Here, $O$ is the center of mass of the unconstrained rigid bar: none of the points, including $O$, are affixed.
According to Salman Khan, due to $\vec{F}$ the bar will start to rotate about $O$.
However, according to @Asad, the rotation will not be about $O$.
Who is correct?
Answer: No
Explanation:
A bar that has a non-zero net force applied to it will always have its center of mass translate. This is due to Newton's second law which states the motion of the center of mass depends on the net force on a body only.
Additionally, If there is a net torque about the center of mass (just as in this case), the bar is going to rotate also. To make the point of rotation the center of mass then there must be zero net force. Any rigid body with a pure torque applied (force couple) is going to rotate, and since the center of mass is not going to translate (zero net force), the center of rotation is the COM.
Note: By definition the center of rotation is the point on the body (or the extended frame) that does not translate.
In summary: Net force relates to the translational motion of the center of mass, and net torque relates to the rotation of the object about the center of mass.
In this case, with a single force applied offset from the center of mass, the result is the general case of rotation of the bar and motion of the center of mass.
The center of rotation is going to be on the other side of the center of mass from where the force is applied. For example, with point C above the instantaneous rotation center, the following relationship is true
$$ OC = \frac{\kappa^2}{OB} $$
where $\kappa = \sqrt{\frac{I}{m}}$ is the radius of gyration of the object. For a straight bar $\kappa^2 = \frac{\ell^2}{12}$.
Notice that the closer the force is to the center of mass, the smaller $OB$ is, the larger $OC$ gets. At the limit when $OB=0$, the bar will translate (or rotate about a point at infinity).
This answer assumes the bar is unconstrained as you stated in your question; specifically, the only forces are gravity down at the center of mass $O$ and $F$ up applied at point $B$ at the right end of the bar as shown in your figure.
The center of mass of the unconstrained body will accelerate if the vector sum of the external forces, here gravity down and the applied force up, is not zero. The unconstrained bar will rotate with an increasing angular momentum if there is a net torque due to the force of gravity down and the applied force up. Remember that torque depends on the point taken about which the torque is evaluated. For motion of an unconstrained rigid body, the point used for evaluating the rotational motion is almost always taken to be the center of mass of the body, and in that sense the rotation is about the point $O$, the center of mass, in your question.
The translational motion is given by $m\vec a_{CM} = \vec F_{ext}$ where $\vec a_{CM}$ is the acceleration of the center of mass, $m$ is the mass of the bar, and $\vec F_{ext}$ is the net applied external force. Here, $ma_{CM} = F - mg$ where $a_{CM}$ is taken as positive up, $F$ is the force up applied at the end of the bar, and $g$ is the acceleration of gravity downward.
You can describe the rotational motion with respect to any point $Q$ in the body, but this is easiest when $Q$ is taken as the center of mass of the body, in which case (as stated in elementary physics textbooks) $\vec \tau_{CM} = {d \vec L_{CM} \over dt}$, where $\vec \tau_{CM}$ is the net external torque about the center of mass and $\vec L_{CM}$ is the angular momentum about the center of mass; this is true even if the center of mass is accelerating. If $Q$ is not taken as the center of mass, the relationship to be solved is more complicated. See a good textbook on intermediate/advanced mechanics such as Mechanics, by Symon, for the details.
The total motion is best described as: (a) translation of the center of mass with (b) rotation about the center of mass. This motion can be evaluated using the relationships previously discussed.
If the body is constrained, for example by a force keeping the left end $A$ in your diagram fixed, the motion is best described as (a) translation of the center of mass with (b) rotation about the fixed point.
The above discussion assumes that you, the observer, are in an inertial reference frame. If you yourself are accelerating, then fictitious forces and torques must be considered. Again, see a textbook such as Mechanics by Symon for more details.
I didn't know the formula used by John Alexiou, so I tried a way to prove it.
If we suppose that there is a inertial point C in the bar, the torque calculated from this point is:
$$T = I\alpha \implies F.(CB) = I\frac{a_b}{(CB)}$$
Where $I$ is the moment of inertia of the bar around C. The COM is accelerating: $$F = ma_o \implies I\frac{a_b}{(CB)^2} = ma_o$$
But the acceleration of the points of the bar must follow a proportionality: $$\frac{a_o}{a_b} = \frac{(CO)}{(CB)} \implies I\frac{a_b}{(CB)^2} = ma_b\frac{(CO)}{(CB)} \implies \frac{I}{m} = (CB)(CO)$$
The moment of inertia around C for a thin bar must be calculated by integration:
$$I = m(CO)^2 + m\frac{l^2}{12}$$
And as $$(CB) = (CO) + \frac{l}{2}$$
we get the same answer of John Alexiou for the situation of the drawing: $$(CO) = \frac{l}{6}$$
After some useful conversations with @Asad, I think this question is ill-posed, in the sense that it can be answered in different ways depending on one's point of view.
