Optimizing Stick Rotation in Space: Ideal Center of Mass Placement

[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?

 

[kuruman]

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.

 

[Erunanethiel]

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?

 

[kuruman]

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?

 

[Erunanethiel]

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

 

[kuruman]

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?

 

[Erunanethiel]

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

 

[kuruman]

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".

 

[Erunanethiel]

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"

 

[PeroK]

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?

 

[Erunanethiel]

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

 

[PeroK]

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.

 

[Erunanethiel]

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

 

[jbriggs444]

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?

 

[PeroK]

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?

 

[Erunanethiel]

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

 

[Erunanethiel]

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?

 

[PeroK]

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.

 

[Erunanethiel]

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?

 

[jbriggs444]

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.

 

[Erunanethiel]

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?

 

[jbriggs444]

Which one has the greater angular acceleration given an impulse on one end?

Which one has the greater moment of inertia?

 

[Erunanethiel]

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.

 

[jbriggs444]

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/

 

[Erunanethiel]

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

 

[jbriggs444]

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.

 

[Erunanethiel]

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

 

[jbriggs444]

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"?

 

[Erunanethiel]

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"?

 

[jbriggs444]

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".

 

[Erunanethiel]

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".

PeroK put it way better than I did: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?

 

[jbriggs444]

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

Answer: undefined. There is no optimum. The distribution with all of the mass exactly in the center results in an undefined speed and, accordingly, is ineligible.

 

[Erunanethiel]

Answer: undefined. There is no optimum. The distribution with all of the mass exactly in the center results in an undefined speed and, accordingly, is ineligible.

Do you mean that the location of center of mass does not matter for this, the only thing that does is how "centralized" the mass is to CoM, irrespective of it's location. In short: Make every where that is not CoM make as light as possible.

Correct?

 

[jbriggs444]

Do you mean that the location of center of mass does not matter for this, the only thing that does is how "centralized" the mass is to CoM, irrespective of it's location. In short: Make every where that is not CoM make as light as possible.

The location of the center of mass relative to the ends matters. But so does the degree to which the mass is centralized. Both are relevant. One can even write an equation.

Try it. Apply an impulse "p" at right angles to one end of a rod of length "l" and mass "m" with center of mass offset "r" from the end where the impulse will be applied. If the rod has a moment of inertia "I", what speed will the other end have as a result?

Edit: Take it a step at a time. For instance, what rotation rate will result from the applied impulse?

 

[Erunanethiel]

Answer: undefined. There is no optimum. The distribution with all of the mass exactly in the center results in an undefined speed and, accordingly, is ineligible.

Well, not that the other bits have no mass, they are as I said made as light as possible