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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- ganeshie8

|dw:1443166269719:dw|

- inkyvoyd

EWWWW I might have this problem on my test!

- Jhannybean

|dw:1443166523518:dw|\[\sum F_y = ma_{cp} = F_{N} + F_{g} = N + mg=\frac{mv^2}{r}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Jhannybean

Well that's technically a little wrong because the vector for the normal force should be longer than the force of gravity, but still....

- inkyvoyd

I think it's more wrong because your bottom circle is elliptical and thus does not obey rules of circular motion ;)

- ParthKohli

haha, vertical circle
looks like you've fallen in the trap that physics is :)

- ParthKohli

Try 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

in the course of a few years parth has learned like 4 times as much physics as me.

- Jhannybean

\[N = \frac{mv^2}{r} -mg\]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

why do we need to know the mass :-)

- Jhannybean

I wasnt there yet haha

- ParthKohli

Exactly, 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

anyway the answer is \(h_{min} = 5R/2\)

- ganeshie8

that 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

yeah, Galileo and Neil Armstrong!!

- ganeshie8

is that tesla in ur dp

- IrishBoy123

yes
if anyone loads up an Edison avatar, there's gonna be blood!!

- ParthKohli

no this is a dog

- IrishBoy123

lol!

- ganeshie8

so a naive q, do roller coasters really work like this?
without any chains hooked to the tracks?

- Jhannybean

\[g=\frac{v^2}{r} \implies v^2 = gr\] 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

Just from the back of my mind, i think Walter Lewin did this in one of the 8.01 lectures on MIT OCW

- DanJS

been a couple years since watching those

- ganeshie8

they had removed all his lectures from MIT after some controversy

- ParthKohli

in ideal conditions, they can work like that

- DanJS

:-O, well good thing i dl them all, hehe

- ParthKohli

yeah, 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

hell yes, his lectures and feynman spiked my interest

- ganeshie8

watching top 5 rolller coaster accidents
https://www.youtube.com/watch?v=mM0HE7DKTvg

- DanJS

i am pretty sure the acceleration at the top of the circle is zero for one instant to 'just' make it over... maybe not

- ParthKohli

https://www.youtube.com/watch?v=HN8nv4tVFuA

- DanJS

maybe a way to do it with just energy conservations

- IrishBoy123

|dw:1443168272771:dw|

- ParthKohli

yes, 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

in practise i think roller coasters go faster and you are pushed into your seat (upwards) by your inertia

- Jhannybean

\[U_i + K_i = U_f + K_f \\K_i = 0 \]\[mgh = \frac{1}{2}mgr +2mgr\]\[h=\frac{r}{2}+r\]\[h=\frac{3}{2}r\]

- ParthKohli

of course they go faster because here we're calculating the minimum velocity

- Jhannybean

I just figured out why \(U_f = mg(2r)\) lol... -_-

- ParthKohli

\(r/2 + 2r = 5r/2\)

- IrishBoy123

|dw:1443168775694:dw|

- Jhannybean

Yah I cant either. That's not eve important. Pfft.

- Jhannybean

Oh I just realized something

- DanJS

ah gravity doesn't matter

- DanJS

isn't that convenient

- Jhannybean

We had \(v^2=gr\) when we simplified our forces in the y-direction. We could have multiplied both sides by \(\dfrac{1}{2}m\) to associate the forms of kinetic and potential energies!! Whoops.

- IrishBoy123

@DanJS so if g = 0 ??!!

- ParthKohli

ganeshie try solving this question about vertical circles... it'd really test your knowledge.

- DanJS

The 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

2.5R min height to start

- Jhannybean

\[h=\frac{5}{2}r \qquad r=\frac{D}{2} = \frac{100}{2} = 50~m\]\[h_{min} =\frac{5}{2}(50) = 125~m \]

- Jhannybean

Yeah 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

Yes PK, post...
today it is going to be a gravity/uniform circular motion day for me!

- Jhannybean

Maybe you can also post the breaking-of-the-plate momentum question, another classical physics question involving vectors and momentum. I always had trouble with those.

- ParthKohli

can I have that extension, ganeshie? you know, the drawing one.

- ParthKohli

yeah I have a huge collection of questions so no worries

- ganeshie8

send me meeting id in teamviewer
it would be simpler if i install it

- Jhannybean

Back to eating my kichadi.

- ParthKohli

whatever... I found an image that does suit my purpose.

##### 1 Attachment

- DanJS

https://www.youtube.com/channel/UCliSRiiRVQuDfgxI_QN_Fmw/playlists
someone uploaded em

- Jhannybean

I 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

in 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

treasure trove
https://www.youtube.com/watch?v=7E34VOj8Elc&index=182&list=PLERGeJGfknBR3pXCPlV3bgb_qHCSNOdBf

- ganeshie8

that is a very useful collection, i hope they don't delete his lectures from youtube too
thanks @DanJS

- Jhannybean

@DanJS Im saving this.

