Optimizing Stick Rotation in Space: Ideal Center of Mass Placement

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Discussion Overview

The discussion revolves around the optimal placement of the center of mass (COM) of a stick in space to maximize the backward rotation of one end when a force is applied to the opposite end. Participants explore various configurations and the implications of mass distribution on motion, considering both theoretical and practical aspects.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that for a uniform stick, the center of mass is at the midpoint (5 cm), and any push will result in motion and rotation about this point.
  • Others question the implications of having the center of mass offset, suggesting that different configurations could lead to varying rotational dynamics.
  • A participant proposes a scenario with three sticks having different center of mass placements, seeking to determine which configuration would yield the most backward motion of one end when pushed.
  • Some participants explore the effects of mass distribution, suggesting that concentrating mass at one end may enhance backward acceleration of the opposite end.
  • There is a discussion about the relationship between torque, mass distribution, and resultant velocities, with some participants expressing uncertainty about the calculations involved.
  • Participants express differing views on whether the ideal position of the center of mass is at the midpoint or closer to the end where the force is applied, with some suggesting that a closer position may yield faster rearward acceleration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the optimal position of the center of mass. There are multiple competing views regarding the effects of mass distribution and the resulting motion of the stick, indicating an unresolved discussion.

Contextual Notes

Participants mention the need to calculate torque and resultant velocities, but the specific mathematical steps and assumptions involved remain unresolved. The discussion also highlights the complexity of the problem, with varying interpretations of the center of mass and its implications for motion.

Who May Find This Useful

This discussion may be of interest to those studying physics, particularly in areas related to mechanics, rotational dynamics, and mass distribution effects on motion.

Erunanethiel
Let's say I have a 10 cm long stick in space.

If I want it's right side to rotate as much as possible (even go backwards) by pushing on it's left end, where would the ideal position of the center of mass of the stick be?

Would it be the middle, or closer to the right or left? And by how many centimeters?
 
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The motion of the stick is governed by the laws of conservation of angular and linear momentum.. If the stick is uniform, the center of mass of the stick is at 5 cm from the ends, at the midpoint. There is no ideal position for it, that's where it is. If you give the stick a kick, in general it will move in the same direction as the kick and spin about the midpoint (center of mass) as it moves. If you give the stick a kick right at its center it will move without spinning. Try it yourself by putting a plastic ruler on a table and giving it sharp kicks with your finger at different points along its length. Friction will eventually stop the motion, but observe what happens immediately after you deliver the kick.
 
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kuruman said:
The motion of the stick is governed by the laws of conservation of angular and linear momentum.. If the stick is uniform, the center of mass of the stick is at 5 cm from the ends, at the midpoint. There is no ideal position for it, that's where it is. If you give the stick a kick, in general it will move in the same direction as the kick and spin about the midpoint (center of mass) as it moves. If you give the stick a kick right at its center it will move without spinning. Try it yourself by putting a plastic ruler on a table and giving it sharp kicks with your finger at different points along its length. Friction will eventually stop the motion, but observe what happens immediately after you deliver the kick.
Lets say i have three sticks. One has it's center of mass in the middle, the other has it offset to the right and the last one has it offset to the left. Which one of the three sticks has the most likelihood of it's right side going backwards when the stick is pushed on it's left side?
 
Erunanethiel said:
Lets say i have three sticks. One has it's center of mass in the middle, the other has it offset to the right and the last one has it offset to the left.
It seems that we have a different understanding what "center of mass" means. Can you draw a picture of the three sticks that you mentioned?
 
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kuruman said:
It seems that we have a different understanding what "center of mass" means. Can you draw a picture of the three sticks that you mentioned?
I can't upload pictures because I am on mobile but I can explain.

The sticks that have their centers of gravity off to one side could be made with two different materials on each side, one denser than the other, or have a metal ball on the side where the COM is, whilst the stick with COM in the middle is uniform
 
OK, let's say you have a spherical lollipop with a very light handle so that the CM is essentially at the center of the sphere. You want to set it in motion so that it does what?
 
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kuruman said:
OK, let's say you have a spherical lollipop with a very light handle so that the CM is essentially at the center of the sphere. You want to set it in motion so that it does what?
I want the handle bit to go backwards, but we need a bit of the handle to stick above the lollipop as well to be able to push on that bit
 
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I am not sure what you mean by "backwards". Here is a picture of a lollipop in space. Please explain how you define the forward and backward direction in terms of "up", "down", "left" and "right" as related to the figure. I have labeled the tip of the stick as "A" and the center of mass as "CM".

