What causes unhinged objects to rotate?

[Frigus]

Suppose their is a unhinged rod lying on table and someone applied force at some point then due to it object start rotating.i tried to find why it rotates and came to know that if line of force is not passing through the centre of mass then force will produce torque around the centre of mass and this torque will cause the object to rotate.

Now,
•why do we took centre of mass as Axis of rotation.
- is it's answer is that because it rotates around centre of mass.

So,
•why do object moves around centre of mass?

 

[jbriggs444]

•why do object moves around centre of mass?

Because that is a convenient way to think about it.

One can decompose the motion of a rigid object into a combination of translation of a reference point and rotation about that point in infinitely many ways -- one way for every choice of reference point.

It is convenient to choose the center of mass because, for an object that is free from external forces, the center of mass will translate smoothly instead of accelerating in a circle (or worse). It is easier to describe a motion as an unchanging translation and a rotation than as a continuously changing translation and a rotation.

 

[Frigus]

Because that is a convenient way to think about it.

One can decompose the motion of a rigid object into a combination of translation of a reference point and rotation about that point in infinitely many ways -- one way for every choice of reference point.

It is convenient to choose the center of mass because, for an object that is free from external forces, the center of mass will translate smoothly instead of accelerating in a circle (or worse). It is easier to describe motion as an unchanging translation and a rotation than as a continuously changing translation and a rotation.

How can we explain this if we take some other point as a reference point and apply force on the center of mass, in that situation, there should be some torque on the object about the reference point we have taken but if we apply force on the center of mass we know it undergoes only rectilinear motion.

 

[jbriggs444]

How can we explain this if we take some other point as a reference point and apply force on the center of mass, in that situation, there should be some torque on the object about the reference point we have taken but if we apply force on the center of mass we know it undergoes only rectilinear motion.

If you apply torque, you expect angular momentum to change, right?

But a body that is translating and not rotating about its center of mass does possesses angular momentum about the reference axis you chose.

 

[Lnewqban]

... So,
•why do object moves around centre of mass?

Copied from
https://en.wikipedia.org/wiki/Center_of_mass

"In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration."

"In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid."

"A body's center of gravity is the point around which the resultant torque due to gravity forces vanishes. Where a gravity field can be considered to be uniform, the mass-center and the center-of-gravity will be the same."

Force applied out of line with CM: The balance of gravity forces around the CM is disturbed by that additional force.
Due to the torque (force x perpendicular distance to CM) introduced by that out-of-line force, the object could start rotation, accelerate previous rotation, decelerate previous rotation, stop previous rotation or change direction of previous rotation.

Force applied in line with CM: Regardless of the magnitude of the new additional force, the balance of gravity forces around the CM is not disturbed.
Due to the absence of a new torque, the object could keep previous rotation or lack of it.
The new force (directly applied over the CM) will accelerate or decelerate the whole body (previously rotating or not) in the direction it is applied.

 

[Frigus]

If you apply torque, you expect angular momentum to change, right?

But a body that is translating and not rotating about its center of mass does possesses angular momentum about the reference axis you chose.

Thanks,
Please also help me in this query also,
If we apply force at centre of mass and take refrence point same as taken in the photo(posted in post #3)
Then angular momentum will change as velocity is changing and using this equation MVR(angular momentum) we know that angular momentum is changing and if angular momentum is changing then there should be some torque similarly if we take centre of mass as refrence point and apply force on some other point (another than COM) then also both variables M and R can't change but only Velocity can change then why does it rotates rather than just moving in rectilinear path as in case I.

 

[tech99]

If you apply a force to an object, other than through the Centre of Mass, there is a couple formed by our force and a Reaction acting through the CM.
To analyse the situation, you need to resolve in (a) the direction of our applied force (b) in a direction at right angles to our applied force and (c) take moments about any point.
Having once found the rotational couple and the linear force on the object, you can use its inertia and moment of inertia to find its motion.