The one invariant statement that everyone should agree on is that the motion is not purely a rotation about the center of mass, nor is is purely a translation. At this point, you can make different statements depending on how you interpret the question.
After some discussions with @Asad, I think his point of view (which is reasonable) is that this question is asking whether the motion is a pure rotation about the center of mass. Then, the answer is no. The motion is a pure rotation about the center of rotation, which is a point on the rod which is not moving at all, at least instantaneously. One can say the motion of the rod is undergoing a pure rotation about this point, instantaneously. From this point of view, the equations of motion of the rod are irrelevant for answering the question "what point is the rod rotating around." The question is purely kinematic and geometrical, and should be answered in those terms.
Another point of view (which is the way I would tend to interpret the question, and I think some others in the comments) is that even though the question is, on its face, just about what the rod is doing at the instant the force is applied, in physics we are usually interested in how to solve the dynamical equations of motion for the rod. When we solve for the motion of a freely rotating rigid body, it is useful to decompose the motion into a translation of the center of mass, and a rotation about the center of mass. This decomposition simplifies the equations of motion for the rod. If we perform this decomposition, then the motion is a combination of a translation of the center of mass, and a rotation about the center of mass. The motion of the bar is not simply a translation; there is some component of "rotation about the center of mass" in this decomposition.
To summarize, I would say the answer to the question in the title, "will the bar rotate about the center of mass?", depends a bit on how you interpret the English language. Everyone should agree that the bar will not just rotate about the center of mass, nor will the motion simply be a translation of the bar. Beyond that, it is correct to say both
I want to look at @JohnAlexiou’s answer in a slightly different way and it all hinges on what you understand by the centre of rotation.
I hope that he does not mind me using his diagram?
Suppose that the rod is at rest and then a force is applied on it as shown in the diagram below.
A way of analysing this system (the rod) is to say that the system is under the action of a force $F$ upwards whose line of action is through the centre of mass $O$ and an anticlockwise couple $F\, l$ where $l$ is the length of $AB$.
What this means is that all parts of the rod undergo a vertical acceleration give by $F=m\, a$ where $m$ is the mass of the rod.
At the same time all parts of the rod undergo a rotational acceleration, $\alpha$ given by $F\,l = I_{\rm O} \,\alpha$ where $I_{\rm O}$ is the moment of the rod about its centre of mass, $\dfrac{m\,l^2}{12}$.
This means that the section of the rod $OB$ undergoes an upward linear acceleration of a value which depends on the distance from the centre on mass whilst the section $OA$ undergoes a downward linear acceration.
The magnitude of the linear acceleration due to the rotation at a distance $x$ from the centre of mass being $x\,\alpha$.
Now somewhere to the left of the centre of mass the upward linear acceleration of the rod due to force $F$ is equal to the downward linear acceleration of the rod due to the couple.
So the linear acceleration at that point in the rod is zero - the equivalent of @JohnAlexiou’s definition of centre of rotation which was the center of rotation is the point on the body (or the extended frame) that does not translate.
$F=m\,a = \dfrac{I_{\rm O}\alpha}{x}\Rightarrow x = \dfrac{l}{6}$.
Asad is peculiarly defining the center of rotation as the point on the body that isn't moving, and in this case there is such a point immediately following an impulsive application of force. But, of course, that point isn't fixed in the unconstrained body, and doesn't generally exist. Thus, we don't normally use this definition in unconstrained motion. But if you have an axle...
To my understanding, your question is flawed. If a single force is applied to a rigid body under the influence of no other forces, either:
- The line of action of the force passes through the center of mass, causing a pure translation and no rotation
- The line of action of force does not pass through the center of mass, in which case you end up with a pure rotation about an axis which does not pass through the center of mass. In other words the instantaneous axis of zero velocity induced by a single force can never be the center of mass.
If you apply an eccentric force, the center of mass of the body will undergo a linear acceleration, and the body itself will undergo an angular acceleration. In a fixed reference frame, this can be viewed as a pure rotation about a certain point, but this point will never be the center of mass of the body.
The whole answer is misleading in my opinion. The question of whether an object is rotating or not depends on what point you ask it of. It makes sense to ask if an object is rotating about a point, in the answer I've quoted, it's implicitly taken that it is rotating about COM. A similar point is mentioned in this answer about torque I wrote.
