B Does a body behave as a point mass even at rest?

[mark2142]

Hi,
A body with center of mass behaves as a point mass when a force is applied. So when $F_{ext}=0$ then does it also behave as a point mass with $a_{com}=0$, at rest. If yes, How can we prove this?

(And can somebody please answer my other question I posted a week ago https://www.physicsforums.com/threa...ting-definitions-and-laws-in-physics.1016599/)
Thank You.

 

[berkeman]

A body with center of mass behaves as a point mass when a force is applied.

No. Why do you say that? If the force is not applied exactly along the axis of the COM, there will also be a moment, no?

 

[hutchphd]

A body with center of mass behaves as a point mass when a force is applied.

All massive bodies have a center of mass. See above @berkeman. A point mass has no physical extent and therefore no moments of force (i e Torques) can be generated

So when Fext=0 then does it also behave as a point mass with acom=0, at rest. If yes, How can we prove this?

It behaves according to Newtons Laws. In the absence of external force (and Torque) there is no a_com (or a_angular). Physics cannot "prove it". Newton hypothesized it to be true and we have been unable to disprove it for 500 yrs . That's the best we can do
The Feynman thing is not well stated (uncharacteristically) by Prof Feynman. I do not understand his point.

 

[berkeman]

See above @berkeman.

No comprendo...

 

[Nugatory]

I’m reading that as hutch saying to OP “Look at what Berke said above”

 

[hutchphd]

Yep. And that is generally good advice.

 

[jbriggs444]

A body with center of mass behaves as a point mass when a force is applied. So when $F_{ext}=0$ then does it also behave as a point mass with $a_{com}=0$, at rest. If yes, How can we prove this?

If by "behave like a point mass" you mean "Does $\sum F_\text{ext} = m_\text{tot} a_\text{cm}$" then the answer is yes.

The proof is a matter of algebra. Defining your terms. Manipulating nested summations (or integrations). Pulling constant factors out. Putting constant factors in. It is completely tedious and uninteresting.

You start with $\sum F_{i_\text{ext}} = m_i a_i$ for each piece $i$ and sum that over $i$.

 

[mark2142]

No. Why do you say that? If the force is not applied exactly along the axis of the COM, there will also be a moment, no?

Oh sorry. A rigid body does not behave as a point mass because it has rotational motion too.
I meant to say that when a force F is applied to a rigid body the COM moves as if all the mass is concentrated there of the body and the force F is applied at that point.
I am asking can we take the mass of the body to be concentrated at COM even when the force F is zero i.e to treat a rest rigid body as a point mass mainly to solve problems?

 

[jbriggs444]

Oh sorry. A rigid body does not behave as a point mass because it has rotational motion too.
I meant to say that when a force F is applied to a rigid body the COM moves as if all the mass is concentrated there of the body and the force F is applied at that point.
I am asking can we take the mass of the body to be concentrated at COM even when the force F is zero i.e to treat a rest rigid body as a point mass mainly to solve problems?

Yes. We can. The acceleration of the COM is dictated entirely by the sum of the external forces divided by the object's total mass.

If the sum of the external forces happens to be zero then we can immediately conclude that the COM will follow a straight line path at constant speed.

If the sum of the external forces happens to be zero and the COM starts at rest, we can immediately conclude that the COM will remain at rest.

Of course, the object may rotate under the net torque of external forces, even though those forces sum to zero. As long as this rotation is irrelevant to the problem, you can ignore it.

 

[mark2142]

As long as this rotation is irrelevant to the problem, you can ignore it.

That means we can always treat it as point mass regardless of rotation?
So rotation is completely independent of translational motion ?

 

[jbriggs444]

So rotation is completely independent of translational motion ?

I would say "yes". However, it is the same external forces that apply to both.

It is just that for purposes of rotational motion, you are considering torques rather than forces -- $\vec{F}_\text{ext} \times \vec{r}$ rather than $\vec{F}_\text{ext}$ and you are considering the moment of inertia $\sum m r^2$ rather than the total mass $\sum m$

Edit: If the body is non-rigid, you will also need to ditch the idea of a single rotation rate multiplied by a fixed moment of inertia and stick with the sum of the angular momenta of each piece: $L=\sum m \vec{r} \times \vec{v}$.