 

[Frigus]

there is a couple formed by our force and a Reaction acting through the CM.
To analyse the situation, you need to resolve in (a) the direction of our applied force (b) in a direction at right angles to our applied force and (c) take moments about any point.

Sir can you please elaborate it more as I am unable to understand it.

 

[jbriggs444]

Thanks,
Please also help me in this query also,
If we apply force at centre of mass and take refrence point same as taken in the photo(posted in post #3)
Then angular momentum will change as velocity is changing and using this equation MVR(angular momentum) we know that angular momentum is changing and if angular momentum is changing then there should be some torque similarly if we take centre of mass as refrence point and apply force on some other point (another than COM) then also both variables M and R can't change but only Velocity can change then why does it rotates rather than just moving in rectilinear path as in case I.

That is a pretty long sentence. I hope you took a breath while speaking it. Unfortunately, I am having trouble understanding it all in one go. So let's take it a bit at a time.

If we apply force at centre of mass and take refrence point same as taken in the photo(posted in post #3)

So we apply a force at the center of mass. And we have adopted a reference point "as shown in the photo". The photo shows a horizontal rod with a chosen reference point over to the left. It is subject to an upward force at its midpoint.

Then angular momentum will change as velocity is changing and using this equation MVR(angular momentum)

Yes. One can write the equation as L=MVR. Though it would be more descriptive as $$L=\vec{R} \times M\vec{V}$$Note that the vector cross product is not commutative.

we know that angular momentum is changing and if angular momentum is changing then there should be some torque

Yes. And there is some torque.$$\tau=\vec{R} \times \vec{F}$$

similarly if we take centre of mass as refrence point and apply force on some other point (another than COM) then also both variables M and R can't change but only Velocity can change then why does it rotates rather than just moving in rectilinear path as in case I.

It does not rotate because you changed the reference point. It rotates because you changed the setup.

Are you asking for an elaborate explanation of how an object subject to a force on one end can rotate?

 

[sophiecentaur]

Thanks,
Please also help me in this query also,
If we apply force at centre of mass and take refrence point same as taken in the photo(posted in post #3)
Then angular momentum will change as velocity is changing and using this equation MVR(angular momentum) we know that angular momentum is changing and if angular momentum is changing then there should be some torque similarly if we take centre of mass as refrence point and apply force on some other point (another than COM) then also both variables M and R can't change but only Velocity can change then why does it rotates rather than just moving in rectilinear path as in case I.

"If we apply force at centre of mass" there will be no torque and no angular acceleration.

Step back into simple lever calculations. We can always choose what point to take out moments about. Now, if we choose to use the CM of the 'leverless object', we can use Newton 2 to calculate the linear acceleration of the CM and we can choose to use the couple of the applied force and an N3 reaction force through the COM. The resulting equations will produce two answers - the value of linear acceleration and the angular acceleration at the time the force is first applied. You could choose any other point on the object but each of the two equations would contain terms relating to the other equation so it would just be more hard work to solve. So why not choose the easiest pair of equations from the start?

 

[FactChecker]

•why do we took centre of mass as Axis of rotation.
- is it's answer is that because it rotates around centre of mass.

You can think of it as rotating around any point you want, but the Center Of Mass (COM) has a unique advantage: The equation, $F = Ma$ applies to the acceleration of the COM. So it is easiest to know how the COM moves, regardless of where the force is applied.

 

[Frigus]

That is a pretty long sentence. I hope you took a breath while speaking it. Unfortunately, I am having trouble understanding it all in one go. So let's take it a bit at a time.

My apologies

It does not rotate because you changed the reference point

Here I am talking about the case in which we have applied force on Centre of mass and adopted refrence point as the point on left of the rod.

Q. Why it does not rotate?
Answer:- it does not rotate because you changed the refrence point.

My reasoning:- if we have choosen some other point as refrence point then there should some distance between applied force and refrence point due to which there will be some torque.
if there is torque then there should be change in angular momentum.

