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06:40
Deleted
07:36
@JohnRennie good morning
@Zerix morning :-)
@JohnRennie can u explain why normal force is zero in vertical circle at the top
I think we should be more precise.
For example suppose you are considering a stone being whirled around on a string. Then the tension in the string is zero at the top of the circle.
Ah. The normal force at the top is the centripetal force minus the gravitational force, so it is $mv^2/r - mg$.
Whether this is zero or not depends on the velocity
Correct
07:42
@JohnRennie so how do we decide the condition for velocity to complete the circle
Is that a stone on a string? Or an object moving on a circular track? Or something else?
Object moving on circular track
It's just direction of normal I have shown
Though it's direction is changing continuously
OK. For the object to move in a circle the centripetal acceleration must be $v^2/r$. That means an inwards force of $mv^2/r$ must be applied to the object. Yes?
Yes
There are two things exerting a force on the object:
1. gravity
2. the track
When you add the two forces together the resultant must have a radial component of $mv^2/r$. OK so far?
07:46
Ok
In general the forces act in different directions because gravity always acts downwards while the normal force exerted by the track always acts towards the centre.
Yes
But there are two angles for which the two forces line up. At the bottom gravity $mg$ acts downwards while the normal force $N$ acts upwards. Since they have to add up to $mv^2/r$ we get $N - mg = v^2/r$
So $N = mv^2/r + mg$
Okay
At the top the two forces act in the same direction because both act straight down. Then we get $N + mg = mv^2/r$, or $N = mv^2/r - mg$
07:51
So mg can sustain the centripetal avveleration in second case while it cannot in first
The condition for the car to stay on the track is $N \ge 0$ because a negative normal force means the car loses contact with track i.e. the wheels come off the track.
Sorry, I accidentally clicked send :|
At the bottom $N$ is always greater than zero because it's the sum of two positive quantities. But at the top $N$ can be greater or less than zero depending on $v$
The condition that $N=0$ gives us the minimum possible value of $v$ for which we can have circular motion.
@JohnRennie why can't normal be zero. In between these two cases
3 mins ago, by John Rennie
The condition for the car to stay on the track is $N \ge 0$ because a negative normal force means the car loses contact with track i.e. the wheels come off the track.
07:57
@JohnRennie Sorry. I was reffering to the idea of normal being zero between any point between top and bottom
If $N=0$ at any point lower than the top then $N$ will go below zero when the car moves any farther upwards, and that means the car will come off the track.
The only place where $N$ can be zero is right at the top.
If you write down the equation for $N$ that will be obvious ...
@JohnRennie yes I am doing that only. Hope I make sense out of the equation
$$ N = mv^2/r + mg\cos\theta $$
where $\theta$ is the angle measured from the bottom i.e. $\theta = 0$ at the bottom and $\theta = \pi$ at the top.
Just draw a graph of $N$ and it will be obvious ...
A small ball of mass $m$ lies on the superior pole of a frictionless semi-sphere of mass $M$ which lies on a horizontal frictionless surface. The ball is smoothly released from its unstable equilibrium position and starts moving on the surface of the semi-sphere (with no initial velocity). Find the angle $\theta$ for which the ball detaches from the semi-sphere. Solve the problem in the particular case $m=M$.
@Don'tbeax-tripledot is the answer in expression
Or numerical
08:09
I get a numerical answer, but they express it in terms of the radius and solve in the particular case $r=1$ which is weird
This question seems particularly difficult to me
If the sphere was not moving then I knew the anwer but it's question of com
I doubt it's a question of COM this time
I can present my approach which led to an equation in $\theta$ which I could only solve using Wolfram alpha which gave $\approx 61^\circ$
@Don'tbeax-tripledot it's a question similiar to the vertical circle
The author's solution depends on the radius though... For $r=1$ they get $42,9^\circ$
So should I show my method and trouble-shoot together?