 

[mark2142]

So it behaves exactly like a point mass with force or no force regardless of all that rotation.
Thank you @jbriggs444 and all the guys for the help.

 

[Frabjous]

In general, I would say ”no”. In general one must have conservation of energy, linear momentum and angular momentum; however, there is a subset of problems where things can be decoupled.

 

[mark2142]

The Feynman thing is not well stated (uncharacteristically) by Prof Feynman. I do not understand his point.

I will try to explain if that is what you are saying in that thread.

 

[Herman Trivilino]

That means we can always treat it as point mass regardless of rotation?

To be treated as a point mass it would also have to have a size that is small compared to the other distances involved.

So rotation is completely independent of translational motion ?

Yes because the actual motion is a superposition of translational motion of the center of mass and rotation about the center of mass.

 

[hutchphd]

I will try to explain if that is what you are saying in that thread.

I will be happy to discuss the physics should such a question be raised. I am less happy about interpreting Feynman scripture when not well written. Do not misinterpret: I love to read Feynman but in this case I am pretty certain that I know the physics yet cannot really parse the lecture.
So explain if you think it will be useful to you and I will help as I can.

 

[hutchphd]

So it behaves exactly like a point mass with force or no force regardless of all that rotation.

You need to be carefull here. Even if he force is off-center, the net impulse from the external force will equal the change in momentum of the Center of Mass of the object . But some of the original linear kinetic energy will likely go into rotational KE so conservation of total mechanical energy gets more complicated as @caz mentioned

 

[mark2142]

So explain if you think it will be useful to you and I will help as I can.

It is.

 

[mark2142]

You need to be carefull here. Even if he force is off-center, the net impulse from the external force will equal the change in momentum of the Center of Mass of the object . But some of the original linear kinetic energy will likely go into rotational KE so conservation of total mechanical energy gets more complicated as @caz mentioned

Okay. I get it. Superposition of translational and rotational motion.

The acceleration of the COM is dictated entirely by the sum of the external forces divided by the object's total mass.

If the sum of the external forces happens to be zero then we can immediately conclude that the COM will follow a straight line path at constant speed.

If the sum of the external forces happens to be zero and the COM starts at rest, we can immediately conclude that the COM will remain at rest.

As if all the mass of the body is concentrated at the center of mass. Yes?

 

[hutchphd]

Okay. I get it. Superposition of translational and rotational motion.

Superposition is not the word I would choose, but I believe you understand the physics.

 

[jbriggs444]

The acceleration of the COM is dictated entirely by the sum of the external forces divided by the object's total mass.

As if all the mass of the body is concentrated at the center of mass. Yes?

Yes. That aspect of the body's motion (the motion of the center of mass, given the sum of the external forces) is as if the mass of the body were concentrated at the center of mass.

Like a number of others posting in this thread, I worry that you are trying to read more into that "yes" than is actually there.

 

[mark2142]

Like a number of others posting in this thread, I worry that you are trying to read more into that "yes" than is actually there.

Frankly I don't understand those eqns because I have not got there yet. How can if one force is causing the rotational motion can cause translational motion too. Like we can either toss a body high up or rotate it in air. We can't toss and rotate both at the same time. That would require much greater forces. Its not clear ( Is it because of conservation of linear momentum?).
And I have written a lot of time 'acts as a point mass' in hope that someone will clear it if its wrong. Does a body really behave as a point mass?
Please clear these two doubts.
Thank you.

 

[Frabjous]

Frankly I don't understand those eqns because I have not got there yet. How can if one force is causing the rotational motion can cause translational motion too. Like we can either toss a body high up or rotate it in air. We can't toss and rotate both at the same time. That would require much greater forces. Its not clear ( Is it because of conservation of linear momentum?).
And I have written a lot of time 'acts as a point mass' in hope that someone will clear it if its wrong. Does a body really behave as a point mass?
Please clear these two doubts.
Thank you.

Have you ever flipped a coin?

 

[jbriggs444]

Frankly I don't understand those eqns because I have not got there yet. How can if one force is causing the rotational motion can cause translational motion too. Like we can either toss a body high up or rotate it in air. We can't toss and rotate both at the same time. That would require much greater forces. Its not clear ( Is it because of conservation of linear momentum?).

