12 - Oscillations - Physical pendulum

[@majidsedighi5816]

Thank you so much

[@sabinbhandari8106]

absolutely genius ✌✌😍

[@bilalzaman498]

great sir great video subscribeed

[@markkennedy9767]

Hi thank you for this. You might be able to help me. With the physical pendulum I understand that an equation relating torque to angular acceleration times the moment of inertia around the pivot (I_a), is the way to solve it: mlgphi+(I_a)phi double dot (where we use the usual small angle approximation sin(phi) as phi). But could you tell me why a simple equation using Newton's second law on the centre of mass along the phi hat direction of the circular arc through the centre of mass doesn't work. In other words: mgphi+mphi double dot=0. The reason I ask is translational motion for an extended body can be described using Newton's second law as long as we use the mass m and the centre of mass. Then shouldn't the net force applied at every instant through the centre of mass act in the phi hat direction and be equal to the mass times the acceleration in the phi hat direction. I'm obviously overlooking something as the physical pendulum isn't reducible to the simple pendulum. Can you shed light on this. I would appreciate any help. Thanks.

[@CogverseAcademy]

Well, to your point precisely, the translational motion can be described by applying F_net = ma (Newton's Second Law) to the center of mass, but here we are talking about purely rotational motion. In other words, there is no point of the physical pendulum, not even the center of mass, that undergoes translational motion. Therefore, to properly describe the motion at hand you have to apply net torque = I*ang acceleration where the moment of inertia is going to describe how the mass is distributed relative to your axis of rotation (and the mass m all alone fails to do). Make sense?

[@markkennedy9767]

@@CogverseAcademy Hi thanks for getting back. When you say there is no point in the extended body undergoing translational motion, is this actually true. 🙂 This is actually central to my issues understanding this: what exactly is translational motion and what is rotational. (And I know about Chasles theorem- how all motion of a body can be decomposed into translational motion and rotational motion). My rough understanding at the moment is translational motion is any motion that doesn't involve a pivot in the body around which the body moves. So I suppose what you say makes sense: this is purely rotational motion since the pivot is inside the body. I was thinking of this since I posted the question and my understanding is that using the force equation would only describe a situation where somehow the extended body maintained its orientation as the COM travelled along a circular arc (as if somehow the COM was attached by a string to a pivot outside the body; so essentially orbital angular momentum). Which obviously doesn't happen here: If I did this, I would be neglecting the spin angular momentum around the COM that seems to arise because of the pivot here. So I suppose the torque equation in describing the angular acceleration, automatically takes care of these orbital and spin components of the rotation. Does this sound right? I still feel I'm missing something: what is it about the pivot here that gives the missing spin angular momentum that I would miss if I used the force equation: On the face of it, if you think about it, we still have a net force along the phi hat direction (ie the mgphi) that we surely can still put equal to the mass times acceleration of COM along the phi hat direction. Irrespective of whether the overall motion is rotational or translational or a combination. I mean we're just looking at the motion of the COM along the circular arc and equating the net force to the mass times it's angular acceleration using polar coordinates. I know the torque equation works but I can't seem to say why the force equation just described is wrong. Can you comment on this. Even just to comment on why exactly the force equation as described over the phi hat direction (polar unit vector) can't work. Thanks.

[@CogverseAcademy]

@@markkennedy9767 In the case of a solid with a rigid shape, the motion is considered to be translational motion if any line segment between two points A and B of the solid maintains a constant direction throughout the motion of the solid. If you pick any two points of the physical pendulum and draw the line segment that connects them, you should be able to convince yourself that the direction of the line segment at each instant changes. This motion is therefore not translational motion. What it is, is rotational motion because each point of the solid moves along a circular path about a fixed point (the pivot). This fixed point does not have to be inside the solid, it can be anywhere in space. A more practical reason why you would apply net torque = I*alpha is that all the points of the solid undergo the same angular acceleration at any given time, because they all rotate about the pivot. They do not all have the same linear acceleration however so it is far less useful to try to describe what is going on with F_net = ma. Now, what you can do, is derive the linear acceleration of the center of mass once you know the angular acceleration alpha of the solid. You would write a_com = R*alpha where R is the distance between the pivot and the COM. Does that make sense?

[@markkennedy9767]

@@CogverseAcademy Again thanks for getting back. Since my last post I realise I'm confused about translational motion. Translational motion is as you just described and the pivot can be inside or outside for rotational motion. However! However I'm re-reading my Kleppner and Kolenkow and on p. 262 they derive how the angular momentum of an extended body around a fixed axis can be decomposed into its spin angular momentum (e.g. earth spinning on its axis) plus its orbital angular momentum (e.g. earth spinning around sun). Now annoyingly, in their derivation they talk about how the orbital angular momentum term R X mv comes from the "translational motion" of the COM. So they correlate orbital angular momentum with what they call the translational motion of the COM. In other words e.g. they regard the earth's orbital motion as translational in this sense. This is actually why I have been confused about translational motion so far. I actually agree with your definition. But I'm left wondering then with the earth's rotation about the sun and it's spinning on its axis: does the earth not have any translational motion? Edit: after further reading, the COM translates, everything else rotates I think. Anyway, after thinking more on this, even if the pivot was outside the extended body here, the force equation wouldn't really be applicable would it. The torque equation would still work yeah? Edit: no, after thinking about it, I think you could use the force equation in the case of the extended body undergoing orbital rotation around a pivot outside the body as long as there is no spin angular momentum around any point inside the body. Does this sound right? Lastly, I still can't see why we can't zoom into the COM, and using a polar coordinate system, look at the phi hat component of the net force acting at the COM (ie mgphi), and equate it to the mass times rphi double dot. I know it contradicts the torque equation but I can't justify to myself why it doesn't hold. Everything's inertial, we have a force on one side and a kinematic expression in phi hat times mass on the other side. Can you pick holes in it. Thanks.

[@CogverseAcademy]

@@markkennedy9767 I wouldn't state that they consider the motion to be translational in the case of the earth on its orbit (it isn't). However, yes, you construct angular momentum from linear momentum p = mv by taking the cross-product r x mv where r is the position vector of a point mass, in this case the center of mass. This is the same mathematical step that you find in torque where you create torque created by a force F by taking the cross-product r x F where r is the vector that extends from the pivot to the point of application of the force. While that yields a quantity that measures the amount of rotation created by a force, it has no more relationship to any linear motion. You now have a quantity that lives in the "rotational motion world" and no longer in the "translational motion world". Regarding the "zooming in" part, you cannot claim that the frame is inertial: it is not. Precisely because the COM follows a circular path which therefore requires an acceleration.

[@altugeren]

I've freaked out that u can write backwards :o

[@DebasishSarker]

The video is flipped -_-

[@JeffD-z3g]

@@DebasishSarker even if you flip the video he is still writing backwards relative to himself

[@DebasishSarker]

@@JeffD-z3g that's only because the camera was in opposite side of the glass. I took a screenshot, mirrored it and it's verified :)

[@JeffD-z3g]

@@DebasishSarker really? I saw that he was writing the 2 with the curved side away from his nose, which i then thought that he was writing backwards because we normally write 2 with the curved side towards our noses.