You are wonderful! You have been so helpful to me this entire semester! You have allowed me to get a much higher grade in physics than I ever imagined! Thank you!
At first I didn’t quite understand the moment of inertia of the sandbag after contact being mR^2 but then I realized that the model for the sandbag is a mass at the end of a string of length R and I understood. Thank you Prof. van Biezen, I have watched scores of your videos with much gratitude. Thank you, sir. ⭐️
Hi sir. What if it was elastic collision? We know that kinetic energy is conserved in this case. This would mean that also linear or translational kinetic energy is included in the equation of conservation of kinetic energy together with the rotational kinetic energy? Or translation kinetic and rotational kinetic are separated equations?
professor can we assume that when the weight has been dropped onto the cylinder it has become a part of it's mass and thus we can treat both masses as being one (m+M) and the corresponding Rotational Inertia (1/2)(m+M)(R squared)
You are correct with the first part, but not with the second part. You have to calculate the moment of inertia separately. I total = I disk + I bag = (1/2) MR^2 + mR^2
Looking at the example which is a Completely inelastic example for conservation of angular momentum, I just wondering is there any example of a completely elastic case for conservation of angular momentum?
if the sandbag is dropped on the center of the disk , the moment of inertia for the bag will be zero , but concerning the disk can it be 0.5(M+m)R^2 ? because i think that there would be a change in the motion of the disk , and we should be able to figure it by calculations
If you drop the bag in the middle so that R = 0, then the final angular velocity will be equal to the initial angular velocity. (Note: let R = 0 for the moment of inertial of the bag).
@@MichelvanBiezen The reason this seems implausible is because if we did this in real life, a bag is not a point mass, and part of it will jut over the centre, thus having some angular momentum. But if we assume it is a point mass, then R=0 and angular momentum is zero.
Could you possibly do an example involving centripetal acceleration and angular momentum? Like two masses connected via massless string to a massless rotating cylinder?
The disk will continue to rotate (slower) about its center of mass. The sandbag will remain motionless (relative to the disk) on the edge of the disk and will not rotate.
Hello, do you have any rotational mechanics for astronomy type stuff or can all of this be applied to planetary body's?? I've noticed a few videos where you have G(Mm/r²) for finding masses and such but I was curious to whether these can also be applied to astro and gravity subjects
NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO: the bag, having been dropped from a certain height, lands on the edge of the disk with a certain velocity (the bag does not materialize on the disk from nothing with zero velocity!). The bag's angular momentum IMMEDIATELY BEFORE THE COLLISION is, with respect to the center of the disk, the vector product of its position vector from that center and its linear momentum - mass x velocity immediately before the collision -, the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation. The total angular momentum of the system disk + bag, IMMEDIATELY BEFORE THE COLLISION - a vector! -, has therefore two components: one along the axis of rotation of the disk, contributed only by the disk, and one along an axis perpendicular to it, contributed only by the bag. The total angular momentum (disk + bag), IMMEDIATELY AFTER THE COLLISION, has only one component, the one along the axis of rotation (the disk and the bag rotate as a single body about the original axis). The total angular momentum of the system is therefore not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the center of the disk has to generate an EXTERNAL (to the system) torque of "impulsive" nature, i.e., intense and of very short duration, to maintain the rotation about the original axis! What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis! With this premise, your calculation is then correct but the reasoning and the logic behind it conceptually and pedagogically misleading.
Shouldn't the initial momentum also include the linear momentum of the bag? So wouldn't it be Iw + mv (both momenta have the same vector)? After all isn't the sandbag part of the system, would it not affect the final momentum?
@@MichelvanBiezen NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO. It is, with respect to the center of the disk, the vector product of the position vector of the bag from the center of the disk and the bag's linear momentum immediately before the collision (mass x velocity immediately before the collision), the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation. The total angular momentum of the system disk + bag, IMMEDIATELY BEFORE THE COLLISION - a vector! -, has therefore two components: one along the axis of rotation of the disk, contributed only by the disk, and one along an axis perpendicular to it, contributed only by the bag. The total angular momentum (disk + bag), IMMEDIATELY AFTER THE COLLISION, has only one component, the one along the axis of rotation (the disk and the bag rotate as a single body about the original axis). The total angular momentum of the system is therefore not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the disk has to generate an external torque of impulsive nature, i.e., intense and of very short duration, to maintain the rotation about the original axis! What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis! With this premise, your calculation is then correct.