Now in both of the cases in which we have i.)applied to force on Centre of mass and adopted point on left as refrence point and ii)applied force on some other point than centre of mass nd taken centre of mass as refrence point what is difference between them as In one case rod is rotating and in another rod is not rotating.

'leverless object'

I understand leverless object as point object.

we can choose to use the couple of the applied force and an N3 reaction force through the COM.

If leverless object is point object how can couple(two equal and opposite forces which are separated by some distance) be formed.

Please correct me in these points so I can understand why object rotates In one case and don't rotate in another case.

 

[jbriggs444]

Here I am talking about the case in which we have applied force on Centre of mass and adopted refrence point as the point on left of the rod.

No. That is not what you were talking about. I quoted what you were talking about. That was not it.

Here, I will quote it for you again.

if we take centre of mass as refrence point and apply force on some other point (another than COM) then also both variables M and R can't change but only Velocity can change then why does it rotates rather than just moving in rectilinear path as in case I.

You proceed to [mis]quote yourself and my response:

Q. Why it does not rotate?
Answer:- it does not rotate because you changed the refrence point.

Here you have taken my response out of context, because you had forgotten what question you'd asked.

Remainder of your previous posting left unread.

 

[Frigus]

Sorry to revisit this old thread

Here you have taken my response out of context, because you had forgotten what question you'd asked.

I was too much confused back then so please forgive me for my mistake.
I was trying to ask this ask this question

Are you asking for an elaborate explanation of how an object subject to a force on one end can rotate?

And also this question,
Why objects Don't rotate when force is applied to centre of mass.
Thanks.

 

[etotheipi]

Why objects Don't rotate when force is applied to centre of mass.

Consider a coordinate system with its origin at the centre of mass. The total torque on the rigid body w.r.t. this coordinate system equals the rate of change of angular momentum of the body about the centre of mass. Even if the centre of mass is moving non-inertially, any fictitious forces (which are body forces, and act through the centre of mass) have zero torque about the centre of mass and thus do not cause a change in the angular momentum in that frame.

So if you apply a force whose line of action passes through the centre of mass, the force has no torque about the centre of mass and you will only end up with translational acceleration of the centre of mass. There will be no rotation in the centre of mass frame.

Contrast this with applying a force to a point on the rigid body that is not the centre of mass. The force has a non-zero torque about the centre of mass, and we end up with angular acceleration in this frame.

 

[sophiecentaur]

This thread is lacking a good labelled diagram for us all to talk around. Some of the remarks imply that we may well not be talking about the exact same situation.
Everyone has access to a simple drawing (not sketching) utility.

 

[etotheipi]

This thread is lacking a good labelled diagram for us all to talk around. Some of the remarks imply that we may well not be talking about the exact same situation.
Everyone has access to a simple drawing (not sketching) utility.

I will try and oblige... I hope you enjoy my illustration of two cases, featuring pancake!

 

[sophiecentaur]

Thanks

pancake

For me it was an asteroid.
So the applied force F will rotate the pancake in the right hand diagram. The diagram seems to suggest not?? I'm assuming that O and O' are the position of the CM after the same interval of time in each diagram.

 

[jbriggs444]

I will try and oblige... I hope you enjoy my illustration of two cases, featuring pancake!

View attachment 266394

It seems intuitive that a force $\vec{F}$ applied on the upper left would produce a clockwise rotation and that a force $\vec{F}$ applied on the lower right would product a counter-clockwise rotation. It seems perfectly reasonable that the same force applied in the middle would produce no rotation.

Certainly if you apply the mathematics of torque, angular momentum, rotation rate and moment of inertia that the result of "no rotation" comes out of the calculations when a force is applied at the center of mass. It does not even matter what axis you choose to use about which to calculate angular momentum.

 

[etotheipi]

So the applied force F will rotate the pancake in the right hand diagram. The diagram seems to suggest not?? I'm assuming that O and O' are the position of the CM after the same interval of time in each diagram.