The same force can and does have both effects -- translation and rotation. Clear your desk. Lay down a pencil. Strike one end at a right angle. The pencil moves off and rotates.

Newton's second law remains in full effect for linear motion. No matter where you apply a force on that pencil, the acceleration of the pencil's center of mass times the total mass of the pencil will be equal to the applied force. The linear effect of a given force is not reduced because a rotation results along with the translation.

The corresponding law for torques is also in full effect. The same force affects both linear momentum and angular momentum.

If you look at the pencil tip where the force is applied, you can see that because the pencil both translates and rotates that the pencil tip moves farther than it would under either effect alone. So the work done and the kinetic energy gained will be greater than it would be under either effect alone. A force with the same magnitude can have greater effect because it acts over a greater distance. [This had already been pointed out up-thread]

And I have written a lot of time 'acts as a point mass' in hope that someone will clear it if its wrong. Does a body really behave as a point mass?

Yes, in one respect. No, in other respects.

Yes: the center of the mass will accelerate based on the sum of the external forces and the object's total mass.

No: the object can rotate and deform, unlike a point mass which can do neither. Nor can a point mass be subject to off-center forces.

Edit: In a uniform gravitational field, the center of gravity will coincide with the center of mass and one can also treat the object as a point mass for purposes of calculating the total gravitational force on the object. [For spherically symmetric objects, this holds for non-uniform fields as well, though tidal deformations and the resulting departure from spherical symmetry may remain problematic]

 

[hutchphd]

I would add that in real life the time profile of forces supplied by an external agent often depends upon the detailed characteristics of the object and the (Newtonian) reaction forces it supplies. The fact that I can toss a bowling ball at, say, 5 m/s does not mean I can throw a golf ball at 500 m/s even though the mass is 100 times less.
But the "acts like a point mass" concept is very useful as you clearly realize.

 

[mark2142]

Yes, in one respect. No, in other respects.

Yes: the center of the mass will accelerate based on the sum of the external forces and the object's total mass.

No: the object can rotate and deform, unlike a point mass which can do neither. Nor can a point mass be subject to off-center forces.

So as I understand we can imagine the whole mass as point mass at the COM point if we can ignore all that rotation. Imagine that it is just a point mass with same mass accelerating under the effect of force. But if rotation has to be taken into account we have to treat it as rigid body.

The linear effect of a given force is not reduced because a rotation results along with the translation.

Yes. This is the problem that concerns me. I am flipping coins but the coin either rotates fast in air at lower height or it travels a greater height but with less rotation speed. To both rotate the coin fast and make it go higher to same height we need more force. Yes?

 

[jbriggs444]

This is the problem that concerns me. I am flipping coins but the coin either rotates fast in air at lower height or it travels a greater height but with less rotation speed. To both rotate the coin fast and make it go higher to same height we need more force. Yes?

We need more energy. We do not need more force.

Say that we have a point mass and an equally massive pencil. We apply the same force to each for the same time. The resulting momentum will be equal, regardless of where we apply that force to the pencil.

But if we apply the force to the end of the pencil, the same force will be applied to a point on the pencil that moves for a greater distance. That means that the same magnitude of force can do a greater amount of work. It also means that we will have to put more effort into pushing the pencil on its end to maintain the same magnitude of force.

This is what @hutchphd was getting at with his example of the bowling ball and the golf ball.

Intuitively, it is easier to push hard on a bowling ball than it is to push hard on a golf ball. The golf ball moves away fast when you push on it hard. It is difficult to keep up.

If you exert 10 pounds of force on a 10 pound bowling ball for half a second, it will move by about four feet. If you exert 10 pounds of force on a 1.6 ounce golf ball for half a second, it will move 100 times further. About 400 feet. Exerting 10 pounds of force over 400 feet of distance imparts 100 times more energy than exerting 10 pounds of force over 4 feet of distance.

Both bowling ball and golf ball get the same momentum. But the golf ball gets 100 times more energy.

 

[mark2142]

Say that we have a point mass and an equally massive pencil. We apply the same force to each for the same time. The resulting momentum will be equal, regardless of where we apply that force to the pencil.