Because the Weight acts downwards, which is parallel to the axis of rotation, therefore it wouldn't have any impact on the net torque. Alo the formula says that Torque=Frsin(alpha), and when alpha=0, then sin(alpha)=0, hence no new torque :))
@@aleksandertrakul1446 The moment of a force - torque - is, like the angular momentum, calculated with RESPECT TO A POINT NOT TO AN AXIS. It is defined as the vector product of the position vector of the point where the force is applied from the reference point and the force. In our case, since the angular momentum is referenced to the center of the disk, also the moment (torque) of any externally acting force, including the weight of the bag, is to be calculated with respect to that point. It this clear, then, that such a moment IS NOT ZERO; in particular, it is not zero IMMEDIATELY BEFORE and IMMEDIATELY AFTER THE IMPACT. The position vector of the bag from the center of the disk is indeed perpendicular (it is in the plane of the disk) to the weight of the bag (directed vertically downwards). So the sin of the angle in between (90 deg.) is 1, different from zero, like the torque, which is perpendicular to the plane defined by the position vector and the weight. Although negligeable with respect to the "impulsive" torque developed by the constraint at the center of the disk during the impact time to maintain the rotation about the original axis, it exists, non-zero, for all the duration of the process. AFTER THE IMPACT, it gets counterbalanced by the constraint which has to develop a torque equal and opposite to that of the weight of the bag, now attached to the disk, to keep the component of the total angular momentum (disk + bag) perpendicular to the axis of rotation constantly equal to zero.
sir if this bag is having some initial linear momentum then will it not affect the conservation of angular momentum .or it is something like while applying conservation of angular momentum we dont include linear momentum
If the bag has initial linear momentum it will indeed affect the disk's angular momentum. Take a look at video # 3 in this playlist: PHYSICS 13.5 ANGULAR MOMENTUM th-cam.com/users/ilectureonlineplaylists?view=50&shelf_id=4&sort=dd
J On the left side of the equation you have the angular momentum BEFORE the collision. (the bag does not have any angular momentum before the collision). On the right side of the equation you have the angular momentum AFTER the collision. (the bag is now rotating so it has angular momentum)
@@MichelvanBiezen NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO: the bag, having been dropped by a certain height, lands at the edge of the disk with a certain velocity (it - the bag - does not materialize on the disk from nothing!). The bag's angular momentum IMMEDIATELY BEFORE THE COLLISION, is, with respect to the center of the disk, the vector product of its position vector of from that center its linear momentum (mass x velocity immediately before the collision), the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation.
How angular momentum is conserved in this problem?if we consider torque about the axis of rotation ....there is certainly some external torque for the bag
@@MichelvanBiezen The total angular momentum of the system (disk + bag) is not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the disk has to generate an external torque of impulsive nature, i.e., intense and of very short duration, to maintain the rotation about the original axis! What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis!
You are wonderful! You have been so helpful to me this entire semester! You have allowed me to get a much higher grade in physics than I ever imagined! Thank you!
Elizateth. That is great! Congratulations! I am sure that was due to you studying very hard, but I am glad to read that these videos are helping.
At first I didn’t quite understand the moment of inertia of the sandbag after contact being mR^2 but then I realized that the model for the sandbag is a mass at the end of a string of length R and I understood. Thank you Prof. van Biezen, I have watched scores of your videos with much gratitude. Thank you, sir. ⭐️
You are welcome. Physics sometimes takes time to simmer and sink in. Glad you got the concept.
Fantastic playlists. Very well done, saving my finals!
Clearly an amazing person, thanks for making sense of this and taking time to have numerous examples.
holy cow thank you so much for helping me. I needed this for my final exam in Physics 1
Glad we could help! Good luck on your final exam! 🙂
You are simply a god. THANKS. YOU ARE AMAZING.
