4:30 AM
@GaurangTandon This is that question

@Tanuj always up the incline

@AvnishKabaj yes , but why .

Needs to counter mgsin(theta)

@AvnishKabaj okay , could you elaborate ?

@Tanuj don't be confused by the velocity
friction acts independently of velocity

4:40 AM
@GaurangTandon okay , there is static friction only , right ?

just derive the value for friction the way you always do
yes only static

@GaurangTandon no idea

@Tanuj just keep the cylinder at rest on the plane and make its FBD

@GaurangTandon yea , mgsin(theta) downwards , mgcos(theta) balancing Normal reaction , what about the friction ?

@Tanuj there's some acceleration
write the net torque and net force

4:43 AM
umgR=I.alpha

(sorry for the pause)
actually f is not equal to umg
because it is static friction

Why stars are having different colors?
How temperature influence their color?

4:58 AM
Better Google before asking here, or after Googling if you don't get , be specific.
2 days ago, by John Rennie

Is there a possibilities of another big bang?

Thug lyf

5:20 AM
Hello
@Akash.B Maybe I'm not seeing the whole picture because I'm not a regular in this room, but that question seems more like a speculative science question that like something would fit in a room about "problem-solving stragies."

5:54 AM
@JohnRennie Hi

Hi, what's up?

1 hour ago, by Tanuj
@JohnRennie I did get the answer to that question as B , which is correct using this logic
since during the upward journey , the rotation of the cylinder eventually dies out , friction must be acting in the upward direction , to provide an anticlockwise torque (for the clockwise rotation) . Also, when the cylinder moves down the plane , the anticlockwise rotation of cylinder keeps increasing , so the friction would again act in the upward direction , to provide an anticlockwise torque , supporting the rotation this time.

Yes, that seems a good argument
Actually that's a very good argument - better than the one I was about to use :-)

@JohnRennie cool , but now I need to ask you a few things.
@JohnRennie how do I decide the direction of 1) linear acceleration . 2) Angular acceleration

Acceleration is just dv/dt
Look at the direction of the velocity and ask yourself how it's changing
Likewise look at the angular velocity and ask how it's changing

6:01 AM
okay , so it's along the incline and upwards for upward journey and down the incline for downward journey ?

We need to choose axes to determine vector quantites

let x axis be along the incline and y axes to be perpendicular to the incline

Remember that a vector like acceleration can be positive or negative
OK, take x to be positive in the direction up the incline

@JohnRennie yea

Then on the upward journey $v$ is positive and decreasing i.e. becoming more negative.

6:04 AM
@JohnRennie cool , acceleration is down the incline for both upward and downward journeys

And on the downwards journet $v$ is negative and becoming more negative, so it's still decreasing

ok that's sorted. About angular acceleration , how can I decide the directions?

The angular velocity vector points out of the page. Again we need to choose a sign convention.

outwards positive
@JohnRennie sorry , we can take inside as positive

I only suggested inwards as positive because that would make $\omega$ initially positive.

6:09 AM
@JohnRennie cool. So $\omega$ decreases , meaning angular acceleration is negative during upward journey .

Yes

And it still stays outwards and negative for downward journey

This is how I remember the direction of $\omega$

@JohnRennie cool.
@JohnRennie one more question

@Tanuj yes, with the sign convention we have chosen $d\omega/dt < 0$ throughout

6:12 AM
For this , conservation of angular momentum can be used as there is no external torque.

Yes, that's the way I would do it.

Now , initially , $L_i=(MR^2/2+mR^2) \omega$ , m is mass of tortoise and M of disc

Well, that's $I$ you've written not $L$, but yes ...

Now as tortoise starts moving along a chord , it must be nearing the center of disc , so
@JohnRennie oh yea let me correct it
$L_f=(MR^2/2+mr^2)\omega_f$
$\omega_f$ increases here.
now the tortoise would continue moving and would be getting away from the centrer again. So , $\omega$ would once again decrease , as I would increase.
Is my logic correct ?

6:19 AM
@JohnRennie cool

Better with labels ...

yea.