Aha! How same linear momentum? The equally massive point mass will have more linear velocity and so it will cover more linear distance compared to a pencil on which force is applied at one end. The pencil cannot travel the same linear distance because it will rotate too. Yes?
('resulting momentum will be equal' means if masses are equal then linear velocity will be equal too).
Do you see pencil traveling the same distance when force is applied at one side as compared to when force is applied at the COM?

If you exert 10 pounds of force on a 10 pound bowling ball for half a second, it will move by about four feet. If you exert 10 pounds of force on a 1.6 ounce golf ball for half a second, it will move 100 times further. About 400 feet. Exerting 10 pounds of force over 400 feet of distance imparts 100 times more energy than exerting 10 pounds of force over 4 feet of distance.

Both bowling ball and golf ball get the same momentum. But the golf ball gets 100 times more energy.

That is true for balls. Both of them will have same changes in momentum but one will have more K.E due to more work done.

 

[jbriggs444]

How same linear momentum? The equally massive point mass will have more linear velocity and so it will cover more linear distance compared to a pencil on which force is applied at one end.

No. It will not. Both the point and the center of mass of the pencil will move with identical linear velocities when subject to equal forces for equal times. This applies regardless of where the force is applied on the pencil.

Do you see pencil traveling the same distance when force is applied at one side as compared to when force is applied at the COM?

Yes. Have you done a careful experiment that shows otherwise? How have you controlled or measured the applied force?

 

[mark2142]

No. It will not. Both the point and the center of mass of the pencil will move with identical linear velocity when subject to equal forces for equal times.

Okay! Maybe the example of pencil on a table is not appropriate. In space where there is no obstacle, where there is no other forces like friction I think you are right. Both equally massive point mass and COM will cover the same distance.

I tried to do an experiment with coin and pencil but I couldn't see the required phenomenon.
But I have done an experiment with glass cone on a glass table also. I put water and so the surface became frictionless but I was unable to apply same force everytime. I am thinking how can I achieve this. I will try with balloon in air also.

 

[PeroK]

How can if one force is causing the rotational motion can cause translational motion too. Like we can either toss a body high up or rotate it in air. We can't toss and rotate both at the same time.

Have you never played or watched any sport?

In particular, if you go to "Highland Games" in Scotland, you'll find a contest called "tossing the caber", where a tree trunk is projected with both linear and rotational motion. Look on YouTube.

You seem to be unable to relate physics to the real world. If you went out into the street and gave someone a ball, they could throw it and spin it at the same time. You don't have to study physics to know that.

 

[mark2142]

Have you never played or watched any sport?

In particular, if you go to "Highland Games" in Scotland, you'll find a contest called "tossing the caber", where a tree trunk is projected with both linear and rotational motion. Look on YouTube.

You seem to be unable to relate physics to the real world. If you went out into the street and gave someone a ball, they could throw it and spin it at the same time. You don't have to study physics to know that.

No. That's not what I meant. I am doubtful in the independence of both types of motion. I am trying an experiment but I don't know how can I apply constant force each time. Plus when I apply force offcenter the body deflects from the original path.

 

[PeroK]

No. That's not what I meant. I am doubtful in the independence of both types of motion. I am trying an experiment but I don't know how can I apply constant force each time. Plus when I apply force offcenter the body deflects from the original path.

What does "independence" mean in this context?

In practice, a force applied over time to a rotating body may tend to vary in direction. Unless you carefully control the direction of the force.

 

[mark2142]

What does "independence" mean in this context?

Means no effect of rotation on translational motion in case of a rigid body.

 

[hutchphd]

No one has said that they are independent in the sense you are trying to prove. They can often be separated for convenience of thought and calculation. In general if one imparts some force to a free body (and conversely a la Newton a free body imparts an opposite force on you) this will couple to both rotational and translational degrees of freedom in a complicated fashion. After the interaction the the two motions are independent for a rigid body. That is why the Center of Mass is useful (and for gravity too)

 

[jbriggs444]

I am trying an experiment but I don't know how can I apply constant force each time.

Possibly it is time to actually learn the math. It is not like $\sum F_\text{ext} = m_\text{tot} a_\text{cm}$ is terribly difficult.