No, just a simple man with lots of faults.
Hi sir. What if it was elastic collision? We know that kinetic energy is conserved in this case. This would mean that also linear or translational kinetic energy is included in the equation of conservation of kinetic energy together with the rotational kinetic energy? Or translation kinetic and rotational kinetic are separated equations?
This man is truely great
THANK U SO MUCH U MADE ME REALIZED IVE BEEN USING THE WRONGG FORMULA THANK U SO MUCH FPR DERIVING THAT FORMULA OMG THANKS THANKS THANKK❤❤❤❤🎉🎉🎉
professor
can we assume that when the weight has been dropped onto the cylinder it has become a part of it's mass and thus we can treat both masses as being one (m+M) and the corresponding Rotational Inertia (1/2)(m+M)(R squared)
You are correct with the first part, but not with the second part. You have to calculate the moment of inertia separately. I total = I disk + I bag = (1/2) MR^2 + mR^2
@@MichelvanBiezen Do we calculate the moment of inertia for the bag of sand as we would a point mass? - which correlates with MR^2?
Looking at the example which is a Completely inelastic example for conservation of angular momentum, I just wondering is there any example of a completely elastic case for conservation of angular momentum?
In the case where the two object stick together after the collision, (which is usually the case), you don't expect the kinetic energy to be conserved.
if the sandbag is dropped on the center of the disk , the moment of inertia for the bag will be zero , but concerning the disk can it be 0.5(M+m)R^2 ?
because i think that there would be a change in the motion of the disk , and we should be able to figure it by calculations
If you drop the bag in the middle so that R = 0, then the final angular velocity will be equal to the initial angular velocity. (Note: let R = 0 for the moment of inertial of the bag).
@@MichelvanBiezen The reason this seems implausible is because if we did this in real life, a bag is not a point mass, and part of it will jut over the centre, thus having some angular momentum. But if we assume it is a point mass, then R=0 and angular momentum is zero.
Love u sir your a great teacher and changing life of many people.💞
💕💕Love from India💕💕
Thank you and welcome to the channel!
Say you didn't know the sandbag would stick; how can you determine if the bag will be propelled off the disk or not? Thanks.
This is just a hypothetical problem assuming the sandbag will stay in place after it drops.
Could you possibly do an example involving centripetal acceleration and angular momentum? Like two masses connected via massless string to a massless rotating cylinder?
Hmm...I bag=mR^2? It is interesting! Thanks professor; this is helpful.
Yup considered as a point mass
Wouldn't the disk and the sandbag rotate around their center of mass after the collision though?
The disk will continue to rotate (slower) about its center of mass. The sandbag will remain motionless (relative to the disk) on the edge of the disk and will not rotate.
So helpful, thank very much Prof
Hello, do you have any rotational mechanics for astronomy type stuff or can all of this be applied to planetary body's?? I've noticed a few videos where you have G(Mm/r²) for finding masses and such but I was curious to whether these can also be applied to astro and gravity subjects
The equations for orbital mechanics and gravity in astronomy are the same as they are for physics examples in its simpler form.
NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO: the bag, having been dropped from a certain height, lands on the edge of the disk with a certain velocity (the bag does not materialize on the disk from nothing with zero velocity!).
The bag's angular momentum IMMEDIATELY BEFORE THE COLLISION is, with respect to the center of the disk, the vector product of its position vector from that center and its linear momentum - mass x velocity immediately before the collision -, the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation.
The total angular momentum of the system disk + bag, IMMEDIATELY BEFORE THE COLLISION - a vector! -, has therefore two components: one along the axis of rotation of the disk, contributed only by the disk, and one along an axis perpendicular to it, contributed only by the bag.
The total angular momentum (disk + bag), IMMEDIATELY AFTER THE COLLISION, has only one component, the one along the axis of rotation (the disk and the bag rotate as a single body about the original axis).
The total angular momentum of the system is therefore not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the center of the disk has to generate an EXTERNAL (to the system) torque of "impulsive" nature, i.e., intense and of very short duration, to maintain the rotation about the original axis!
What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis!