$\omega(t) = L_0/I(t)$
So yu just need to write down $I$ as a function of time.

okay. Cool

But it's obvious that $I$ decreases then increases again, so it's obvious that $\omega$ increases then decreases again. The only question is whether it's linear i.e. B or D

6:24 AM
Yeah.
It can't be linear.
got it . Thanks :)

I haven't worked through the problem, but yes I'd say it's non-linear.

6:36 AM
More? :-)

@JohnRennie sorry , are you busy ?
yea :)

No. I have to work at 07:30 UTC but for now I'm just messing around.

cool.
$$V_{com}=\dfrac{10\cdot 14+ 4\cdot v_2}{14+v_2}$$
how do I figure out $v_2$ ? I can't even use conservation of linear momentum.

Calculate the total momentum in our initial frame ...

0 ?

6:40 AM
Total momentum = m1v1 + m2v2

yea.

So it's ...

hi

Ahh not again

@Akash.B Morning :-)
@Tanuj Focus grasshopper

6:41 AM
@JohnRennie Oh , sorry . yes

Total momentum is ... ?

Yeah I hope it would be good mornig
I have a question

@JohnRennie before impulse imparted its $0$ but after that , it becomes $140+4v_2$

Is there a possibilities of another big bang?

@Tanuj At the instant that the impulse is imparted to m1 the velocity of the 4kg block is zero
@Akash.B be with you in a tick

6:43 AM
okay

@JohnRennie hmm. Just at that instant.

So the total momentum is ... ?

@JohnRennie $140$ only ?

Correct! :-)

cool.

6:44 AM
Now the COM frame is the one in which the total momentum is zero

@JohnRennie why

That's the definition of the centre of mass frame

okay.

In physics, the center-of-momentum frame (also zero-momentum frame or COM frame) of a system is the unique (up to velocity but not origin) inertial frame in which the total momentum of the system vanishes. The center of momentum of a system is not a location (but a collection of relative momenta/velocities). Thus "center of momentum" means "center-of-momentum frame" and is a short form of this phrase. A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the...

oh I got it !

6:46 AM
Is that 30

@Akash.B I think it should be 10

Okay then you may be right
It was a guess

@Tanuj Yes

cool
@JohnRennie you can discuss the big bang now ;)

:-)
@Akash.B we describe the geometry of the universe by a function called the metric, which is a function of time. At time zero this function becomes infinite, and it's that moment that we call the Big Bang.
So the question is whether the metric can become infinite again.

6:50 AM
@JohnRennie I think he is 14.

And the answer is that as far as we know it cannot become infinite again so there cannot be another Big Bang.
@Tanuj I'm giving the simplified version :-)

@JohnRennie haha , way too over my head , still.

@JohnRennie tbh , I'll leave it for some other time.

What is this ?
Is this a novel?
Too long

6:57 AM
@Akash.B the real answer is many times longer :-)

@JohnRennie do you have time for another question

@AvnishKabaj yes, I have about half an hour before I need to work

Just a second
Finally
@JohnRennie I have managed to solve this question using integration and buoyancy
But we we're taught a projected area method for finding the force

OK ... ?

1. Find the centroid of projected shape you want to find the force for calculate the pressure at that point.
2. Multiply it by the area of the projected shape
I wasn't able to apply it here

7:07 AM
I'm not familiar with that method, but I think I can see what it is getting at.

Oh
It's used a lot for finding the force applied at a plane surface

Suppose the total buoyant force is $F$, then this force is the sum of all the forces acting on the cone. Yes?

Rectangle
@JohnRennie yes

And the force acting on the base of the cone is pressure times area i.e. $\rho g H \pi R^2$

@JohnRennie yes

7:10 AM
The total buoyant force is $\rho g V$ where $V$ is the volume of the cone (I can't remember the equation for the volume of a cone)

@JohnRennie $\pi r^2 h$

So if we call the net force on the slanted surface $F'$ we must have $$\rho g V = \rho g H \pi R^2 + F'$$
($F'$ is going to be negative with this sign convention)

Yup

@AvnishKabaj that's the volume of a cylinder isn't it?