Yes. If the sphere was not moving then also the result depended on radius so. It makes sense it does
@Don'tbeax-tripledot I gtg in few minutes. Kinda busy. Sorry
You can discuss with JR
@JohnRennie after analysing the equation it makes sense for normal to be zero at the top
It can't be zero for first 0 to 90 from bottom
In 90 to 180,the normal is decreasing with minimum at the top so it's the only safe spot to assume
Finally understood this. Thanks for your help
08:19
@Zerix :-)
I tried using energy conservation: The initial energy is $mgR$ and the final energy is $mgR\cos\theta+\dfrac{1}{2}mu^2+\dfrac{1}{2}mv_a^2$, where $u$ is the velocity of the semi-sphere at the final moment, and $v_a$ the absolute velocity of the ball at the end, which I think is given by $v_a^2=gR\cos\theta+u^2-2u\sqrt{gR\cos\theta}\cos\theta$.
I am pretty sure this approach is correct so far, but here's the part I am unsure of: let $dx$ be a small change in the position of the semi-sphere and $d\varphi$ be a small change in the center angle. I've written $dx=Rd\varphi\cos\varphi$ and $du/dt=N\sin\varphi/m=g\cos\varphi\sin\varphi$, then I multiplied by $dx$ to get:$$du\cdot\dfrac{dx}{dt}=udu=gR\cos^2\varphi\sin\varphi d\varphi$$ which after integration gives $u^2/2=gR(1-\cos^3\theta)/3$
@Don'tbeax-tripledot Is that the answer?
08:36
@Dante Which that?
That is my approach which is seemingly incorrect
right above my message
ok
conservation of momentum is better ig
I'll try it after I finish my lunch. Do you know the right answer ?
31 mins ago, by Don't be a x-triple dot
The author's solution depends on the radius though... For $r=1$ they get $42,9^\circ$
i see
08:59
I am getting an equation of the form $$\frac 1 3=\frac 3 2 \cos\theta-\frac 2 3\cos^3\theta -\sqrt{\frac 2 3 \cos^3\theta(1-\cos^3\theta)}\longrightarrow \theta\approx 69^\circ$$
I get $Cos^{3}\theta+3Cos\theta-2$=0
53.5 degrees :/
58.4 you mean :P
@JohnRennie Any idea?
@Don'tbeax-tripledot not really
I used cubic equation solver
I used my CASIO and Wolfram Alpha :P
Got cosx= 0.5960716379833214
09:06
Ah whoops
I thought you also had the leading coefficient $3$
Anyway, the correct answer seems to depend on the radius so we're both wrong :)
yeah..
The author has $\cos^3\theta-3\cos\theta+2=0$ for $r\to\infty$
i see.
I give up, gotta prepare for JEE also :p
 
2 hours later…
10:52
@JohnRennie are you here?
@Nobodyrecognizeable morning :-)
@JohnRennie can you help me with c?
Are you uploading the question?
@JohnRennie i was wanting to know how to write a matrix in its form in c .
Oh, you mean the programming language C?
10:55
@JohnRennie yep. Sorry had to mention.
So you have a 2d array x[m][n] and you want to write it out as a matrix?
@JohnRennie poor me. I hope you thought it a question.
@JohnRennie specifically we are using two for loops and two dimensional array as well but the matrix does not comes out as the form of matrix so I was asking you.
for (i = 0; i < m; i++)
{
  for (j = 0; j < n; j++)
    printf("%8f", x[i][j]);
  printf("\n");
}
Something like that?
@JohnRennie yes. But the output didn't came as a matrix
You'll have to post your code and the output it produced for me to see what's going on.
11:00
Ok @JohnRennie itll take some time. Once i write the code . Ill post it and ping you.
@JohnRennie the user used spaces and enters to have the format.
@JohnRennie can't we use readymade programme of two matrix addition so that the user gives 18 elements of the matrix and the sum comes out as a matrix by itself?
@Nobodyrecognizeable C doesn't have any built in way of adding two arrays together.
@JohnRennie i see so the space and enters are only options.
I don't understand what you are asking. Are you asking about how the user enters the numbers that are going into the matrix?
@JohnRennie when the user puts the two matrix then while putting eventually the whole numbers of the two cross two Matrixes hi give spaces and enters to put the number so when the output comes out it's in the form of a matrix but can we build a program so that you just put the numbers one by one giving the enters and the final result means the matrix addition comes out as a form of a matrix as a form of a matrix which i mean to be 2×2 matrix.
Like suppose the output should be 4 2
6 7
Without putting A and B matrix via spaces and enters. Can you understand my problem now?
The code you posted allows the user to enter the numbers any way they want: 1 at a time, 2 at a time or all 4 numbers on a single line.