My apologies, I should have been more clear. The red arrows are just two sets of coordinate axes corresponding to two different non-rotating (as opposed to body fixed) reference frames. We have $\vec{r} = \vec{R} + \vec{r}'$ if $\vec{r}$ is the position measured in the lab frame, $\vec{r}'$ is the position measured in the CM frame, and $\vec{R}$ is the vector between the origins of the frames.

We have that in the centre of mass frame,$$\vec{\tau} = \vec{r}' \times \vec{F} = I_{cm}\vec{\alpha}$$in the left example $\vec{r}' = \vec{0}$ so the angular acceleration of the rigid body is zero, on the right diagram $\vec{r}'$ is non-zero and $\vec{\alpha}$ will also be non-zero. In both cases, the acceleration of the centre of mass is $\vec{a}_{cm} = \frac{1}{m}\vec{F}$. We might also note that the acceleration of an arbitrary point on the rigid body at position $\vec{r}_i$ is $\vec{a}_i = \vec{a}_{cm} + \vec{\omega} \times (\vec{\omega} \times \vec{r}') + \vec{\alpha} \times \vec{r}'$.

If you keep applying a constant force $\vec{F}$ to one particular point on the pancake in the right example, I think it will result in an oscillatory motion

It does not even matter what axis you choose to use about which to calculate angular momentum.

You are of course correct, it is probably easiest demonstrated with the König theorem. If you don't mind that I refer once again to the pancakes, w.r.t. our lab coordinate system we can write$$\vec{r} \times \vec{F} = d_t \vec{L} = d_t \vec{L}_{cm} + d_t \vec{L}'$$where $\vec{L}_{cm}$ is the angular momentum of the centre of mass, and $\vec{L}'$ that w.r.t. the centre of mass. We can also say$$\vec{L}_{cm} = m\vec{R} \times \vec{v}_{cm} \implies d_t \vec{L}_{cm} = m \left[ d_t \vec{R} \times \vec{v}_{cm} + \vec{R} \times d_t \vec{v}_{cm} \right] = m\vec{R} \times \vec{a}_{cm}$$and that w.r.t. the CM$$\vec{L}' = I_{cm} \vec{\omega} \implies d_t \vec{L}' = I_{cm} \vec{\alpha}$$So back to the original relation$$\vec{r} \times \vec{F} = m\vec{R} \times \vec{a}_{cm} + I_{cm} \vec{\alpha}$$But $\vec{r} = \vec{r}' + \vec{R}$, so $\vec{r}\times \vec{F} = \vec{r}' \times \vec{F} + \vec{R} \times \vec{F} = \vec{r}' \times \vec{F} + m \vec{R} \times \vec{a}_{cm}$, so we recover$$\vec{r}' \times \vec{F} = I_{cm}\vec{\alpha}$$which is the same as in our first analysis, and to be expected since the above essentially amounts to "un-deriving" the Konig theorem

 

[Frigus]

Certainly if you apply the mathematics of torque, angular momentum, rotation rate and moment of inertia that the result of "no rotation" comes out of the calculations when a force is applied at the center of mass. It does not even matter what axis you choose to use about

In case-I we can see that their is torque about centre of mass and thus it rotates.

In case-II their is torque about point P as we have taken it refrence point and we say that due to torque angular momentum changes as $\vec V$ term in MVR changes but I can't understand why mode of change of angular momentum is different in both cases as in case-I it changes by rotating and in case-II it changes by change in magnitude of velocity.

It seems intuitive that a force F→ applied on the upper left would produce a clockwise rotation and that a force F→ applied on the lower right would product a counter-clockwise rotation. It seems perfectly reasonable that the same force applied in the middle would produce no rotation.

I also have gut feeling that this would happen but I can't figure out why it rotates.
Mathematically this will happen but please tell me if their is any non mathematical answer to this or is it in higher studies and I have to accept this and go on using maths.

