Thank you for this amazing video! As a high school student, your content is invaluable for my preparation for the physics olympiad. I truly appreciate it! I would also love to see a video discussing a conducting hemisphere placed on an infinite conducting plate under a uniform electric field.
Wow! Thank you, professor! I hope one day I get such a physical intuition as yourself, so that I could solve hard problems like this. If not for this video, it would've taken me months to come out with your solution! Are there any tips you would want to share to this end? Or is it just time and doing lots and lots of physics? Haha! I send you my highest regards! I love your channel!
Thanks for your kind words, I'm glad you're enjoying the videos! It really is just a result of many years of studying (and teaching) Physics - keep going, practise solving problems and maintain your curiosity - you'll get there!
It was a very interesting problem! You should be able to solve the final part of the question (about the minimum angular velocity required) using a method outlined by one of the other commenters on this video.
It seems to me that the fact about the time interval related to the contact with the ball and the ground should be mentioned in the problem's heading too. Something like "the time the ball is in contact with the ground is very small" . Anyways, great video, it is a very challenging problem 😊😊
Interesting that the situation naturally imposes constraints on k. E= hmg KE (post) = ½m(v²+w²) v²= 2ghk; w² =μ²2gh(1+sqrtk)² KE = mgh(k+μ²+2μ²sqrtk+μ²k) k(1+μ²)+2μ²sqrt(k)+μ²‐1 sqrt(k)
There's also rotational KE that should really be included in those equations! The angular velocity after the bounce will depend on the angular impulse the ball receives, and hence the ball's radius will end up making an appearance too.
This is all in the absence of drag/air forces right? Could be interesting to include the Magnus force which propels the object forwards as it rotates. Does there exist an angular velocity where the magnus force pulls it back to where it was dropped from? Also, by assuming negligible contact time with the ground, does this mean the angular velocity of the ball doesn’t change? What’s the minimum angular velocity for skidding? So many questions sorry! It’s an interesting problem as always. Let me try answering my own qs: If contact time is zero, there is a horizontal impulse on the ball at the ground equal to mw. Therefore there is an angular impulse (change in angular momentum) of mwr, where r is the radius of the ball. So using angular momentum, mwr = I(omega_0 - omega), and we can solve for omega after the bounce so yes it changes. My guess: if we let omega equal zero, we can solve for the minimum allowable omega_0 to stay skidding.
Questions are always welcome, it's part of what makes posting videos interesting! Including the Magnus force would certainly complicate the equation of motion as you now have an additional force which is always perpendicular to the velocity. I'm not sure if this ends up being possible to solve analytically, but will give it a try when I get a chance. For your angular velocity question - I mostly agree with your solution, but should the condition to remain skidding perhaps be rω>w instead of ω>0?
@@DrBenYelverton Thanks for the response. For the magnus force, I think there is a closed form solution and I spent quite some time working it out, but the answers I got made no sense. Frustratingly, other people who have tried to work it out online have also made mistakes. I find it hard to believe nobody has done this analysis before, but I couldn't find any! In cartesian coordinates, the equations of motion I got for a falling anticlockwise-rotating cylinder are: mx'' = -ky' and my'' = kx' - mg where the magnitude of the Magnus force is |F| = k|v|. This can be solved on wolfram alpha but gives a solution with trig functions, rather than exponentials. Changing the sign of the first equation to +ky' gives exponential solutions that seem more realistic, but why the sign should be positive confuses me. Honestly, this might be worth a video!
Another problem could be to properly solve for kh, I remember doing it 2 years back, and the results surprised me. Both components of velocity after collision are independant of radius of the ball, so any problem where a ball is spinning and collides with a surface, or even not spinning and collides obliquely with a surface has to be dealt with properly by finding angular velocities after collision and taking care of rotational energy and energy loss when it is slipping against the surface during collision, NO MATTER HOW SMALL THE BALL IS. Any problem that tells you to ignore these effects by assuming the ball is small is essentially made in the wrong light. Very interesting I might say.
@@DrBenYelverton As far as I remember, only h is given, coeffecient of friction mu and rotational velocity omega. I think we may also need coeffecient of restitution. It could be that I am remembering incorrectly. I will whip up my past notes and repost here.
You could model the ball as a spring and use its effective spring constant to find the time taken for all KE to be converted to EPE, then back into KE. As for how to estimate the effective spring constant, I assume it could be done if you knew the radius and thickness of the ball and the Young's modulus of the material, but it's not obvious to me how you'd put this all together!
Still wrestling with the fact that initial angular velocity or acceleration does not change the horizontal distance.( In tennis for example is obvious that it does(?))
We're assuming that the angular velocity is high enough that the surface of the ball scrapes against the surface for the entire time it's in contact. Presumably tennis balls are not usually spinning fast enough to enter this regime, so the result from the video doesn't apply. If you want to make this quantitative - have a look through the video comments and you'll find an overview of a method we could use to estimate the minimum angular velocity for which the result is valid.
Nice problem but the whole "ignore \delta t" thing seems a little deus ex machina. Don't you need some kind of estimate of \delta t to justify your neglecting of it? And its magnitude would strongly depend on the value of \mu, I think, as well as maybe trickier stuff like the modulus of compressibility.
Thank you for this amazing video! As a high school student, your content is invaluable for my preparation for the physics olympiad. I truly appreciate it!
I would also love to see a video discussing a conducting hemisphere placed on an infinite conducting plate under a uniform electric field.
Interesting suggestion..
@@nako7569 Thanks! I hope Dr. Yelverton will appreciate it!