11:14
@JohnRennie that doesn't have to do with the output regarding its form ?
The code always prints the output as a matrix. The input can be entered in any format.
@JohnRennie ah ok . Sorry for bothering. Thanks for the help professor. Btw these things always happen when you don't have a pc.
@JohnRennie have a nice day professor, goodbye.
 
1 hour later…
12:19
4 hours ago, by Don't be a x-triple dot
A small ball of mass $m$ lies on the superior pole of a frictionless semi-sphere of mass $M$ which lies on a horizontal frictionless surface. The ball is smoothly released from its unstable equilibrium position and starts moving on the surface of the semi-sphere (with no initial velocity). Find the angle $\theta$ for which the ball detaches from the semi-sphere. Solve the problem in the particular case $m=M$.
Regarding this problem, I've solved it using momentum and energy conservation.
I don't need assistance anymore
A
A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligible push is givej to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to r/2
Any1 any ideas?
12:40
@Pranav Answer?
Root 14rg/3
I mean i know how to get e answer via energy consv. But i cant underatand how come both the angular vel. And the linear vel. of the carpet increases
Since it is pure rolling both have to increase
And I'm getting answer twice of what you've mentioned
 
4 hours later…
17:18
@Dante Hey there!
@Tanuj Hello
@Dante Could you possibly explain me this titration in chemistry
@Tanuj I suck at chemistry
@Dante Ah , alright
Sorry
17:20
I don't really know if I could get any help here regarding chemistry
@Dante Ah , alright
You can ask someone in chemistry group
For basic stuff it's best to watch youtube videos about it.
@Dante Alright , thanks mate
17:52
@JohnRennie Can you help please?
And also this one
Please explain what the question means in the second one. I want to solve it by myself but I can't inderstand the question
@tatan the first question is just silly. In principle a guinea pig could survive contact with a 220kV conductor if it wasn't earthed, or was earthed through a very, very high resistance. In practice you'd end up with an oily smear where the guinea pig used to be.
In the second question you just need to figure out how many protons there are in the ring then multiply this number by 24GeV to get the total energy of the protons.
@JohnRennie Is this the same this as in van de graaff generator?
You can work out how many protons there are in the ring from the current of 30A. That means 30 coulombs of charge have to pass a point in the ring every second, and a proton has a charge of $1.602 \times 10^{-19}$ coulombs.
@tatan it depends how you interpret the question.
Is option 2 reasonable?
Option 2 is certainly possible. Whether it's reasonable or not is debatable.
18:06
So 950m circumference is useless data?
In the second question
The 950m circumference is a key piece of data. Assuming the protons move at the speed of light (which they almost do) then you can work out how long it takes all the protons to pass a point in the ring i.e. 950m divided by $c$.
If the number of protons is $n$ then the charge passing some point in the time $t$ you calculated above is $ne$ and the current is therefore $ne/t$. And we're told this is equal to 30A.
yes
@JohnRennie Hi ! :)
@Tanuj hi
@JohnRennie I'm actually thinking of learning web development and Machine learning this year
18:11
@Tanuj Both highly marketable skills! :-)
@JohnRennie Well , you deserve a lot of credit for where I am today !
3
I constantly pinged you for all my doubts
and you never left me hanging
Thanks a lot :)
You're welcome :-) I have a lot of free time these days (being partly retired) so I'm happy to help.
I'll be eating your brains out for web development and machine learning possibly
lets see
@Tanuj Hi! Are you a +2 student right now?
@tatan No , B.tech first year . What kinda plus 2 student aims of learning machine learning xD
18:17
@Tanuj true. Where are you studying? May i know please?
He mentioned "where I am today ". It's obvious he's not +2 ;p
;-D
@tatan Computer Science and Engineering at NIT Uttarakhand
Oh nice ;-)
@Tanuj actually I know nothing about machine learning, and my web development skills are all badly out of date. Web development moves so fast. It feels like every time I look round there's a new Javascript framework.
18:19
Yeah , well , looking back at my preparation days , I think its safe to say I deserved not less and not more
There are some machine learning guys who hang out in the h Bar.
@JohnRennie Whatever , you're my self proclaimed mentor now !