Lollipop.png
 

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kuruman said:
I am not sure what you mean by "backwards". Here is a picture of a lollipop in space. Please explain how you define the forward and backward direction in terms of "up", "down", "left" and "right" as related to the figure. I have labeled the tip of the stick as "A" and the center of mass as "CM".

View attachment 217292
I want to know the optimal position for cm in order for point "B" to initially go "down" when force is applied at point "A"
Screenshot_20171223-003559.jpg
 

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  • #10
Erunanethiel said:
Let's say I have a 10 cm long stick in space.

If I want it's right side to rotate as much as possible (even go backwards) by pushing on it's left end, where would the ideal position of the center of mass of the stick be?

Would it be the middle, or closer to the right or left? And by how many centimeters?
Let me rephrase your question:

When you give an impulse to one end of the rod, the COM moves forward and the object rotates around the COM. The initial velocity of the other end if the rod is the sum of the forward velocity of the COM and a backwards velocity from the rotation.

Question: what mass distribution will result in the other end of the rod initially moving backwards from its starting position with the greatest speed?

Suggested simpler version: you have a uniform rod of mass ##M## and a mass ##m## that can be attached to the rod at any point. Where do you attach the mass ##m## to maximise the backwards speed, as above?

How much of this can you work out for yourself? Do you know about the moment of inertia?
 
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  • #11
PeroK said:
Let me rephrase your question:

When you give an impulse to one end of the rod, the COM moves forward and the object rotates around the COM. The initial velocity of the other end if the rod is the sum of the forward velocity of the COM and a backwards velocity from the rotation.

Question: what mass distribution will result in the other end of the rod initially moving backwards from its starting position with the greatest speed?

Suggested simpler version: you have a uniform rod of mass ##M## and a mass ##m## that can be attached to the rod at any point. Where do you attach the mass ##m## to maximise the backwards speed, as above?

How much of this can you work out for yourself? Do you know about the moment of inertia?
Thank you for rephrasing it for me.

I think middle would be the answer, but I am most likely wrong
 
  • #12
Erunanethiel said:
Thank you for rephrasing it for me.

I think middle would be the answer, but I am most likely wrong

Even the simpler version is quite complicated I think. Interesting question.

In the middle is as good a guess as any.
 
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  • #13
PeroK said:
Even the simpler version is quite complicated I think. Interesting question.

In the middle is as good a guess as any.
The more I think about it, it makes more sense to me that center of mass being closer to the side where we applied the impulse would result in the fastest rearward acceleration of the other side, but up to a point
 
  • #14
Erunanethiel said:
The more I think about it, it makes more sense to me that center of mass being closer to the side where we applied the impulse would result in the fastest rearward acceleration of the other side, but up to a point
How much of the stick's mass can you concentrate at the center of mass? Can you see what sort of effect concentrating the mass versus spreading it out would make?
 
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  • #15
Erunanethiel said:
The more I think about it, it makes more sense to me that center of mass being closer to the side where we applied the impulse would result in the fastest rearward acceleration of the other side, but up to a point

Actually, the general problem is simpler. You need almost all the mass concentrated at one point.

You can't just guess that point. You need to calculate the torque and the resultant velocity of the end point.

What's your level of physics knowledge?
 
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  • #16
jbriggs444 said:
How much of the stick's mass can you concentrate at the center of mass? Can you see what sort of effect concentrating the mass versus spreading it out would make?
I guess we need as little mass as possible on the side "B" so it would rotate "backwards" with the greatest possible velocity
 
  • #17
PeroK said:
Actually, the general problem is simpler. You need almost all the mass concentrated at one point.

You can't just guess that point. You need to calculate the torque and the resultant velocity of the end point.

What's your level of physics knowledge?
My physics education is not enough to tackle this.

Can we calculate it using the 10 cm stick with the force applied at point "zero"?Do you reckon it would be closer to the edge the force is applied, in the middle, or closer to the side which the force isn't applied?
 
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  • #18
Erunanethiel said:
My physics education is not enough to tackle this.