Then you can do a thought experiment. Consider a bar-bell type object. A thin and nearly massless rod connecting two massive balls that are small compared to the length of the rod.

Apply (on paper) a small force on the right-hand ball for a small time.
Consider the resulting motion of the assembly. How fast does the center of mass move? In what direction does the center of mass move?

 

[vanhees71]

One should also be careful, which external force you consider. All you discuss here is about the motion in the homogeneous gravitational field close to Earth.

To that end consider the body under consideration (it can be an elastic or rigid solid body or even a fluid) as a discrete set of point particles, held together by interaction forces, which can be described as "pair forces", i.e., on the particle with the label $j$ the total force is given by
$$\vec{F}_j=\sum_{k \neq j} \vec{F}_{jk} + m_j \vec{g},$$
where $\vec{g}$ is the constant gravitational acceleration on due to the Earth (an approximation valid for motions close to Earth). The equations of motion for the particles thus read
$$m_j \ddot{\vec{x}}_j = \sum_{k \neq j} \vec{F}_{jk} + m_j \vec{g}.$$
Now we sum this equation over $j$ and use Newton's Third Law ("lex tertia"), according to which $\vec{F}_{jk}=-\vec{F}_{kj}$, i.e., if the interaction force on particle $j$ due to the presence of particle $k$ is $\vec{F}_{jk}$ then the interaction force on particle $k$ due to the presence of particle $j$ is opposite. This means that by summing over $j$ all the interaction forces cancel and you get
$$\sum_j m_j \ddot{\vec{x}}_j=M \ddot{\vec{x}}_{\text{cm}} =M \vec{g},$$
where
$$M=\sum_j m_j, \quad \vec{x}_{\text{cm}}=\frac{1}{M} \sum_j m_j \vec{x}_j.$$
Our calculation shows that indeed the center of mass moves like a point particle in the gravitational field of the Earth,
$$\ddot{\vec{x}}_{\text{cm}}=\vec{g}=\text{const}.$$
For all other forces, and even the gravitational interaction with the Earth when considering the position dependence, is not as simple, because then you have (treating the Earth as a "point particle" sitting fixed at the origin of the coordinate frame),
$$m \ddot{\vec{x}}_j = \sum_{k \neq j} \vec{F}_{jk} -\gamma m_j m_{\text{Earth}} \frac{\vec{x}_j}{r_j^3} \quad \text{with} \quad r_j=|\vec{x}_j|.$$
Then summing over $j$ gives
$$M \ddot{\vec{x}}_{\text{cm}} = -\gamma m_{\text{Earth}} \sum_j \frac{m_j \vec{x}_j}{r_j^3}=\vec{F}_{\text{ext}},$$
and $\vec{F}_{\text{ext}}$ cannot be expressed in terms of $\vec{x}_{\text{cm}}$!

 

[mark2142]

If you look at the pencil tip where the force is applied, you can see that because the pencil both translates and rotates that the pencil tip moves farther than it would under either effect alone. So the work done and the kinetic energy gained will be greater than it would be under either effect alone. A force with the same magnitude can have greater effect because it acts over a greater distance.

We need more energy. We do not need more force.

But if we apply the force to the end of the pencil, the same force will be applied to a point on the pencil that moves for a greater distance. That means that the same magnitude of force can do a greater amount of work. It also means that we will have to put more effort into pushing the pencil on its end to maintain the same magnitude of force.

Intuitively, it is easier to push hard on a bowling ball than it is to push hard on a golf ball. The golf ball moves away fast when you push on it hard. It is difficult to keep up.

Both bowling ball and golf ball get the same momentum. But the golf ball gets 100 times more energy.

I think I got you. Its because its difficult to keep up at the end of a pencil(less mass) than at COM(more mass) when applying the force on its end that we are unable to provide it same momentum. We are unable to apply force for the required time. That's is why my glass cone on water experiment is failing. It is not covering the same distance when applying force off center.
Similarly the work done also becomes less since force is applied for lesser distance and so reduced KE for rotating object. Difficult to do. Yes?

 

[mark2142]

It takes time to connect the dots. Thank you guys @hutchphd @jbriggs444 and everyone else. That was some new and interesting information not given in a book.