How can you be this good? I feel like everything just fell into one place. You used a great tactic. Firstly, you introduced the formulas and laws that we are all familiar with and then you explained everything step by step perfectly (leaving nothing unexplained). Just magnificent. Thank you for this gem! 🔥
Correct me if I am wrong, but we don't even need to think about what happens after poking the stick. We just need to think what happens at the time of poking. The only acceleration the center of mass can have is that of the impulse, meaning it must move forward in the direction of that acceleration. If its motion deviates from this direction (e.g. if it was to rotate about some point), then Newton's 2nd law would be violated.
This was a really well-made video so thank you for that but I have two questions that i would be delighted if you could answer. 1. The thing I didn't understand is that if the stick were to rotate around another point on the stick, the COM would need a centrifugal force while the net force on the object is 0 which is impossible. Yes, this is true but when I look to some other point on the upper part of the stick when its rotating around its center of mass that point is making a circular motion around the COM so it should have a centrifugal force. First it seemed like the force was coming from the electrostatical forces between the atoms of the stick but the point we're talking about is pulled exactly the same amount on both sides because the stick is a rigid body. Where is this force coming from? 2. Suppose we have a rigid stick in space that has zero velocity. If I were to apply two equal forces that face each other to the upper part of the stick while one is a little lower that the other like this: >| |< | | | When i think about it it feels like the stick would rotate around the middle point between those two forces. The COM needs a centrifugal force in this example for it to be true while the net force is 0 I know but I just can't simulate the stick rotating around the COM in my mind. I can't comprehend why this is not true so if you could show me why it does not rotate like that it would mean a lot to me since I couldn't have gotten this question out of my mind for quite some time now. These are my questions and I know this was long sorry about that but I just wanted to make my thought process clear.
Ooooo, these are interesting questions, I will do my best to answer :3 1) Yes, there will absolutely be centripetal forces provided by the electrostatic forces between the molecules. You kind of answered it yourself, the force of interaction is between the particles. Think about it like this: we have this constraint that the rod is a rigid body- this is no trivial constraint. There are plenty of materials that lack the structural integrity to hold together when being spun, and even "rigid" objects will break when spun at a high enough RPM. However, as long as we accept that this constraint is true (as part of our effective model), we can ignore all the internal forces which cause this constraint to be satisfied (remember, from Newton's 3rd law all of these internal forces come in action/reaction pairs and will not accelerate the system's center of mass). I've found in my experiences with physics so far that a lot of behaviors that seem confusing at first have to do with the fact that we're "hiding" or just ignoring the internal forces of constraint such that we can keep things simple: analyze net external forces on the center of mass only + net external torques. 2) Hmm, so in my example I give a very quick impulse that then gives a quick change in angular momentum. I don’t see why this would be any different, if the net torque is in the + direction from the brief combination of those 2 forces, we’ll get a + angular momentum afterwards (and vice versa if -). As you said, there’s nothing to provide a centripetal force to let the center of mass circle around, so we should just have a net rotation about the center of mass after all is said and done like before. I think it feels like the point should be somewhere other than the center of mass because that force you indicated on the bottom we would think of as like a physical object which takes the role of a pivot point, for example like if I shove the stick into the edge of a table for a continual amount of time, the pivot ends up being the contact point between the stick and the table (this is very slippy contact point, it would be difficult to maintain this rotation without the stick inevitably sliding up or down). And thank you for letting me know you enjoyed the video!
I have two doubts: does an object thrown in the air(having both translation and rotation), would want less resistance to rotation or translation? Next doubt is whether centre of mass always has only transitional acceleration when any 3d object is thrown in the air Also, is the translational resistance constant for a particular object?
when a simple rigid body (such as a rod) experiences a torque about its center of mass, it will have more energy (same linear but also rotational) than one where the force was along the center of mass. how is this possible? what's the mechanism behind it having more energy? how does the force hitting a different part of the same object assuming no energy loss result in this?
Remember, work requires forces acting over displacements. It’s not fair to say “the same force” adds more or less energy. To find the amount of energy gained with the rotating body you’d have to look at the angular displacements and the center of mass displacement together. Those transient details (which are usually ignored if the impulse time is small, like the “poking” I showed in the video) must follow the work energy theorem
But one question, how can we have an external torque working? After the initial impulse, there are no net forces acting on any point on the stick. So there may be an angular momentum but should be conserved and no external torque should apply. Please correct me if I am wrong. Thanks for your presentation.
Hi Sudipto, yes the initial “poke” as I call it provides a net external torque *in that moment* to cause the rotation. After that angular momentum remains constant, thus the stick continues to rotate about its center of mass with a constant angular velocity thereafter. Good question!
Hello, first of all thanks for this video. Can we still conclude that the rigid body rotate around its center of mass even if we take into account gravity (lets say constant gravitational field)? If yes, what is the proof, I have difficulty proving it.