Can we calculate it using the 10 cm stick with the force applied at point "zero"?Do you reckon it would be closer to the edge the force is applied, in the middle, or closer to the side which the force isn't applied?
It turns out you want all the mass in the middle of the rod.
 
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  • #19
PeroK said:
It turns out you want all the mass in the middle of the rod.
With mass concentrated all in the middle with as little weight as possible on either side of the stick right? Does how heavy the end of the sticks make a difference as long as the Com is in the middle?
 
  • #20
Erunanethiel said:
With mass concentrated all in the middle with as little weight as possible on either side of the stick right? Does how heavy the end of the sticks make a difference as long as the Com is in the middle?
Take two extremes -- a heavy ball with two sticks on opposite sides or two heavy balls with a stick between. The moment of inertia and, therefore, the amount of angular acceleration for a given torque is different for the two.
 
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  • #21
jbriggs444 said:
Take two extremes -- a heavy ball with two sticks on opposite sides or two heavy balls with a stick between. The moment of inertia and, therefore, the amount of angular acceleration for a given torque is different for the two.
Which one has the greater angular acceleration given an impulse on one end?
 
  • #22
Erunanethiel said:
Which one has the greater angular acceleration given an impulse on one end?
Which one has the greater moment of inertia?
 
  • #23
jbriggs444 said:
Which one has the greater moment of inertia?
As far as I know, moment of inertia is the distance between the point the force is applied and the point where the object rotates around. In that case, it wouldn't be any different between the two.
 
  • #24
Erunanethiel said:
As far as I know, moment of inertia is the distance between the point the force is applied and the point where the object rotates around. In that case, it wouldn't be any different between the two.
The [perpendicular] distance between the point where a force is applied and the axis of rotation is the "moment arm". If you multiply the magnitude of the force by the length of the moment arm, the result is "torque".

The moment of inertia is a measure of how an object will accelerate under an applied torque. The greater the moment of inertia, the more torque (or the more time) it will take go achieve a particular rotation rate. It can be computed by considering an object as a bunch of tiny pieces, taking the mass of each piece multiplied by the square of its distance from the axis of rotation and adding up all of those results.

An object with a large mass or with a mass that is very spread out will have a high moment of inertia. It will take a lot of torque to change its rotation speed.

You would be well served to try to educate yourself on rotational mechanics. https://physics.info/rotational-dynamics/
 
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  • #25
jbriggs444 said:
The [perpendicular] distance between the point where a force is applied and the axis of rotation is the "moment arm". If you multiply the magnitude of the force by the length of the moment arm, the result is "torque".

The moment of inertia is a measure of how an object will accelerate under an applied torque. The greater the moment of inertia, the more torque (or the more time) it will take go achieve a particular rotation rate. It can be computed by considering an object as a bunch of tiny pieces, taking the mass of each piece multiplied by the square of its distance from the axis of rotation and adding up all of those results.

An object with a large mass or with a mass that is very spread out will have a high moment of inertia. It will take a lot of torque to change its rotation speed.

You would be well served to try to educate yourself on rotational mechanics. https://physics.info/rotational-dynamics/
Yes and that is not different between the heavy ball with two sticks and the two heavy balls joined by a stick is it?

The link is not working, it gives and error
 
  • #26
Erunanethiel said:
Yes and that is not different between the heavy ball with two sticks and the two heavy balls joined by a stick is it?
Excuse me? What is not different? And the link works fine for me.
 
  • #27
jbriggs444 said:
Excuse me? What is not different? And the link works fine for me.
The perpendicular distance between the point the force is applied and the point the objects rotates around (center of mass) given the centers of mass of the objects are in the same place
 
  • #28
Erunanethiel said:
The perpendicular distance between the point the force is applied and the point the objects rotates around (center of mass) given the centers of mass of the objects are in the same place
That is the "moment arm" and is indeed the same. But what about the "moment of inertia"?
 
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  • #29
jbriggs444 said:
That is the "moment arm" and is indeed the same. But what about the "moment of inertia"?
Okay I get it know thank you!

By the way, do we agree the answer to the original question is "the middle"?
 
  • #30
Erunanethiel said:
By the way, do we agree the answer to the original question is "the middle"?
The original question is not well enough posed to have a definite answer. But yes, for a reasonable understanding of the intended question, the answer is "the middle".
 
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