 

[sophiecentaur]

any non mathematical answer

Arm waving but: any acceleration of the COM will involve a Reaction Force through the COM. That Reaction Force and the Applied Force constitute a couple with the radius being the distance of the Applied Force from the COM.
To work out the numbers you need some Maths but the distribution of the mass around the object will affect the Moment of Inertia and that will affect the initial rate of angular acceleration.

 

[etotheipi]

Arm waving but: any acceleration of the COM will involve a Reaction Force through the COM.

Wouldn't it be an inertial force through the COM, $-m\vec{a}_p$, that we would need to account for in the frame of reference of P? We would only get a reaction force through the centre of mass if it were physically hinged about the centre of mass, but in that case the centre of mass would not translate.

 

[sophiecentaur]

The mass is not hinged. The mass will accelerate if you push it anywhere. If you push it through a point that's not the COM, the reaction force cannot disappear, Its value ( yet to be calculated) will be less than F because there is also some angular acceleration, due to the difference in the reaction force and F. You don't want any Maths here so this is as far as we can go. But:
Consider a very simple body consisting of two balls (mass m) on a light rigid rod (length 2r). You push one ball with force F perpendicular to the rod. Reaction force on the ball will be F, initially and taking moments about the other ball, the force on the COM will be F/2. So the couple will be Fr/2. (F-F/2)r That couple acts to turn the object with an angular acceleration ωdot of (Fr/2) / (MI about the COM). MI = 2m r² (Should I use MI about the other ball?)
So ωdot = Fr/2mr²
= F/2mr
The initial linear acceleration of the COM will be F/2m (the other ball hasn't started moving yet)

 

[etotheipi]

Consider a very simple body consisting of two balls (mass m) on a light rigid rod (length 2r). You push one ball with force F perpendicular to the rod. Reaction force on the ball will be F, initially and taking moments about the other ball, the force on the COM will be F/2. So the couple will be Fr/2. (F-F/2)r That couple acts to turn the object with an angular acceleration ωdot of (Fr/2) / (MI about the COM). MI = 2m r² (Should I use MI about the other ball?)

I don't know what you mean by force on the CM. There is only one real force, that which acts on the ball that you are pushing. If you transform into the CM frame, we will have a fictitious force $-\vec{F}$ (but not $-\frac{1}{2} \vec{F}$ ) acting through the centre of mass in addition to the force $\vec{F}$ applied to the ball. That constrains the CM to have zero acceleration in the CM frame and would also indeed result in a force couple. We would have that$$F\frac{r}{2} = I_{cm}\alpha_z = \frac{mr^2}{2} \alpha_z$$If you instead transform into the frame of the other ball, whose acceleration is initially along the line of centres (there is only initially a horizontal tension), the torque of the inertial force through the centre of mass is zero and we would only have a single torque to deal with, that of the real force$$Fr = mr^2 \alpha_z$$Thankfully the same equations end up being the same so long as we are careful in accounting for the inertial forces in each frame

 

[jbriggs444]

In case-I we can see that their is torque about centre of mass and thus it rotates.

In case-II their is torque about point P as we have taken it refrence point and we say that due to torque angular momentum changes as $\vec V$ term in MVR changes but I can't understand why mode of change of angular momentum is different in both cases as in case-I it changes by rotating and in case-II it changes by change in magnitude of velocity.

Let me be sure that I understand the two cases.

In case-I we have an off-center force and are using an on-center reference axis.
In case-II we have an on-center force and are using an off-center reference axis.

Right?

Let us clearly identify our variable names. We have a force $\vec{F}$ applied for time $\Delta t$ which gives rise to a change in momentum which we can write as $m\vec{v}$. We can use $\vec{r}$ to denote the offset of the applied force from the reference axis. We can use $\omega$ to denote the resulting rotation rate and $I$ to denote the moment of inertia of the pancake.

In case-I we have a non-zero torque from the force not at the reference axis.

