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ganeshie8
 one year ago
classic yet interesting question about roller coaster :
find the minimum height from which the car needs to fall so as to complete the loop safely. The diameter of loop is 100m and the mass of car is 1000kg.
ganeshie8
 one year ago
classic yet interesting question about roller coaster : find the minimum height from which the car needs to fall so as to complete the loop safely. The diameter of loop is 100m and the mass of car is 1000kg.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443166269719:dw

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0EWWWW I might have this problem on my test!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443166523518:dw\[\sum F_y = ma_{cp} = F_{N} + F_{g} = N + mg=\frac{mv^2}{r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well that's technically a little wrong because the vector for the normal force should be longer than the force of gravity, but still....

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0I think it's more wrong because your bottom circle is elliptical and thus does not obey rules of circular motion ;)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2haha, vertical circle looks like you've fallen in the trap that physics is :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Try to prove that the minimum velocity at the bottom is \(v_{min} = \sqrt{5gR}\) and thus the minimum kinetic energy is \(\frac{5}{2}mgR\) which can be equated to the initial potential energy to find the minimum height.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0in the course of a few years parth has learned like 4 times as much physics as me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[N = \frac{mv^2}{r} mg\]um... at the top of the roller coaster the normal force is neither more than the gravitational force, and nt less, but equal to it, therefore \(N=0\) \[mg = \frac{mv^2}{r}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2why do we need to know the mass :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wasnt there yet haha

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Exactly, no need to know the mass. Jhanny is almost there. To be precise, \(N \ge 0\) to complete the vertical circle. At the minimum value of \(N\), we also get the minimum value of \(v\) at the top.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2anyway the answer is \(h_{min} = 5R/2\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that looks really nice and simpler! i think mass is given so that we can assume that newton mechanics works just fine but it is amazing how mass disappears after the calculations in most of these problems!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yeah, Galileo and Neil Armstrong!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0is that tesla in ur dp

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes if anyone loads up an Edison avatar, there's gonna be blood!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so a naive q, do roller coasters really work like this? without any chains hooked to the tracks?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[g=\frac{v^2}{r} \implies v^2 = gr\] Yeah the mass is relevant when we're finding the change in work done, but this is where I get a little lost every time. :( \[U_i +K_i = U_f + K_f\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Just from the back of my mind, i think Walter Lewin did this in one of the 8.01 lectures on MIT OCW

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1been a couple years since watching those

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0they had removed all his lectures from MIT after some controversy

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2in ideal conditions, they can work like that

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1:O, well good thing i dl them all, hehe

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yeah, he harassed a young female student on one of his MOOCs it's still a shame, because he did know his physics pretty well

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1hell yes, his lectures and feynman spiked my interest

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0watching top 5 rolller coaster accidents https://www.youtube.com/watch?v=mM0HE7DKTvg

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i am pretty sure the acceleration at the top of the circle is zero for one instant to 'just' make it over... maybe not

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1maybe a way to do it with just energy conservations

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443168272771:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yes, as Jhanny said  the velocity at the top of the roller coast is \(\sqrt{gR}\) at least so if we look at the energies at the top and at the bottom at the top: \(mg(2R) + \frac{1}{2}m(gR)\) at the bottom: \(\frac{1}{2}mv_{\rm min, \rm bottom}^2 \)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2in practise i think roller coasters go faster and you are pushed into your seat (upwards) by your inertia

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[U_i + K_i = U_f + K_f \\K_i = 0 \]\[mgh = \frac{1}{2}mgr +2mgr\]\[h=\frac{r}{2}+r\]\[h=\frac{3}{2}r\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2of course they go faster because here we're calculating the minimum velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just figured out why \(U_f = mg(2r)\) lol... _

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\(r/2 + 2r = 5r/2\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443168775694:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yah I cant either. That's not eve important. Pfft.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I just realized something