@@JohnDoe-xc2nzIf the gravitational field is uniform then think of this: the force vector distribution will be solely dependent on the *mass distribution* of your object. Define a “center of gravity” using this distribution as the point on the object where the net torque about it is 0. The center of mass location is *also* solely dependent on the mass distribution of your object. Thus, what you will prove is that the center of gravity is identical to the center of mass point in a uniform gravitational field. Thus, there is no torque about the center of mass just like in this example in the video.
@@madaydude_physics Thank you very much for your time and response. What I don't understand is that the reason you gave in this video to justify the fact that the rigid body was rotating around its center of mass was that if it was not the case, then that would mean that the center of mass has an acceleration not equal to zero, which contradict the initial hypothesis. However, since now we have a gravitational field, center of mass has an acceleration so we cannot use this argument anymore. What argument should I chose to justify that rigid body is still rotating around center of mass despite the fact that now the center of mass is accelerated by gravitational acceleration. Sorry if I misunderstood something trivial.
@@JohnDoe-xc2nz Here’s the difference: the gravitational field would cause a net *linear* acceleration on the center of mass. This is different than the requirement to have an object’s center of mass move in a circle (like in an object spinning at a point other than its center of mass) which would require a *centripetal* acceleration.
Question: what about the angular acceleration of a free body while we are applying the force? What makes it suffer angular acceleration about the center of mass (or why does T=Icm*alfa only apply to CoM)? E.g. positive linear acceleration the same for all points, but then I calculate torque relative to another point (that is not the CoM) and say the motion is the sum of translational acceleration and rotational acceleration about that point. Thanks in advance.
Yeah, you can totally calculate torque about any point you desire, as long as you're careful. The center of mass is very useful here (and in many problems) because Newton's 2nd law tells us that net external forces cause accelerations on the center of mass. So it makes things very intuitive to lock into the center of mass and measure torque about that point. You don't necessarily HAVE to do this: the more general expression for torque on a point particle is r cross F. For the stick example in this video, if you picked some arbitrary point other than the center of mass to measure torque about while I was poking it, you would get effectively 2 terms once you sum together all r_i cross F_i, where i refers to the ith particle on the stick: one term would be a torque about the center of mass, and the other would be a torque due to the center of mass of the stick accelerating (but this term doesn't really help us analyze any new physics). And, because we have these 2 separate terms, we can't just define some "I*alpha", the total, combined motion is now more complex than that. You can write the T_p = I_p*alpha only in situations where all the particles actually are spinning together around some axis through point p, all with unified angular acceleration, because now there's only one term to worry about. TLDR: you can always measure torque about any point, not necessarily the center of mass, but it will generally just give you an extra constant term which is not important. We should really just stick to analyzing this part of the motion with Newton's 2nd law. Although, I will add one last comment to this! Consider Earth orbiting around the sun. In this problem, it turns out to be very useful to measure the angular *momentum* (not torque, but close enough!) about the sun, and not just the Earth's center of mass. If you do this, similar to what I said earlier you would get the 2 terms: 1 SPIN angular momentum from the Earth rotating around its axis, and 1 ORBITAL angular momentum from its circulation around the sun... so indeed, in certain cases it does become useful to measure these angular quantities about points other than the center of mass, or a point where all the indivdual particles share some common angular velocity/acceleration!
I have a question: Since we can define an instantaneous axis of rotation for any rigid body in motion, which is a point about which the entire body is rotating about at that instant; doesn't that mean that even the center of mass is rotating about that point at that instant? If yes, this implies that it experiences a centripetal acceleration towards that point even though there is no external force.
Hopefully I'm correctly understanding your question: The only way the center of mass, in an intertial frame of reference, will experience a centripetal force, is if there is something to provide that net external force. Instantaneous cases are slightly more subtle, but this still holds. Think rolling without slipping on a flat ground at constant velocity, that instantaneous axis of rotation is at the base of the wheel. Does the center of mass accelerate? Well, it WOULD if the the axis of rotation at the base of the wheel was actually a fixed pivot locked in place, and then that pivot would provide us the centripetal force. But of course, we can't actually complete any sizable rotation about that point before we, well, run into the ground again, and then establish the next contact point on the wheel. The net result of this, is that the center of mass just moves in a straight horizontal line, it has no acceleration, even though the point on the ground is an instantaneous point of rotation.
How can you be this good? I feel like everything just fell into one place. You used a great tactic. Firstly, you introduced the formulas and laws that we are all familiar with and then you explained everything step by step perfectly (leaving nothing unexplained). Just magnificent. Thank you for this gem! 🔥
Excellent! Love to hear when these ideas just “click” for you all :)
Bro are you fuckin kidding me I was trying to understand this and now I understood and find a gem channel.