One third of it

7:14 AM
That would give: $$\rho g \tfrac{1}{3}\pi R^2 H = \rho g H \pi R^2 + F'$$

Yup

$$F' = -\tfrac{2}{3} \rho g \pi R^2 H$$

@JohnRennie yes

Negative because it's acting downwards ...

I've solved the question using this method
I was actually enquiing about another method

7:19 AM
Ah, OK, when you said you solved it using integration I thought you meant you had integrated the force over the slanted area, which is quite a calculation!
But anyway, we are in effect projecting the force on the slanted surface onto the area of the base.
So we're kind of doing the projection you talked about.

Just a second
This was mainly taught to us by some examples
It's very useful for curved surfaces

@AvnishKabaj Isn't that just treating the cylinder as two half cylinders i.e. you slice it along the middle then equate the pressures on the curved surface and the flat slice?
Which is exactly what we did with the cone

@JohnRennie we did?
I thought we calculated the force at the flat part of the cone(base).
Then used Archimedes principle for calculating the net force on the slant.

@AvnishKabaj yes, you get the pressure on the curved part by calculating the pressure on a flat surface and comparing the two.
Suppose you want the pressure on the upper (curved) surface of the cylinder in your example.

7:34 AM
Ok

You slice the cylinder horizontally to get a half cylinder with a flat lower surface.
Then do the same calculation we just did with the cone

@JohnRennie that makes sense
Thanks!
A lot

@AvnishKabaj the surface through the centroid that you're projecting onto is the same as the surface created by our slice.

@JohnRennie yup

7:52 AM
@JohnRennie
The directions (vector) just don't make sense to me. Does the direction of B not depend on where the point of consideration is in the x-y plane ?

We're told the current is flowing downwards along the $z$ axis.

yea

That creates a circular magnetic field, and the direction of the field lines must be clockwise or anticlockwise.

yea

To get the direction you use ... erm ... can't remember

7:55 AM
its clockwise

Right hand screw rule?
@Tanuj OK, so consider the point (1,0). At this point the field lines point in the negative $y$ direction i.e. $-\hat{j}$. Yes?

@JohnRennie yes

OK, so feed $x=1$ and $y=0$ into the options. Which give you you a field in the $-\hat{j}$ direction?

@JohnRennie I see what you did there ;)

And as a sanity check take (0,1), which should give you a field in the $+\hat{i}$ direction.

8:00 AM
yea , got it.

If there's a sneaky but easy way to answer a question I recommend you take it :-)

hmm why not !

8:37 AM
why the infrared rays are not visible to naked eyes?

@Akash.B
Hello

hi

8:41 AM

@Akash.B we see visible light because the light excites an electron in a molecule called rhodopsin that exists in our eyes.
The excited electron triggers a wave of reactions that end up with a signal being sent down our optic nerve to the brain.

Why it splits into vibgyor when it passes through a prism?

But those rhodopsin molecules are only sensitive to certain photon energies. IR light has too little energy to cause a transition and UV has too much energy to cause a transition. So both of those cannot be seen.
@Akash.B that is a phenomenon called optical dispersion. It happens because the refractive index of the prism is not constant but changes with wavelength.

Look at this picture
I Am having a confusion

8:53 AM
@JohnRennie @Abcd @Tanuj @GaurangTandon Please look at the Fission Problem posted above
(why option 1 is not correct?)

@MadhuchhandaMandal well, we find that by experiment. The capture cross section of a neutron by e.g. U-235 decreases with increasing energy.

Just imagine a plane flying from India to San Francisco If the plane going is in straight path why doesn't it move to the space ?

@MadhuchhandaMandal Are you asking about the fundamental physics behind the experimental observation? If so, that's a complicated question.
@Akash.B the plane isn't going in a straight line
The plane flies at (approximately) constant altitude i.e. constant distance from the centre of the Earth.

@JohnRennie Isn't earth round?