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1ah gravity doesn't matter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We had \(v^2=gr\) when we simplified our forces in the ydirection. We could have multiplied both sides by \(\dfrac{1}{2}m\) to associate the forms of kinetic and potential energies!! Whoops.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@DanJS so if g = 0 ??!!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2ganeshie try solving this question about vertical circles... it'd really test your knowledge.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1The initial potential energy from the minimum height will be equal to the sum of the potential energy at height of 2R + the kinetic energy from the min velocity to make it around

DanJS
 one year ago
Best ResponseYou've already chosen the best response.12.5R min height to start

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[h=\frac{5}{2}r \qquad r=\frac{D}{2} = \frac{100}{2} = 50~m\]\[h_{min} =\frac{5}{2}(50) = 125~m \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that's another way to look at it , @DanJS , and sorry I got a little mixed up in the Kinetic energies part... i'm not as practiced in physics anymore :\

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes PK, post... today it is going to be a gravity/uniform circular motion day for me!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe you can also post the breakingoftheplate momentum question, another classical physics question involving vectors and momentum. I always had trouble with those.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2can I have that extension, ganeshie? you know, the drawing one.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yeah I have a huge collection of questions so no worries

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0send me meeting id in teamviewer it would be simpler if i install it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Back to eating my kichadi.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2whatever... I found an image that does suit my purpose.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1https://www.youtube.com/channel/UCliSRiiRVQuDfgxI_QN_Fmw/playlists someone uploaded em

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I asked my physics professor that problem parth, he solved in probably a minute tops. then gave me the death stare as to why I had forgotten how to solve it using component methods.... =_=

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2in that image, let \(d = L/2\) and \(\theta = 90^{\circ}\) now if I release it that way, find the maximum height reached by the pendulum bob before the string becomes slack

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2treasure trove https://www.youtube.com/watch?v=7E34VOj8Elc&index=182&list=PLERGeJGfknBR3pXCPlV3bgb_qHCSNOdBf

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that is a very useful collection, i hope they don't delete his lectures from youtube too thanks @DanJS

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@DanJS Im saving this.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Walter Lewin himself has uploaded it on his own channel BTW https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1download those playlists to disk

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443169569982:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have a friend attending MIT that's met him.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2this thread is really slow for me so can we move to another one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, never knew it was called a trifecta.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1@irishboy123 Where is gravity ever zero in this universe .. infinite range

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well gravity is practically zero at the center of the earth or in free fall :D

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1hmmm, free fall means you do not sense the gravity, it is still there and it matches your acceleration of falling

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[F \propto \dfrac{mM}{r^2} \] so the Force is not defined at \(r=0\) right

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1man i discussed this awhile back, about how you can not have zero distance.. been too long

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1strong force repulsion or something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, yes. that's true, @ganeshie8

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1you can't be sure r=0 by uncertainties in position

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2nah, inside the earth gravity is pulling you outwards. zero at centre  theoretically

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, I mean an astronaut living in the space station is for all intents and purposes living with no gravity.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the r assumes a point mass and works outside the sphere that is the earth or any body

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And that's why their bone structure slowly decomposes over time  he gravity that acts on them!

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1microgravity, they are moving very fast and falling around the earth

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1like a super cannon that fires so fast the ball falls slower than the horizon curves away

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 Is the gravity at the center of a uniform Ball zero?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Jhannybean that's 2 degrees of separation, i think it's called ie w/ Lewin

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes, if you look at it classically.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1In the real universe there is gravity everywhere, i think, i can idealize a empty space void, no gravity

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 When you have the \(1/r^2\) dependence and end up at the center of the earth, this is actually a kind of false division by zero since your model was assuming we could just look as though the center of mass was good approximation. But once you're inside the earth, the outer mass doesn't affect you anymore, so this model only applies for distance for r>Radius of the earth. Just imagine yourself at the center of the earth, look at the contribution of a tiny bit of force from each tiny bit of mass, and integrate over the sphere which we can approximately see to cancel out since all the little vectors will be in equilibrium! ;D

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK if we are playing with words, the gravity *due to the sphere* is zero at the centre of the sphere according to Newton's law of gravitation :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't like playing with words one bit.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2but i agree totally that a force that has unlimited reach much reach everywhere :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Can we measure the gravitational pull of planets outside of our solar system?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Does that have something to do with a metal sphere and charge concentrated on the surface...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@DanJS but i only know classical physics so if it reaches back on itself .... :p i dunno