Correct me if I am wrong, but we don't even need to think about what happens after poking the stick. We just need to think what happens at the time of poking. The only acceleration the center of mass can have is that of the impulse, meaning it must move forward in the direction of that acceleration. If its motion deviates from this direction (e.g. if it was to rotate about some point), then Newton's 2nd law would be violated.
Yup, in both cases, during and after poking the stick, the center of mass motion must obey Newtons 2nd law
ive had this question in my head for months, thanks :D
Of course, glad this video helped!
This was a really well-made video so thank you for that but I have two questions that i would be delighted if you could answer.
1.
The thing I didn't understand is that if the stick were to rotate around another point on the stick, the COM would need a centrifugal force while the net force on the object is 0 which is impossible. Yes, this is true but when I look to some other point on the upper part of the stick when its rotating around its center of mass that point is making a circular motion around the COM so it should have a centrifugal force. First it seemed like the force was coming from the electrostatical forces between the atoms of the stick but the point we're talking about is pulled exactly the same amount on both sides because the stick is a rigid body. Where is this force coming from?
2.
Suppose we have a rigid stick in space that has zero velocity. If I were to apply two equal forces that face each other to the upper part of the stick while one is a little lower that the other like this:
>|
|<
|
|
|
When i think about it it feels like the stick would rotate around the middle point between those two forces. The COM needs a centrifugal force in this example for it to be true while the net force is 0 I know but I just can't simulate the stick rotating around the COM in my mind. I can't comprehend why this is not true so if you could show me why it does not rotate like that it would mean a lot to me since I couldn't have gotten this question out of my mind for quite some time now.
These are my questions and I know this was long sorry about that but I just wanted to make my thought process clear.
Ooooo, these are interesting questions, I will do my best to answer :3
1) Yes, there will absolutely be centripetal forces provided by the electrostatic forces between the molecules. You kind of answered it yourself, the force of interaction is between the particles. Think about it like this: we have this constraint that the rod is a rigid body- this is no trivial constraint. There are plenty of materials that lack the structural integrity to hold together when being spun, and even "rigid" objects will break when spun at a high enough RPM. However, as long as we accept that this constraint is true (as part of our effective model), we can ignore all the internal forces which cause this constraint to be satisfied (remember, from Newton's 3rd law all of these internal forces come in action/reaction pairs and will not accelerate the system's center of mass). I've found in my experiences with physics so far that a lot of behaviors that seem confusing at first have to do with the fact that we're "hiding" or just ignoring the internal forces of constraint such that we can keep things simple: analyze net external forces on the center of mass only + net external torques.
2) Hmm, so in my example I give a very quick impulse that then gives a quick change in angular momentum. I don’t see why this would be any different, if the net torque is in the + direction from the brief combination of those 2 forces, we’ll get a + angular momentum afterwards (and vice versa if -). As you said, there’s nothing to provide a centripetal force to let the center of mass circle around, so we should just have a net rotation about the center of mass after all is said and done like before. I think it feels like the point should be somewhere other than the center of mass because that force you indicated on the bottom we would think of as like a physical object which takes the role of a pivot point, for example like if I shove the stick into the edge of a table for a continual amount of time, the pivot ends up being the contact point between the stick and the table (this is very slippy contact point, it would be difficult to maintain this rotation without the stick inevitably sliding up or down). And thank you for letting me know you enjoyed the video!
5:30 if you came just for the answer and don't require any of the context of rigid bodies/Newton's 2nd law of linear and rotational motion :)
I have two doubts: does an object thrown in the air(having both translation and rotation), would want less resistance to rotation or translation?
Next doubt is whether centre of mass always has only transitional acceleration when any 3d object is thrown in the air
Also, is the translational resistance constant for a particular object?
when a simple rigid body (such as a rod) experiences a torque about its center of mass, it will have more energy (same linear but also rotational) than one where the force was along the center of mass. how is this possible? what's the mechanism behind it having more energy? how does the force hitting a different part of the same object assuming no energy loss result in this?
Remember, work requires forces acting over displacements. It’s not fair to say “the same force” adds more or less energy. To find the amount of energy gained with the rotating body you’d have to look at the angular displacements and the center of mass displacement together. Those transient details (which are usually ignored if the impulse time is small, like the “poking” I showed in the video) must follow the work energy theorem
But one question, how can we have an external torque working? After the initial impulse, there are no net forces acting on any point on the stick. So there may be an angular momentum but should be conserved and no external torque should apply. Please correct me if I am wrong. Thanks for your presentation.
Hi Sudipto, yes the initial “poke” as I call it provides a net external torque *in that moment* to cause the rotation. After that angular momentum remains constant, thus the stick continues to rotate about its center of mass with a constant angular velocity thereafter. Good question!
@@madaydude_physics Thank you. :-)
Hello, first of all thanks for this video. Can we still conclude that the rigid body rotate around its center of mass even if we take into account gravity (lets say constant gravitational field)? If yes, what is the proof, I have difficulty proving it.