Thanks :-)
@Tanuj Whatever. If you enjoy it then the whole business is worth it. ;-)
I have to go. See you all tomorrow.
18:21
Yup , well atleast I know what I have to do now . Last year , I didn't even know if engineering was really what I wanted to do.
goodbye sir ;-) Thanks for you help!@JohnRennie
Have a good one @JohnRennie
@Tanuj Are you enjoying it now?
@Tanuj Curious why you wanted to know what titration is?
And since you're NITian, shouldn't you be knowing it already?
well , I haven't really started it yet , but the prospects look good
@Dante I've just entered the second semester , and I have this Chemistry titration to tackle
18:24
Oh, I see, which stream have you chosen?
Another bit of a personal question. Are you interested into research or looking forward to a good placement after 4 years?@Tanuj
I can basically do every bit of it , I just wanted to know how Diphenylamine combines with K2Cr2O7
@Dante CSE
@tatan I don't really know atm
@Tanuj Oh, so even CSE peps have to study Chem?
@Tanuj oh i see
@Dante stupid isn't it , in the first year of the 4 year course , every student from every stream has to study every stream
18:27
Dang, I still don't understand why do I have to study chem to get into NIT even though I'm sure I'm not taking any chem strem :(
India really need educational reforms
@Tanuj Could you name the subjects?
Which one has to study in first year
Are you supposed to study metallurgy,ceramic tech and shit also?
@Dante Why do you think mettulurgy and material sc is shit?
Just because after 4 years there will be no 20 lpa placements?
One of the reason, yes. Main reason is it's irrelevant to someone interested in CSE.
I called it shit because it's chem and I hate it ;p
Sorry if it's your fav subject
nothing personal
actually i used to think the same
but someone made me realise that its not the right way to think.
then what changed your mind?
@tatan Oh I see.
@tatan That's effect of my surrounding ig.
Actually i used to think that its physics who's the king and chem is useless
18:35
All my teachers say NIT/IIT is worth it only if you get CSE/EE
@Blue hi. Are you usually free at this time
@tatan Then?
Depends on how you want to approach your life. if salary's everything then its kinda true@Dante
@Dante you should add bits too in that list
We have studied electromagnetism in +2 right?
18:37
@tatan Yeah
Have you ever wondered what gives rise to magnetism at the atomic level. i mean at the fundamental level?@Dante
Anonymous
@Zerix Hi. I'm doing my homework now and am a bit busy. If you want to say something, please go ahead! I usually get around to replying within a day.
@tatan Maybe some stuff related to quantum physcis.
@Blue ohkay. Actually I just wanted to know when you are free to ask my semiconductor doubts. No doubts for today though.
Chemistry actually was my favourite subject actually , specially organic. But , no scope is why CSE.
18:41
@tatan I don't think it's necessary to know what happens at atomic level for everyone. Take example of a person who invented a useful machine based on electromagnetism principle, was it necessary for him to know what happens at atomic level? no ;p Everyone has their own interest and field of study.
I used to think the same... now then some knowledgeable person made me realise that its the concept of "spin". This is where physics bows to so called chemistry. The concept of spin angular momentum etc. is borrowed from so called convetional "chemistry". That's when he made me realise that after a certain level theres no physcis and chemstry distinction. its actually two slightly differenct approaches that merges after some time. so we should not draw a line ;-)
@Tanuj Exactly, I'm actually interested in engineering physics. Gotta choose CSE for same reason.
@tatan Oh, that's a nice one!
@Dante I mean that really was a turning point
Maybe that's why JEE makes us study all the three subjects PCM
i never looked down upon chemistry after that
18:43
To get basic idea of all maths and sciences
Gotta go guys , I'mma watch this football match and go to sleep
So long
@Tanuj Cya!
@Dante Yep. an enginner ofcourse does not need to get into the pure physics so much... Another interesting thing i came across in this respect.
Yeah
@tatan Goodnight! gtg
An engineer does not need to know coloumbs law at the very detailed scale . But the parallel plate capacitor formula \frac{A\epsilon}{d} is of much more use to him... Simplying that formula was a very important reason why the constant in coloumbs law is expressed in $\frac{1}{4\pi\epsilon}$ form
@Dante Bye ;-D
if only our country provided opportunity in all fields. You want to study engineering physics but theres no surety of what will happen after 4 years ;-(

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