With this premise, your calculation is then correct but the reasoning and the logic behind it conceptually and pedagogically misleading.
The angular momentum of the bag is zero with respect to the rotating disk.
Hi,
Your videos are excellent!
Do you have an example of angular momentum with changing mass (dm/dt)? let's say a sand falling on a disc
That's an interesting question.
How did you calculate the radius of the bag being 2m? Is it because the bag was dropped on the outer part of the disk?
akvolkswagen
That is correct. The bag of sand is like a "point" mass at a distance of 2 m from the point of rotation.
I have the same question
The bag could be taken as a sphere. Could've been a better approximation. 👍
If the bag had fallen on the center of the disk, how would we figure out the final momentum?
The moment of inertia of the bag would be zero.
Thank you so much, you're great for answering so quickly!
Question, why did you halve the moment of inertia of the disk? What happened in the bag-disk system that required us to halve the moment of inertia?
The moment of inertia of a solid disk is indeed: I = (1/2) mR^2
+Michel van Biezen ah, yes I see. Thank you.
Shouldn't the initial momentum also include the linear momentum of the bag? So wouldn't it be Iw + mv (both momenta have the same vector)? After all isn't the sandbag part of the system, would it not affect the final momentum?
Since the initial angular momentum of the bag is zero, the bag's initial momentum will not affect the final angular momentum.
Michel van Biezen
Oh I thought the sandbag was being dropped on top of the disk.
BTW, thank you for these videos!
@@MichelvanBiezen NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO. It is, with respect to the center of the disk, the vector product of the position vector of the bag from the center of the disk and the bag's linear momentum immediately before the collision (mass x velocity immediately before the collision), the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation.
The total angular momentum of the system disk + bag, IMMEDIATELY BEFORE THE COLLISION - a vector! -, has therefore two components: one along the axis of rotation of the disk, contributed only by the disk, and one along an axis perpendicular to it, contributed only by the bag.
The total angular momentum (disk + bag), IMMEDIATELY AFTER THE COLLISION, has only one component, the one along the axis of rotation (the disk and the bag rotate as a single body about the original axis). The total angular momentum of the system is therefore not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the disk has to generate an external torque of impulsive nature, i.e., intense and of very short duration, to maintain the rotation about the original axis! What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis!
With this premise, your calculation is then correct.
why is there no external torque due to the weight of sand bag ?
Because the Weight acts downwards, which is parallel to the axis of rotation, therefore it wouldn't have any impact on the net torque. Alo the formula says that Torque=Frsin(alpha), and when alpha=0, then sin(alpha)=0, hence no new torque :))
The bag was dropped vertically down. If it were thrown at an angle, it would exert a torque.
@@aleksandertrakul1446 The moment of a force - torque - is, like the angular momentum, calculated with RESPECT TO A POINT NOT TO AN AXIS. It is defined as the vector product of the position vector of the point where the force is applied from the reference point and the force.
In our case, since the angular momentum is referenced to the center of the disk, also the moment (torque) of any externally acting force, including the weight of the bag, is to be calculated with respect to that point.
It this clear, then, that such a moment IS NOT ZERO; in particular, it is not zero IMMEDIATELY BEFORE and IMMEDIATELY AFTER THE IMPACT. The position vector of the bag from the center of the disk is indeed perpendicular (it is in the plane of the disk) to the weight of the bag (directed vertically downwards). So the sin of the angle in between (90 deg.) is 1, different from zero, like the torque, which is perpendicular to the plane defined by the position vector and the weight. Although negligeable with respect to the "impulsive" torque developed by the constraint at the center of the disk during the impact time to maintain the rotation about the original axis, it exists, non-zero, for all the duration of the process.
AFTER THE IMPACT, it gets counterbalanced by the constraint which has to develop a torque equal and opposite to that of the weight of the bag, now attached to the disk, to keep the component of the total angular momentum (disk + bag) perpendicular to the axis of rotation constantly equal to zero.
Would make any difference if you drop the 50kg bag in the center instead of on the edge of the disk?
+valladolid0711
Yes, that would make a big difference. If the bag is placed in the middle, then its moment of inertia would be near zero.
