Latest Veritasium Video

Discussion in 'Physics' started by Georacer, Aug 30, 2013.

  1. Georacer

    Thread Starter Moderator

    Nov 25, 2009
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    So, there's this channel of Youtube, where Derek creates educational videos about physics. Quite interesting and made with passion.

    Here is the latest one:
    https://www.youtube.com/watch?v=BLYoyLcdGPc

    This is a response to the previous video, which is linked in this one.

    Watch the last (and the previous one if you like) and then read further.

    Derek invokes the conservation of momentum to explain why the two blocks of wood end up with different energy states. However, in my opinion, then fails to demonstrate the energy equilibrium.

    What's your opinion? Are you satisfied with the explanation? Do you think he missed something?
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Thank you for your post, which let me know about this site.
    Most interesting.

    Yes I think the explanation was watered down, since it did not mention the impulsive forces which act when two bodies collide.
    The presenter did hint about this when suggesting that the off-centre bullet did not penetrate quite as far.

    It should be noted that kinetic energy is not preserved (is lost) during such an impact. This is accounted for by the work of the impulsive forces.

    I noted that the blocks were initially supported on a spindle through their centre of gravity.

    There will therefore be an impulsive reaction generated at this point, and in the off-centre case an impulsive moment. Momentum and moment of momentum balances in these cases yield the correct mechanics.

    There is an apparatus, known as the ballistic pendulum, that is used to study this type of phenomenon. In this apparatus the bullet is fired at a suspended heavy pole that swings to one side as a result of the impact and the angle of swing is measured.
     
    Last edited: Aug 31, 2013
  3. Georacer

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    Nov 25, 2009
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    Part of the kinetic energy of the bullet is expected to be lost on plastic deformation while the bullet is lodged in the wood block. I felt it was the right direction to look at the depth of the bullet hole, but the results came the same, which surprised me.

    If the wood block support counteracted momentarily the rotation of the block and hence allowed the bullet to penetrate as deep, then why did the block finally end with more total energy? Shouldn't the offset bullet indeed penetrate less deep, if it was to transfer more kinetic energy to the block?

    Can you lay out in more detail what you think happened there? I don't catch what you 're trying to say.
     
  4. studiot

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    OK it works like this.

    In the initial impact kinetic enery is not conserved.

    See here

    http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

    or here

    http://ocw.mit.edu/courses/physics/...echanics-fall-1999/video-lectures/lecture-17/

    However momentum is.

    The youtube video correctly identified that the linear upward momentum and the angular momentum are separate.

    In both cases exactly the same upward momentum of the bullet is transferred to the (block+bullet).

    So both combined (block+bullet)s have the same upward momentum and are subject to the same constant acceleration (deceleration) due to gravity so will rise to sensibly the same height, since acceleration edit: at constant mass is rate of change of momentum divided by mass.

    I attribute the differences in the various heights reached over several experiments to the blocks being of nearly but not identical mass. Since they were of simple wooden stock this is the most likely source of variation. The video did not make this point.

    So yes I agreed with them up to this point (and yes I got the height prediction correct)

    Are you with me thus far?
     
    Last edited: Aug 31, 2013
  5. Georacer

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    Yes, I can follow you. Go on.
     
  6. studiot

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    Now any object with mass that is in motion posseses momentum, which is a vector.

    This vector exerts a property called moment of momentum about any point not in its line of path.
    The value of this moment of momentum is the product of the momentum and the perpendicular distance from the point to the line of path.

    compare with the definition of an ordinary moment of a force about a point.

    If the path of the bullet is through the centre of gravity of the block the moment of momentum is zero.
    This zero moment of momentum is conserved after the collision and the combined (block+bullet) does not rotate.

    If the path of the bullet is to one side of the COG then the moment of momentum will not be zero and this value will be conserved into the combined (block+bullet). So the combined (block+bullet) rotates about the COG, wobbling slightly since the old COG is slightly displaced compared to the COG for the combination.

    Of course the modern name for moment of momentum is angular momentum.
     
  7. Georacer

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    So you are saying that the system initially had angular momentum because the speed vector of the bullet was off-center of the system COM?

    That is an interesting concept which I was not aware of.
     
  8. studiot

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    Yes, but nothing was rotating at that time, just as a force has a moment about any point, even if it is not exerting it on a material object at any particular time.

    If you do an angular momentum balance you come up with the same result as the linear momentum balance that was used in the videos I posted.

    This must be so since linear momentum, mv = mrω

    The angular momentum = moment of this = mv times r = mr^2ω

    Whatever momentum balance you do, the normal analysis makes the assumption that the bullet strike is so short that nothing changes position during the momentum transfer.