Newton's second law says that the center of mass acquires linear velocity in the direction of the applied force. The linear momentum is given by $$m\vec{v}=\vec{F}\Delta t$$.
Since the center of mass is on the reference axis, there is no angular momentum associated with this linear momentum. Any angular momentum in the pancake must instead manifest as rotation.

We know the supplied angular momentum is given by torque($\vec{r} \times \vec{F}$) times time ($\Delta t$). We know that the resulting angular momentum is equal to moment of inertia ($I$) times rotation rate ($\omega$). So we can write:$$I \omega=\vec{r} \times \vec{F} \Delta t$$
Solve for $\omega$ and see that it is non-zero.

But no one was confused about case-IIn case-II we still have a non-zero torque from a force not at the reference axis.

Newton's second law says that the center of mass acquires linear velocity in the direction of the applied force. The linear momentum is given by $$m\vec{v}=\vec{F}\Delta t$$.
We know that the supplied angular momentum is given by torque ($\vec{r} \times \vec{F}$) times time ($\Delta t$). We know that we have some resulting angular momentum due to the linear momentum of the pancake. This contribution to angular momentum is given by $\vec{r} \times m\vec{v}$. If we have any rotation, this contributes an additional $I\omega$ to the total angular momentum of the pancake. So we can write:$$I\omega + \vec{r} \times m\vec{v} = \vec{r} \times \vec{F} \Delta t$$
But we already know that $m\vec{v} = \vec{F} \Delta t$. So it is clear that $\omega$ = 0 and no rotation results from the on-center force.

Just as we should have expected. On center force = no rotation.

 

[sophiecentaur]

I don't know what you mean by force on the CM

You would accept that, as the other ball is not subjected to the input F, there is a reaction force from the struck ball, that's equal to F. That is equivalent to a force of F/2. applied at the COM.

 

[jbriggs444]

If you push it through a point that's not the COM, the reaction force cannot disappear, Its value ( yet to be calculated) will be less than F because there is also some angular acceleration, due to the difference in the reaction force and F.

You need to do some serious work to clarify what you are talking about here. As written it is false, false, false.

Newton's third law applies always. The action force and the reaction force are equal and opposite. If you are going to use the terms "action" and "reaction" otherwise you need to clarify.

 

[etotheipi]

You would accept that, as the other ball is not subjected to the input F, there is a reaction force from the struck ball, that's equal to F. That is equivalent to a force of F/2. applied at the COM.

I would definitely not! So that we can ignore the added complexity of fictitious forces, we shall analyse this situation in the lab frame of reference and the centre of mass frame of reference only. Let us also restrict ourselves to applying the force over a very short time period, so that we can ignore matters of changing geometry.

We exert an external force $\vec{F}$ on one ball, in the $\hat{\theta}$ direction perpendicular to the line of centres. This results in the centre of mass of the configuration accelerating at $\frac{1}{M}\vec{F}$, where $M = 2m$ is the total mass of the configuration. About the centre of mass, the force $\vec{F}$ has a torque $\vec{\tau} = \vec{r} \times \vec{F} = \frac{Fd}{2} \hat{z}$, and as such$$\frac{Fd}{2} = I_{cm} \alpha_z = \frac{md^2}{2} \alpha_z \implies \alpha_z = \frac{F}{md}$$We determine that the total initial acceleration of the ball on the right, with $\vec{\omega}$ initially zero, is $$\vec{a}_{r} = \vec{a}_{cm} + \vec{\alpha} \times \vec{r}_r = \frac{1}{M}\vec{F} + \frac{F}{2m} \hat{\theta} = \frac{F}{m}\hat{\theta}$$On the other hand, the initial acceleration of the ball on the left is (noting that the $\hat{\theta}$ basis vector now points in the other direction)$$\vec{a}_{l} = \vec{a}_{cm} + \vec{\alpha} \times \vec{r}_{l} = \frac{1}{M} \vec{F} + \frac{F}{2m} \hat{\theta} = \vec{0}$$We might increase the complexity by dealing with the tension forces that arise between the balls as soon as the string goes taut, et cetera, but the basic idea is here.