@Akash.B you mean a sphere? Yes, the Earth is roughly spherical.

8:58 AM
Okay I think You didn't understand my question well
I will update that picture
@JohnRennie Okay just leave that question it is aching my head
How did Einstein's equation helped in the progress of atomic bomb?

@Akash.B the plane doesn't fly off into space because it isn't moving in a straight line. It's actually flying slightly downwards so it curves round the Earth instead of moving straight off into space.

@JohnRennie well I have googled it
It is due to earth lining
hmm
Anyone?

9:38 AM
@JohnRennie Yeah I was asking the Physics behind this.. Anyway I will remember the fact

@MadhuchhandaMandal as a general rule it's hard for a fast moving particle to be captured because it has too much energy to form a bound state. When the particle encounters the nucleus it's total energy will be whatever binding energy there is plus the original kinetic energy.
If that original KE is very large there is no chance of forming even a transiently bound state.

@JohnRennie Okay I see.. But I kinda tried to apply Le Chatelier's Principle : As Absorption of Neutron leads to radioactive decay , so the System becomes more stable due to release of Fission energy. So when the Energy of the Neutrons is increased more Absorption will take place (because Absorption leads to decrease of Energy of the system).

9:54 AM
(a) Le Chatelier's principle is just a phenomenological rule that is unreliable and (b) it applies only to the sort statistical systems we get in thermodynamics and not so single nuclei.
Le Chatelier's principle? Really? Do they teach that to students these days?
I though science education had moved on from that sort of thing.

10:07 AM
It is still taught in India
:-/
If you dive deep into how Indian Education System works, you will be astonished more !!!
@JohnRennie

@MadhuchhandaMandal there is a lot wrong with the way that students are treated in India, and I'm sure everyone suffering through JEE preparation would agree. But I have to say that you emerge from the far side of the JEE process with some pretty amazing skills at doing physics problems.

@JohnRennie Thanks a lot !!! :-)ðŸ˜Š

10:36 AM
@JohnRennie I don't get why in elementary reactions can equate pressure to moles.

@Abcd for gases. Specifically for ideal gases.

But why?
How does it work?

Because $P = nRT/V$
For constant T and V we get $P \propto n$

Okay, but then we should use $P=kn$, k being a constant
But we use directly $P=n$

It's obviously not true that $P=n$ but in many cases when we're doing calculations we'll find the constant $k$ can be cancelled on both sides of the equation so we can just omit it.

10:56 AM
@MadhuchhandaMandal i'm almost sure Le Chatelier's principle wasn't given to us to be applied to fission of neutrons, or was it?
i'm sure if we apply the right thing in the wrong place, the answer would obviously be wrong :) @MadhuchhandaMandal

11:22 AM
@GaurangTandon JEE is well known for asking conceptual questions and the closest concept we were taught in the regard to predicting direction of reactions is Le Charlie's . So I thought they are asking for that concept rather than being factual

3 hours later…
1:55 PM
@JohnRennie I'll get back to you on this after Sunday.
@JohnRennie What's the gamma of air?
(assume its a mixture of nitrogen and oxygen only)

@Abcd gamma?
Oh, as in the adiabatic coefficient?

@JohnRennie Gamma= Poisson's ratio
@JohnRennie Yes
31 secs ago, by Abcd
(assume its a mixture of nitrogen and oxygen only)

Well N2 and O2 are both diatomics and pretty close to ideal

So you'd take it to be $7/5$ :/ ?

So they'll have very close to the ideal diatomic specific heats
Basically 1.4 at regular temperatures

1:59 PM
The actual question is:
In adiabatic expansion of air, (assume its mixture of $\ce{N2, O2}$) , the volume increases by $5\%$ , the $\%$ increase in pressure is?
So I was trying to calculate $C_{v_{eq}}$

$1.05^{-1.4}P_0$ presumably

@JohnRennie They haven't mentioned anything in the question.
The gamma of air
or anything
But Yeah, I do know them by heart
But how to find the gamma of mixture

I think they expect you to know that air is close to an ideal diatomic gas ...

