #123
A.T.
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But only the center of mass will precisely follow the trajectory of a thrown and spinning object.
Yes, the CM follows the motion resulting from linear Newtons 2nd Law. That makes it often a convenient choice for a reference point.
 
  • #124
hutchphd
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I feel like I'm watching Groundhog Day. Given a set of questions, is it really necessary to ask all possible permutations of the same set of questions?
 
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  • #125
kinsler33
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Good question. I will admit that I didn't look at all the replies to this inquiry, principally because there were five or six pages of them. The solution to that administrative quandry is left as an exercise for the site owner.

M Kinsler
 
  • #126
Shane Kennedy
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Or around any other reference point you choose to describe its motion.
If you do that, then the reference point won't be stationary.
 
  • #127
A.T.
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Turn off the boat's engine and lift it out of the water. Now fling it into the air. It will then rotate around its center of mass.
Or around any other reference point you choose to describe its motion.
If you do that, then the reference point won't be stationary.
Neither is the center of mass, if you fling something it into the air.
 
  • #128
Digcoal
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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.




When you have multiple variables, placing one of them at the origin of your frame of reference makes the calculations easier. Some of the videos you posted, that I watched, are simply placing the CoM as the origin of the grid space used to conduct calculations because zeroes are easy to deal with in calculations. It is an unconscious decision for most people to define their grid space in this way so when people see these explanations, they come to incorrect conclusions.

You see this, sometimes, when dealing with simple gravity problems because common language does not distinguish gravitational acceleration "downwards" as negative acceleration while altitude "above" a reference plane is considered positive. Without realizing that the origin is set to the "ground," a ball thrown "up" has a positive displacement, a positive velocity, and a negative acceleration. Once it hits the apex of its travel, its displacement is at its maximum positive displacement, its velocity is zero, and its negative acceleration remains constant (relatively speaking). As it begins to "fall", its displacement decreases, its velocity continues to decrease into negative values, and its acceleration remains constant. Just before it hits the "ground," its displacement is almost zero, its velocity has the same magnitude but negative of its launch velocity, and its negative acceleration remains constant.

However, if you chose the apex of flight as the origin, the calculations are different but give the same results. If you chose the center of the Earth as the origin, the calculations are more complicated still, but the results will still be the same.

Things rotate "around their CoM" because the calculations derived from that choice of origins for your grid space makes all the calculations easier. Basically, it is a short hand way of saying, "I am choosing the CoM" as the origin of my grid space to make my calculations easier." just like saying, "I am choosing the ground as zero altitude to make my calculations easier for the flight of a baseball."
 

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