was no problem at all; when I waited for the object to cross the first mark, it was already clearly going at the terminal speed. So that was fine. Now I want to turn to air. Air, of course, behaves in an extremely different way. The principle is the same, but the values for C1 and C2 are vastly different. If we take air at one atmospheres, and we take it at room temperature, then C1 is about 3.1 times ten to the minus four in our units and C2 is about 0.85. This is very close to the density …
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thank you MIT.
i had taken this class years ago in community college and passed with a C.
i had a terrible instructor from whom i didn’t learn anything.
Have no further arguments then. I appreciate if you send me Lewins email.
I promise you that the time WILL BE LONGER ON THE WAY DOWN. I have actually asked professor Lewin myself and he confirmed this fact. If you want you can look at this as a one-dimensional problem, the time does only depend on the forces and the velocities in the y-direction. If you ever come to the conclusion that t1 is larger than t2 when you try to derive them you can be sure that you have made a mistake in your derivation. If you don’t believe me you can send an email to Lewin and ask him.
I wrote those equations and it will take me a while to solve because of its complexity. But after pre-analysis Ive got a feeling that it may be opposite that it will take more time on the way up then on the way down. I have that opinion because of the shape of the curve, and tanginess (differentials) on the right and left sides.
Well, I would agree to this logic if that was linear dependency. My calculation are based on averaging air drag (making it linear), then you will get those times equals although I see many limitations of that approach. The problem I see that when an object goes up at the certain point air drag is negligible and the speed is very low, on the way back at the stop point the speed is maximum but air drag is maximum.
I have now asked a professor in physics about the correct answer and he told me I am correct. The time for the object to come back down will be longer and my explanation is correct.
If you ever make any calculations on this and show mathematically that the time will be longer, you are very welcome to show them to me, but I don’t think I will spend any time on my own to do that.
It surely is hard to derive t1 and t2 from calculations. I guess we will have to work with differential equations for that, and even if I’m not sure on that I feel that it’s not really my cup of tea. But through pure logical thinking I have become pretty confident that the time will be longer on the way down and I don’t need to do any calculations for that. Since the ball loses energy during the journey, the kinetik energy and speed must be lower and the time must be longer on the way down.
My point is air drag will help gravity to slow down an object on the way up (slow down faster) but on the way down it will work against gravity increasing time that takes the object fall down. (fall down longer). I think we wont get anywhere without formal calculation. I google it, well that is not an easy task to solve. I offer you to cooperate and find or make calculations.
2: Because of friction caused by the airdrag the ball loses energy during the whole journey. This means that all potential energy at point P does not converts into kinetik energy on the way down, but all potential energy at point P comes from kinetik energy on the way up. This means that the speed on any hight h must be larger on the way up than down.
Both ways of thinking make me come to the conclusion that the time on the way down is longer since the mean speed in the y-direction is lower.
I’m afraid that I still believe that the time on the way down will be larger. You can look on this problem in two ways, I think.
1: The airdrag-force is always working in the opposite direction of the velocity. This means that it will co-work with gravity on the way up and it will oppose gravity on the way down this means the force and therefore the acceleration will be less on the way down.
To imagine this process better you can take a tube with air inside of it. Take it into the space where you dont have gravity at all. Throw an object with acceleration =a from left to right and observe the object slowing down. Then reasonable questions arise, would it be different if you throw from right to left (acceleration is the same = a).
Should I be mistaken somewhere in my thoughts I will appreciate if you amend.
the end.
To understand this better thing about the fact that when an object moves up then air drag will work together with gravity to stop the object thereby reduce h but when the object falls down the air drag works against gravity and it will take more time to fall down compare to when you dont have air drag.
Using (1) you can write two equations, first when the object moves up and the second when the object falls. Playing with math you can find that v1t1 = v2t2. Where 1 is up and 2 is down. From (3) you can see that the speed v1=v2 thereby t1=t2
cont below.
All what ive told about was based on my intuition. Your comment made me hesitated so I did calculations.
(1)F=ma=mg-CRV^2
(2)mah=mv^2/2
From above you can derive speed v=mg/(m/2h-CR)(3)
Assume an object is moving up then you can derive h, and if it falls down then you can derive v.
cont below.
Look at the energy conservation principle. At the point P energy is MGH so to reach H will take you the same time as you go down from H to the ground and air drag effect is the same when you go up or down (the last is true only for one up/down cycle).
But the mean-speed should be less from P to S (or O as you call it) in the y-direction, because of the airdrag. This is because the ball loses energy in form of heat, caused by the friction. Since the ball therefore doesn’t come up as high as it else would and that the airdrag opposes the gravitational force on the way down it is obvious that the ball wont have the same speed on the way down than on the way up. The meanspeed will be slower on the way down and the time will be longer.
time has to be the same because time to go up to P is the same to go down from P to O.
Effects of the air resisance will be the same, no matter it travels up or down hence the time must be the same.
Anyone know the answer to the very last question he proposes his class?
I like Dr. Goodwin from Cal Tech better.
natuurkunde!!
the netherlands
I knew it, yay natuurkunde !!!
He’s from the Netherlands!
i red this but it was more in detail in my atpl theory
AWESOME!! =D