@@JohnDoe-xc2nzIf the gravitational field is uniform then think of this: the force vector distribution will be solely dependent on the *mass distribution* of your object. Define a “center of gravity” using this distribution as the point on the object where the net torque about it is 0.
The center of mass location is *also* solely dependent on the mass distribution of your object.
Thus, what you will prove is that the center of gravity is identical to the center of mass point in a uniform gravitational field. Thus, there is no torque about the center of mass just like in this example in the video.
@@madaydude_physics
Thank you very much for your time and response. What I don't understand is that the reason you gave in this video to justify the fact that the rigid body was rotating around its center of mass was that if it was not the case, then that would mean that the center of mass has an acceleration not equal to zero, which contradict the initial hypothesis. However, since now we have a gravitational field, center of mass has an acceleration so we cannot use this argument anymore. What argument should I chose to justify that rigid body is still rotating around center of mass despite the fact that now the center of mass is accelerated by gravitational acceleration. Sorry if I misunderstood something trivial.
@@JohnDoe-xc2nz Here’s the difference: the gravitational field would cause a net *linear* acceleration on the center of mass. This is different than the requirement to have an object’s center of mass move in a circle (like in an object spinning at a point other than its center of mass) which would require a *centripetal* acceleration.
@@madaydude_physicsThank you very much!! Not all heroes wear capes!! Have a good one.
@@JohnDoe-xc2nz Glad this helped! :3
Question: what about the angular acceleration of a free body while we are applying the force? What makes it suffer angular acceleration about the center of mass (or why does T=Icm*alfa only apply to CoM)? E.g. positive linear acceleration the same for all points, but then I calculate torque relative to another point (that is not the CoM) and say the motion is the sum of translational acceleration and rotational acceleration about that point.
Thanks in advance.
Yeah, you can totally calculate torque about any point you desire, as long as you're careful. The center of mass is very useful here (and in many problems) because Newton's 2nd law tells us that net external forces cause accelerations on the center of mass. So it makes things very intuitive to lock into the center of mass and measure torque about that point. You don't necessarily HAVE to do this: the more general expression for torque on a point particle is r cross F. For the stick example in this video, if you picked some arbitrary point other than the center of mass to measure torque about while I was poking it, you would get effectively 2 terms once you sum together all r_i cross F_i, where i refers to the ith particle on the stick: one term would be a torque about the center of mass, and the other would be a torque due to the center of mass of the stick accelerating (but this term doesn't really help us analyze any new physics). And, because we have these 2 separate terms, we can't just define some "I*alpha", the total, combined motion is now more complex than that. You can write the T_p = I_p*alpha only in situations where all the particles actually are spinning together around some axis through point p, all with unified angular acceleration, because now there's only one term to worry about.
TLDR: you can always measure torque about any point, not necessarily the center of mass, but it will generally just give you an extra constant term which is not important. We should really just stick to analyzing this part of the motion with Newton's 2nd law.
Although, I will add one last comment to this! Consider Earth orbiting around the sun. In this problem, it turns out to be very useful to measure the angular *momentum* (not torque, but close enough!) about the sun, and not just the Earth's center of mass. If you do this, similar to what I said earlier you would get the 2 terms: 1 SPIN angular momentum from the Earth rotating around its axis, and 1 ORBITAL angular momentum from its circulation around the sun... so indeed, in certain cases it does become useful to measure these angular quantities about points other than the center of mass, or a point where all the indivdual particles share some common angular velocity/acceleration!
I have a question:
Since we can define an instantaneous axis of rotation for any rigid body in motion, which is a point about which the entire body is rotating about at that instant;
doesn't that mean that even the center of mass is rotating about that point at that instant? If yes, this implies that it experiences a centripetal acceleration towards that point even though there is no external force.
Hopefully I'm correctly understanding your question:
The only way the center of mass, in an intertial frame of reference, will experience a centripetal force, is if there is something to provide that net external force.
Instantaneous cases are slightly more subtle, but this still holds. Think rolling without slipping on a flat ground at constant velocity, that instantaneous axis of rotation is at the base of the wheel. Does the center of mass accelerate? Well, it WOULD if the the axis of rotation at the base of the wheel was actually a fixed pivot locked in place, and then that pivot would provide us the centripetal force. But of course, we can't actually complete any sizable rotation about that point before we, well, run into the ground again, and then establish the next contact point on the wheel. The net result of this, is that the center of mass just moves in a straight horizontal line, it has no acceleration, even though the point on the ground is an instantaneous point of rotation.
@@madaydude_physics makes sense, thank you
Thanks for explanation 🎉
Of course!
fantastic
thanks for the explanaysh
Sure thing!
Bro just 👌🏻
excellent video
Thank you! Glad to hear you enjoyed