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@DanJS Yeah, because Coulomb's law is of a similar form. I think the Divergence formulation of gravity and electric force makes a lot more sense since it naturally gives rise to the 1/r^2 term. Just like a balloon being blown up has the same amount of rubber as you blow it up, it just gets stretched out thinner and thinner with increasing radius 1/r^2 since the surface area is r^2.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the electric field in the center is

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2isit a conducting sphere?!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0it is like, there exists a point between earth and moon where the force would be 0

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1yes, just recalling barely, i have no numbers , just concepts

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Empty hit nail on head, the inverse square law and the surface area of a sphere ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0basic E&MM concepts?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so you get the same stuff happening. \(\dfrac{stuff}{r^2}\)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1omg i have to sleep, the gravity at center of earth from earth, just made me think of a hollow conductive sphere with uniform charge density, and the electric field at the center... bye bye

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Jhannybean basic to anyone with only 2 degrees of separation from Walter Lewin!

Empty
 one year ago
Best ResponseYou've already chosen the best response.2https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity#Differential_form Yeah this is what I am specifically referring to, this should remind you of one of Maxwell's equations I think? :P

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so you can think of earth as all mass distributed around circumference then in that state?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@DanJS as long as you are oustide the radius, it can be a point mass or a shell. the maths works out beautifully.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Empty cracking link!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Is what you're talking about have to do with Poisson's equation @Empty ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1It has to do with gravitational fields with results of electrostatics from Maxwell's equation

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, the Poisson equation is saying that gravity is a conservative vector field because it is the divergence of a scalar potential

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1lol, this all started because i said the universe has gravity throughout..

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Haha well in all fairness I think gravity is technically not a force anymore, but it's still effectively a force to play around with :D

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1and if you could supercompute to unreasonable levels, you can calculate classically the gravity at any point from the net effects of every particle in the universe

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2does gravity "travel" at the speed of light ?!

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@DanJS I thought earlier you were arguing that 0 distance doesn't exist due to uncertainty, but if you were to calculate gravity from every particle, I think that the uncertainty in that would end up throwing away all of the information you'd get since planck's constant is on the order of \(10^{34}\), which basically means anyone who's memorized 100 digits of \(\pi\) is wasting time since that's 3 times as much as we could ever measure xD

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1just some thoughts, i never got to take anything beyond relativity. Need to know quantum mechanics and the basic standard model interactions stuff.. maybe someday

Empty
 one year ago
Best ResponseYou've already chosen the best response.2This has really helped me a lot lately in learning QM: https://www.youtube.com/watch?v=lZ3bPUKo5zc&index=1&list=PLUl4u3cNGP619PEhRognw5vryrSEVLPr

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2+1 Allan Adams is cool

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Most of the time when I have tried to learn QM I always sorta feel skeptical of the teacher and so I can't get very far, but I definitely can watch him, accept what he's saying is as best as we as lil human beings can do at this point in time lol.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1As I just started relativity recently, and haven't got into it too much so I'll probably get this wrong, doesn't Einstein's theory of gravity concerns with space being distort, for example earth going around the sun, it's not because of "gravity" but of space being warped? So how does this relate to gravity and speed of light? Unless you mean space itself is being treated as "gravity" and space time travelling at speed of light, haha but I don't really know if any of this is true, as this has to do with general relativity, and that's a bit advanced.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2As I understand it, objects travel through geodesics in space time, which are effectively 'straight lines'. The mass of the sun is bending space around it in such a way that the earth believes it's going in a straight line, it's just that the space is messed up so there's an effective curve to the path. So 'gravity' travelling at the speed of light is more like curvature of spacetime moving, which can't propagate faster than the speed of light, I think this is a pretty famous thought experiment so you can look into it, like "if the sun disappeared would the earth keep going around the sun for a while or would it instantly feel it?" and gravity waves are two things to look into. I am not an expert in any way for the record lol.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2about 8mins and the lights go out
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