- ParthKohli

Walter Lewin himself has uploaded it on his own channel BTW
https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw

- DanJS

download those playlists to disk

- ganeshie8

|dw:1443169569982:dw|

- Jhannybean

I have a friend attending MIT that's met him.

- ParthKohli

this thread is really slow for me so can we move to another one?

- ganeshie8

okiee

- Empty

http://datagenetics.com/blog/march42014/index.html

- Jhannybean

oh, never knew it was called a trifecta.

- DanJS

@irishboy123
Where is gravity ever zero in this universe .. infinite range

- Empty

Well gravity is practically zero at the center of the earth or in free fall :D

- DanJS

hmmm, free fall means you do not sense the gravity, it is still there and it matches your acceleration of falling

- ganeshie8

\[F \propto \dfrac{mM}{r^2} \]
so the Force is not defined at \(r=0\) right

- DanJS

man i discussed this awhile back, about how you can not have zero distance.. been too long

- DanJS

strong force repulsion or something

- Jhannybean

Ah, yes. that's true, @ganeshie8

- DanJS

EM repulsion

- DanJS

you can't be sure r=0 by uncertainties in position

- IrishBoy123

nah, inside the earth gravity is pulling you outwards. zero at centre - theoretically

- Empty

Yeah, I mean an astronaut living in the space station is for all intents and purposes living with no gravity.

- IrishBoy123

the r assumes a point mass and works outside the sphere that is the earth or any body

- Jhannybean

And that's why their bone structure slowly decomposes over time - he gravity that acts on them!

- DanJS

microgravity, they are moving very fast and falling around the earth

- Empty

Yep

- DanJS

like a super cannon that fires so fast the ball falls slower than the horizon curves away

- DanJS

or the same rate

- DanJS

@IrishBoy123 Is the gravity at the center of a uniform Ball zero?

- IrishBoy123

@Jhannybean that's 2 degrees of separation, i think it's called ie w/ Lewin

- IrishBoy123

yes, if you look at it classically.

- DanJS

In the real universe there is gravity everywhere, i think, i can idealize a empty space void, no gravity

- Empty

@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

OK
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

- Jhannybean

I don't like playing with words one bit.

- IrishBoy123

but i agree totally that a force that has unlimited reach much reach everywhere :-)

- Empty

Can we measure the gravitational pull of planets outside of our solar system?

- DanJS

Does that have something to do with a metal sphere and charge concentrated on the surface...

- IrishBoy123

@DanJS but i only know classical physics so if it reaches back on itself .... :p i dunno

- Empty

@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

the electric field in the center is

- IrishBoy123

isit a conducting sphere?!

- ganeshie8

it is like, there exists a point between earth and moon where the force would be 0

- DanJS

yes, just recalling barely, i have no numbers , just concepts

- IrishBoy123

@Empty hit nail on head, the inverse square law and the surface area of a sphere ...

- Jhannybean

basic E&MM concepts?

- IrishBoy123

so you get the same stuff happening. \(\dfrac{stuff}{r^2}\)

- DanJS

omg 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

@Jhannybean basic to anyone with only 2 degrees of separation from Walter Lewin!

- Empty

https://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

so you can think of earth as all mass distributed around circumference then in that state?

- IrishBoy123

@DanJS as long as you are oustide the radius, it can be a point mass or a shell. the maths works out beautifully.

- IrishBoy123

@Empty cracking link!

- Astrophysics

Is what you're talking about have to do with Poisson's equation @Empty ?

- DanJS

right,

- DanJS

goodnight

- Astrophysics

It has to do with gravitational fields with results of electrostatics from Maxwell's equation

- Empty

Yeah, the Poisson equation is saying that gravity is a conservative vector field because it is the divergence of a scalar potential

- DanJS

lol, this all started because i said the universe has gravity throughout..

- Empty

Haha 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

and 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

does gravity "travel" at the speed of light ?!

- Empty

Yeah

- Empty

@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

newtonian gravity

- DanJS

just 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

This has really helped me a lot lately in learning QM: https://www.youtube.com/watch?v=lZ3bPUKo5zc&index=1&list=PLUl4u3cNGP61-9PEhRognw5vryrSEVLPr

- IrishBoy123

+1
Allan Adams is cool

- Empty

Most 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

As 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

As 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.

- Astrophysics

Ha, right!

- IrishBoy123

about 8mins and the lights go out

Looking for something else?

Not the answer you are looking for? Search for more explanations.