I still dont really understand why you used I = (1/2) mR^2 for the disk?
sir if this bag is having some initial linear momentum then will it not affect the conservation of angular momentum .or it is something like while applying conservation of angular momentum we dont include linear momentum
If the bag has initial linear momentum it will indeed affect the disk's angular momentum. Take a look at video # 3 in this playlist: PHYSICS 13.5 ANGULAR MOMENTUM th-cam.com/users/ilectureonlineplaylists?view=50&shelf_id=4&sort=dd
how did you find the R for the sand bag? as Radius 2m was for the disk and you use them for I1 and I2 ?
Since the sand bag was dropped on the edge of the disk all of the mass is at R away from the point of rotation.
I don't understand why you cancelled out Angular momentum of the bag, yet still use Inertia2 in the problem?
J
On the left side of the equation you have the angular momentum BEFORE the collision. (the bag does not have any angular momentum before the collision).
On the right side of the equation you have the angular momentum AFTER the collision. (the bag is now rotating so it has angular momentum)
Michel van Biezen Got it.
Your videos are great!
@@MichelvanBiezen NO, prof...! The initial (immediately before the collision) angular momentum of the bag IS NOT ZERO: the bag, having been dropped by a certain height, lands at the edge of the disk with a certain velocity (it - the bag - does not materialize on the disk from nothing!).
The bag's angular momentum IMMEDIATELY BEFORE THE COLLISION, is, with respect to the center of the disk, the vector product of its position vector of from that center its linear momentum (mass x velocity immediately before the collision), the latter being, according to the geometry of the problem, parallel to the axis of rotation of the disk and directed downwards. This provides a contribution the total angular momentum (disk + bag), immediately before the collision, perpendicular to the plane defined by the axis of rotation and the position vector of the bag, hence perpendicular to the axis of rotation.
How angular momentum is conserved in this problem?if we consider torque about the axis of rotation ....there is certainly some external torque for the bag
Angular momentum is conserved since no torque is applied.
How can u say total torque is zero... explain in the problem... plz dont write directly ang mnts are equal... plz explain how net torque is zero
@@MichelvanBiezen The total angular momentum of the system (disk + bag) is not conserved through the collision: this non-conservation is explained by the fact that, during the impact, the constraint at the disk has to generate an external torque of impulsive nature, i.e., intense and of very short duration, to maintain the rotation about the original axis! What remains unaltered is actually JUST A COMPONENT of the total angular momentum, the one along the rotation axis!
can i solve this by law of conservation of angular kinetic energy [ 0.5 * i1*(w1)^2=.5*i2*(w2)^2 ]
no because energy may not be conserved.
Sir... explain clearly... how initial and final ang mnt are equal... explain how net torque zero...
Did you ever figure it out?
Thank you so much! This is so helpful.
Why is the bags radius 2 meters? What’s going on?
Once the bag drops onto the edge of the disk, the bag will be 2 meters from the center of the disk (which is the center of the rotational motion).
In which grade do you guys study this concept?
In the US, we either study this in the last year of high school, (but few students take physics in high school), or in the first year of college.
what is the minimum mass that can make the disk completely stop?
Theoretically, the mass of the bag would have to be infinite.
(In real life that would break the equipment of course)
if you were not breathing, i certainly would had drop out physics.
why the I of the disk is not 1/4mr^2. since it has a delta h
I am not sure what you mean by: "since it has a delta h". But the moment of inertia of a solid disk is (1/2)MR^2
Y sir initial moment of inertia of bag is zero
There is no movement tangential to the rotation.
gracias profesor
You are welcome! 🙂
radius of the bag ? 2m ? Confused here
The bag is considered a "point object", and therefore does not have a moment of inertial of its own.
@@MichelvanBiezen thanks a lot sir
Thanks so much!
What is your accent? Curious.
-American.
I grew up in Belgium, so you could call it a Belgian American accent.
legend
can i solve this by law of conservation of angular kinetic energy .5*i1*(w1)^2 = .5*i2*(w2)^2
No, during the collision, energy is lost and that is why we use the conservation of momentum.