Well done, thanks!
Wow! Thank you, professor! I hope one day I get such a physical intuition as yourself, so that I could solve hard problems like this. If not for this video, it would've taken me months to come out with your solution! Are there any tips you would want to share to this end? Or is it just time and doing lots and lots of physics? Haha! I send you my highest regards! I love your channel!
Thanks for your kind words, I'm glad you're enjoying the videos! It really is just a result of many years of studying (and teaching) Physics - keep going, practise solving problems and maintain your curiosity - you'll get there!
Thanks professor Ben for sharing this question physics. From the Brazil here
Thanks for watching!
Thank you so much for the help sir ❤
It was a very interesting problem! You should be able to solve the final part of the question (about the minimum angular velocity required) using a method outlined by one of the other commenters on this video.
It seems to me that the fact about the time interval related to the contact with the ball and the ground should be mentioned in the problem's heading too. Something like "the time the ball is in contact with the ground is very small" .
Anyways, great video, it is a very challenging problem 😊😊
Interesting that the situation naturally imposes constraints on k.
E= hmg
KE (post) = ½m(v²+w²)
v²= 2ghk; w² =μ²2gh(1+sqrtk)²
KE = mgh(k+μ²+2μ²sqrtk+μ²k) k(1+μ²)+2μ²sqrt(k)+μ²‐1 sqrt(k)
There's also rotational KE that should really be included in those equations! The angular velocity after the bounce will depend on the angular impulse the ball receives, and hence the ball's radius will end up making an appearance too.
This is all in the absence of drag/air forces right? Could be interesting to include the Magnus force which propels the object forwards as it rotates. Does there exist an angular velocity where the magnus force pulls it back to where it was dropped from?
Also, by assuming negligible contact time with the ground, does this mean the angular velocity of the ball doesn’t change? What’s the minimum angular velocity for skidding? So many questions sorry! It’s an interesting problem as always.
Let me try answering my own qs:
If contact time is zero, there is a horizontal impulse on the ball at the ground equal to mw. Therefore there is an angular impulse (change in angular momentum) of mwr, where r is the radius of the ball. So using angular momentum, mwr = I(omega_0 - omega), and we can solve for omega after the bounce so yes it changes. My guess: if we let omega equal zero, we can solve for the minimum allowable omega_0 to stay skidding.
Questions are always welcome, it's part of what makes posting videos interesting! Including the Magnus force would certainly complicate the equation of motion as you now have an additional force which is always perpendicular to the velocity. I'm not sure if this ends up being possible to solve analytically, but will give it a try when I get a chance. For your angular velocity question - I mostly agree with your solution, but should the condition to remain skidding perhaps be rω>w instead of ω>0?
@@DrBenYelverton Thanks for the response. For the magnus force, I think there is a closed form solution and I spent quite some time working it out, but the answers I got made no sense. Frustratingly, other people who have tried to work it out online have also made mistakes. I find it hard to believe nobody has done this analysis before, but I couldn't find any!
In cartesian coordinates, the equations of motion I got for a falling anticlockwise-rotating cylinder are:
mx'' = -ky' and my'' = kx' - mg
where the magnitude of the Magnus force is |F| = k|v|.
This can be solved on wolfram alpha but gives a solution with trig functions, rather than exponentials. Changing the sign of the first equation to +ky' gives exponential solutions that seem more realistic, but why the sign should be positive confuses me.
Honestly, this might be worth a video!
Interesting - will put this on my to-do list!
@DrBenYelverton Btw, it would be really amazing if you also did some rigid body dynamics
Another problem could be to properly solve for kh, I remember doing it 2 years back, and the results surprised me. Both components of velocity after collision are independant of radius of the ball, so any problem where a ball is spinning and collides with a surface, or even not spinning and collides obliquely with a surface has to be dealt with properly by finding angular velocities after collision and taking care of rotational energy and energy loss when it is slipping against the surface during collision, NO MATTER HOW SMALL THE BALL IS.
Any problem that tells you to ignore these effects by assuming the ball is small is essentially made in the wrong light. Very interesting I might say.
Interesting - solve for kh given what information?
@@DrBenYelverton As far as I remember, only h is given, coeffecient of friction mu and rotational velocity omega. I think we may also need coeffecient of restitution.
It could be that I am remembering incorrectly. I will whip up my past notes and repost here.
Could you discuss about optics in the upcoming videos? My test on my physics camp is about diffraction 😊
8:24 how can we do this calculation and what information would we need ?
You could model the ball as a spring and use its effective spring constant to find the time taken for all KE to be converted to EPE, then back into KE. As for how to estimate the effective spring constant, I assume it could be done if you knew the radius and thickness of the ball and the Young's modulus of the material, but it's not obvious to me how you'd put this all together!
This is ipho 1990 !!!!
Oh, interesting!
Still wrestling with the fact that initial angular velocity or acceleration does not change the horizontal distance.( In tennis for example is obvious that it does(?))
We're assuming that the angular velocity is high enough that the surface of the ball scrapes against the surface for the entire time it's in contact. Presumably tennis balls are not usually spinning fast enough to enter this regime, so the result from the video doesn't apply. If you want to make this quantitative - have a look through the video comments and you'll find an overview of a method we could use to estimate the minimum angular velocity for which the result is valid.
Nice problem but the whole "ignore \delta t" thing seems a little deus ex machina. Don't you need some kind of estimate of \delta t to justify your neglecting of it?
And its magnitude would strongly depend on the value of \mu, I think, as well as maybe trickier stuff like the modulus of compressibility.