    If we further assume that the mechanism of transfer offers a constant resisting pentration force and therefore constant deceleration we can estimate the deceleration, and therefore the force using the bullet speed of 240m/sc from Lewin's experiment. The final speed of the bullet is zero, relative to the block. The block looked like a piece of four by two to me and allowing a 20mm penetration we can obtain acceleration from V^2 = U^2+2fS (f negative).

    I haven't done that yet but will try to post some actual figuring tomorrow.
     
  9. Georacer

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    Thanks. I learned something today as well.

    About calculating forces and deceleration on impact: This is kind of a constant complaint I have. We were all presented with pictures of feet and footballs in school when talking about forces and F=ma. But if you wanted to simulate the realistic action of kicking a ball, you would find that none of the forces is an impulse, the acceleration either and that you'd have a real hard time to measure the contact time. Sigh...
     
  10. studiot

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    So here is what I have been able to figure out, since all the information is not available.

    The Block looks like a 6 inch piece of 4x2 to me.

    Thus in metric

    size = 0.150 x 0.100 x 0.050, at 0.35kg

    The bullet is stated as about 100J Kinetic energy

    A 3.5 gram bullet travelling at 240m/s fits this (Lewin's bullet)

    KE = 0.5 x 0.0035 x 240 x 240 = 101J

    Upward momentum V(bullet) x 0.0035 = V(block+bullet) x0.353

    V(block+bullet) = 2.38 m/s upwards.

    The block moves freely upwards, retarded by gravity after impact until it stops.

    0 = (2.38x2.38) - (2x9.8xh)

    h = 0.289 m

    PE at apex = KE just after impact = (0.289x0.353x0.98) = (0.5x0.353x2.38x2.38) = 1 J

    So their statement that 97% of the impact energy goes into penetrating the block and only a couple of % goes into raising or spinning it semms to work out.

    If this is the case then the loss of penetration depth to allocate energy to spin the block is indeed negligable.

    I thought the stated rotation rate of the block rather fast at 11 rps. I thought the video was real time not speeded.
     
    Last edited: Sep 1, 2013
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  11. Georacer

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    I see. So all the work goes into lodging the bullet in the wood. In that case, small variance between the blocks, due to the "water" of the wood, could indeed impact the final height of the block.

    Thanks for walking me through!
     
  12. Metalmann

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    Dec 8, 2012
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    Back when I was young, we used to shoot propane cylinders to watch them explode. A "whole jar full" of Physics going on then.;)

    No, we didn't use formulae; but we could predict how far they would fly.:D
     
  13. WBahn

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    We shot some propane tanks (BBQ grill tanks that had the old style valves and couldn't legally be filled any more -- but we talked someone into filling them when they found out what we were doing plus we had some that still had a bunch of gas left) and it was quite disappointing. The higher powered rounds just punched holes in the tanks. The tracer rounds occasionally ignited the propane, at which point it just burned in a jet of flame as it exited the tank. If you hit the tank again if often extinguished the flame because it disturbed the tiny region where the mixture finally had enough air entrained to burn.

    What I always wanted to do -- and have a way-too-developed sense of self-preservation to try -- is to fill a tank with a propane/oxygen (or even just propane/air or, better yet, and oxy/acet) mixture and repeat the experiment. If I were to do that, I would rig a remote fill device so that I could charge the tank from afar.
     
  14. Wendy

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    Mar 24, 2008
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    Look up Brown's Gas (a form of over unity for the nuts). Pretty close.

    Basically you electrolyze a mix of H2 and O2, and fill a tank with it. People have used it with fair success, but personally I don't want to be in the neighborhood if it is.
     
  15. studiot

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    Here is my final installment.

    The moment of inertia, I of a rectangular block, of sides a,b,c about an axis parallel to c through the COG is

    I = \frac{{m\left( {{a^2} + {b^2}} \right)}}{{12}}

    So the moment of inertia of the block about the support pin axis is


    I = \frac{{0.353\left( {{{0.1}^2} + {{0.15}^2}} \right)}}{{12}} = {10^{ - 3}}

    Assume that the bullet impacts 50mm from the centre (this accords with the Xrays).

    Angular momentum = moment of momentum about COG


    I\omega  = {M_{angular}} = {M_{linear}}r = m{V_{linear}}r


    {10^{ - 3}}\omega  = 0.353x2.38x0.05


    \omega  = 42rads/\sec  = 42/2\pi turns/\sec  = 6.7

    Which is again less than their values but quite close -considering I have estimated some of the input values.

    Summary

    Energies

    {E_{rotational}} = I{\omega ^2} = 0.001x{6.7^2} = 0.045J

    {E_{translational}} = 1J

    {E_{original}} = 101J


    So I would say that the rotational energy is insignificant relative to the translational.
     