But we must of course consider the forces acting on the configuration as is. If we pick a non-inertial frame of reference then we will also need to include fictitious forces, which will act through the centre of mass. However these are not reaction forces, not least because fictitious forces do not obey Newton III!

 

[Frigus]

Oh!if I have remembered this property

Newton's second law says that the center of mass acquires linear velocity in the direction of the applied force

Then my problem would have solved.
Now only question left is why object rotates when force is applied?

Consider a very simple body consisting of two balls (mass m) on a light rigid rod (length 2r). You push one ball with force F perpendicular to the rod. Reaction force on the ball will be F, initially and taking moments about the other ball, the force on the COM will be F/2. So the couple will be Fr/2. (F-F/2)r That couple acts to turn the object with an angular acceleration ωdot of (Fr/2) / (MI about the COM). MI = 2m r2 (Should I use MI about the other ball?)

Sir I can't understand how couple can be formed, doesn't all internal forces cancel out and only external force acts.

 

[etotheipi]

I can't understand how couple can be formed, doesn't all internal forces cancel out and only external force acts.

You are correct, it is the external force which determines the angular acceleration as well as the translational acceleration of the configuration, all internal forces cancel pairwise.

In addition to this, if you work in an accelerating frame of reference (for instance the centre of mass frame), you must also include external fictitious forces in your analysis (specifically, assuming the frame is non-rotating, a fictitious body force $-m\vec{a}$ that acts through the centre of mass). Since the centre of mass has zero acceleration in the centre of mass frame, we can determine that an additional fictitious force $-\vec{F}$ will act through the centre of mass in this accelerating frame. But this is not a reaction force, it is a different beast entirely!

 

[jbriggs444]

Now only question left is why object rotates when force is applied?

I thought I'd posted that already in #26. It is a requirement based on torque and angular momentum conservation.

To say it with pompous verbosity...

Given the constraints on the motion of the myriad bits of pancake-stuff imposed by the rigidity of the pancake, the motion of the pancake can be completely characterized by a linear translation rate of the center of mass and a rotation rate of the bits as they orbit that center in lock-step.

Given the magnitude, direction and application point of an external impulse on a pancake that is initially at rest, there is exactly one possible linear motion and exactly one possible rotation rate that can result.

We have calculated said rates and determined that an on-the-center force results in no rotation.

[For simplicity, we have restricted our attention to two dimensions and rotation in the plane]

 

[sophiecentaur]

Sir I can't understand how couple can be formed, doesn't all internal forces cancel out and only external force acts.

Do you really doubt that the object will spin? I'd recommend starting with the fact that it does and then try to explain it by finding whatever flaw there is in your understanding of the situation.
Can you really believe that only one of the balls will move or that the other one will magically move in a straight line parallel to the struck ball? Try it on a flat surface with two masses joined with string.

 

[sophiecentaur]

To say it with pompous verbosity...

I like your style, young man.

 

[sophiecentaur]

You need to do some serious work to clarify what you are talking about here. As written it is false, false, false.

Hah, I fear you may be right there. And with no verbosity, either.

 

[Frigus]

I thought I'd posted that already in #26. It is a requirement based on torque and angular momentum conservation.

I thought I'd posted that already in #26. It is a requirement based on torque and angular momentum conservation.

To say it with pompous verbosity...

Given the constraints on the motion of the myriad bits of pancake-stuff imposed by the rigidity of the pancake, the motion of the pancake can be completely characterized by a linear translation rate of the center of mass and a rotation rate of the bits as they orbit that center in lock-step.

Given the magnitude, direction and application point of an external impulse on a pancake that is initially at rest, there is exactly one possible linear motion and exactly one possible rotation rate that can result.

We have calculated said rates and determined that an on-the-center force results in no rotation.

[For simplicity, we have restricted our attention to two dimensions and rotation in the plane]

Thanks,
I got it!
If we take granted that angular momentum is always conserved then it makes a lot of sense.