But, this issue is there in many physics books. The motivation comes naturally when we try to extend newton's laws from point-particles to massive bodies.
"Newton's laws are applied to bodies which are idealized as single point masses,[19] in the sense that the size and shape of the body are neglected to focus on its motion more easily. This can be done when the line of action of the resultant of all the external forces acts through the center of mass of the body. In this way, even a planet can be idealized as a particle for analysis of its orbital motion around a star."- wiki
To extend it to a rigid body, we can do it with help of chalses theorem.
Chasles’ theorem asserts that it is always possible to represent an arbitrary displacement of a rigid body by a translation of its center of mass plus a rotation around its center of mass. pg-280 , Kleppner and Kolenkow 2nd Ed
We can just take the body to be a point particle for the translation part and use torques about COM to deal with the rotation part.
This link has more details about extending newton's laws to massive bodies
The following, somewhat alternative, approach is offered to show that the rotation of the bar will be about the COM. It is based on the fact that translational and rotational motion are independent of each other.
We assume the force $F$ is continually applied perpendicular to the end of the bar at point B. An example would be a bar in space with a rocket thruster fixed at point B. Given that, we examine the motion of the bar when it is restrained at various points, where the motion of all points on the bar is purely rotational. We then propose that the only change in the motion of the bar when the restraints are removed should be the addition of the translational acceleration of the COM.
If the rod were constrained by a frictionless pin at any of the points in your diagram then the rod would rotate about that point. For example, if the rod were pinned at point A in FIG 1 below we know that the force $F$ applied at $B$ results in a torque about A of $\tau=FL$ where $L$ is the length of the bar. But we also know that if A cannot move, there has to be is a reaction force $F_{A}$ at A as shown in FIG 1 that is always equal and opposite to the applied force $F$. Or, $F_{A}=-F$.
The combination of the reaction force $F_A$ at A and the applied force F, which constitute two equal and opposite parallel forces, can be considered a force couple $M$ where $M=FL$ and $L$ is the length of the bar. A force couple causes pure rotation without translation. And that is in fact what happens here because all other points C, O, D, and B rotate about A but none of the points, including point O (the COM), undergoes translational motion. The same situation would exist if the bar were pinned at points C, O, D and B, except that the magnitude of the force couple $M$ will be different as shown in the figures below.
The figures below assume, for convenience, that the adjacent designated points on the bar are all separated by a distance of one fourth the length of the bar. Note that the average value of the force couples $M$ for the five pinned points equals the value for the couple of the bar pinned at the COM shown in FIG 3, or $M_{ave}=FL/2$. Instead of the five points shown, we can consider an infinite number of equally spaced points with the same result. This observation seems to align with the statement that a rigid body will rotate about the COM because that's where its moment of inertia is a minimum.
Finally, since translational motion is independent of rotational motion, it is suggested that if the constraints are removed one would expect that the only effect on the motion of the bar would be to include translational acceleration of its COM and that the "average" rotation of the bar would still be about the COM.
Hope this helps.
John Alexiou has a great answer above, so I'm not going to reiterate what he said in too much detail.
I will note though that natural language can be confusing sometimes. Suppose someone shows you some planar displacement of a rigid body, and asks you:
What point did the body rotate about?
One way to answer the question is: "all of them". Any given rigid motion can be decomposed into a rotation about some arbitrarily chosen point and some translation.
Needless to say, this is kind of a pointless answer. Of course, in some other hypothetical question you might be given an arbitrary point (say, the center of mass) and asked to interpret a given rigid motion as the composition of a rotation about said point with some translation that you're required to find. The point is that there is nothing special about the center of mass under this interpretation, you could put your finger literally anywhere on the page and that would be a correct answer.
Another way to interpret the question is: which (if any) point is the motion a pure rotation about? In other words:
Find a point so that, if the point was attached to the rigid body, and you (an infinitesimal planar being) were standing on it, you would spin around, but not move
Now we have a more interesting answer: for any motion that is not a pure translation, there is always a unique such point in the plane. In fact, if one takes a projective geometry perspective, there is always such a point, since we can interpret even pure translations as rotations "about infinity".
So, returning to your question, we can take the second interpretation, and ask:
Under the influence of a point force that is not directed at the center of mass, which point (if attached to the bar) would start spinning at a different rate, but not start moving away from its position at a different rate
This second clause crucially disqualifies the center of mass (as explained in John Alexiou's excellent answer above) and (depending on the parameters of the problem), uniquely specifies some point to the left of the CoM as the solution.