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  16. WBahn

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    And this is my biggest criticism of the video.

    Their own analysis revealed how the scale of the variation from one test to another swamped any difference between on-center and off-center heights attained. If your energy calculations are to be used, the difference in energies is so slight that their results neither prove nor effectively demonstrate anything. But it was a nice try.

    What they (or someone) need to do now is devise an experiment in which the fraction of energy transferred is much higher and, more importantly, that the rotational energy is roughly comparable to the translational energy.

    But one thing (recalling from memory) that they said that didn't seem to jive with something they said later. I thought, contrary to your results here, that they claimed that the rotational kinetic of the spinning block was about half of the gravitational potential energy achieved by the either block. Yet later they remarked how the difference in penetration that would account for the difference in energy was only a millimeter or so.

    Also, I don't recall them taking into account the fact that the bullet was embedded in the block and thus part of both the translational and the rotational energies.
     
  17. studiot

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    Please note they did not publish all the figures, so some of my 'figurin' was informed guesswork.


    Yes they did account for conservation of energy, although they made several contradictory statements through the two videos.

    Yes their calculation for the rotational enery was larger than mine.

    Teo points

    1) Their rotational speed was greater. The energy is proportional to the square of the speed so four times for a doubling.

    2) They did not publish details of their moment of inertia calculation. It is very small, perhaps they used the wrong formula and got a larger value (for a rectangualr cross section or lamina)
     
  18. Georacer

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    Nov 25, 2009
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    Thanks for the analysis. I didn't check your numbers but I don't feel I need to.

    Tell me if I get this right:

    The final heights in both experiments should (and were) theoretically be the same. In the case of the off-center shot, the block will rotate as slow as needed to ensure that the energy content of the rotation is low enough so as the previous statement is true.

    Is the above paragraph correct?

    It still isn't apparent to me that the energy equilibrium is maintained through trade-offs in the bullet lodging in the block, but the momentum explanation is clear enough.
     
  19. WBahn

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    The most telling way to analyze it (though not always the easiest way) is to focus on the center of mass of the system (the block and bullet combined). At any given instant, the momentum of the center of mass of the combined system must change only in response to net external forces (F=dp/dt). Similarly, the angular momentum about the center of mass of the combined system much change on in response to net external moments.

    The two are independent constraints.

    Satisfying both will determine the total energy after the collision and that will determine how much of the energy before the collision will be converted to some other form. In this case, there was a LOT of energy to spare.

    Remember that they talked about energies of the block being on the order of 1J or 2J. In order to be legal for hunting in most states, a rifle bullet must have 1000ftlb of energy 15' from the muzzle. That's about 1500J. Even a typical .22 Long Rifle produces about 150J of muzzle energy.

    Assuming they only used a .22 (and I would have in their place) 99% of the bullet's energy was converted to something other than kinetic energy.
     
  20. studiot

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    No, sorry, this is not the case.

    At the outset the block is at rest and the bullet moving so the total energy of the system comprises the kinetic energy of the bullet.

    At the instant of impact the block and bullet are both not moving. The bullet is beginning to embed itself into the block and overcome the increased resistance to motion it thus faces by applying an impulsive force to the block.

    During this impulse most of the kinetic energy of the bullet is transferred to the block into dissipative processes dissipate this energy (they do not destroy it so conservation is preserved).

    However.

    In addition to kinetic energy the bullet also possesses momentum. The block initially possesses zero momentum.

    Energy is not conserved because it is dissipated.

    But momentum is always conserved.

    This momentum conservation allows us to write an equation connecting momentum before and after.
    We cannot describe the momentum during the impulse.

    The block with the bullet embedded then rises with exactly this momentum (momentum cannot be disspiated like energy).

    The equation thus allows us to calculate the consequent upward velocity of the block+bullet.

    Thus the rising block then acquires some kinetic energy, determined by the velocity of rise , in turn determined by the momentum of the bullet.
    This secondary KE comes from the total energy of the system ie the original KE of the bullet.
    This secondary KE and nothing else determines the height the block rises to, at which point the energy has all been transformed into potential energy in the block.

    If the strike is offset from the line through the centroid the block will also be set into rotation about the centroid.
    The rate of rotation is fixed by the properties of the block and the distance the strike is offset. (I have guessed at 50mm).
    This rotation also endows the block with some energy, which also comes from the original total energy.

    So Original total Energy = Dissipated penetration energy + PE + Rotational energy.

    Eo = DE + PE + RE

    Since PE is the same whether the block rotates or not and since Eo is fixed, DE must vary when RE is nonzero.

    However RE is so small that the variation of DE (and thus the penetration depth) is not discernable by this experiment.
     
    Last edited: Sep 2, 2013
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