Agh!!!
Okay :/

@Abcd but N2 and O2 are both diatomics and have the same molar specific heats

2:01 PM
@JohnRennie so?

So the mixture will have the same molar specific heats as well

@JohnRennie Can we prove it.
$\gamma = \dfrac R{C_{Veq}}+1$

Yes. Ideal gases don't interact so the N2 and O2 absorb heat independently of each other.
So $\Delta Q = C\Delta T$ for both gases where $C$ is the same for both gases.
That means $C$ also has to be the same for any mixture.

Okay, thanks!!

2:54 PM
@JohnRennie Hey , I've got something to ask. (If you're free)

@Tanuj Hi

Hi :)

@Tanuj OK ... ?

@JohnRennie Uhm , what does the question mean by dispersion ? Dispersion as in diverging of light rays by concave lens , or dispersion of light by a prism to generate a spectrum

Dispersion means optical dispersion
So yes it is the dispersion of light by a prism to generate a spectrum

3:01 PM
okay , so , how do I start ? I don't even know what to do .

Suppose you take a parallel plate of glass and shine light through it at an angle
When the light strikes the first air-glass interface there is dispersion and the blue light is refracted more than the red light.
OK so far?

yes , just a small question , for dispersion , what kind of surface works ? And does it also depend on the angle ?

Dispersion means that the refractive index is a function of the wavelength rather than being constant.

okay . Oh I remember now . I have done this for a prism

So Snell's law becomes $$\frac{\sin(i)}{\sin(r)} = n(\lambda)$$

3:06 PM
@JohnRennie all good

So for a given incident angle $i$ the angle of refraction varies with $\lambda$.
As a general rule blue light refracts more than red light (I think it's that way round)

Red refracts the least .

So when we shine our ray of light onto the glass plate the light refracts towards the normal, and the blue light refracts towards the normal more than the red light.

@JohnRennie yeah

But ... when the light reaches the glass-air interface on the other side of the glass plate the light refracts away from the normal, and the blue light refracts away from the normal more than the red light does.

3:09 PM
@JohnRennie so does it become white on the other side ?

Yes, exactly!
A glass plate does not exhibit dispersion
That's why we use a prism, i.e. a shape that doesn't have parallel faces, to produce a rainbow.

@JohnRennie ahh , so you mean to say , when radii of curvatures of the two surfaces are equal , there'd be no dispersion ?

Bingo! :-)

@JohnRennie John "Legend" Rennie , idk how to thank you !

Well, there remains some very slight dispersion, but it is minimised when the faces are parallel.

3:13 PM
I really want to meet you in person ! Definitely noting that down in my ''things to do'' list.

If you make it to England there are more exciting things to do than meet me :-)

@JohnRennie I'll come for sure if I can stay at your place. :p

2 hours later…
5:03 PM
@AvnishKabaj that projected area method
you werent taught it's proof?
Here you are

@JohnRennie Quick question.

@Abcd Yes?

@JohnRennie So basically a rod recoils with each end having a different velocity. How do I find the angular velocity of the rod?
One end has velocity $e_1\sqrt{ 2gh}$
Other $e_2 \sqrt {2gh}$

5:19 PM
@Abcd switch to the centre of mass frame

Then?

In the COM frame the rod is stationary and rotating

Yes.

So calculate the velocities of the ends in the COM frame and you should find they are equal and opposite. Just use $v=r\omega$ to calculate $\omega$.

@JohnRennie We dont know COMs velocity
3 mins ago, by Abcd
One end has velocity $e_1\sqrt{ 2gh}$
3 mins ago, by Abcd
Other $e_2 \sqrt {2gh}$

5:22 PM
$e_1$ and $e_2$ are normal unit vectors?

No, just dimensionless constants.
@JohnRennie Author find the relative velocity of the two lends and divides it by $l$
Doesn't make sense to me.

I'd have to think about it, and right now I'm too tired to do physics. Tomorrow morning.

Okay.

@JohnRennie Chemistry maybe ?

Nope, Tomorrow.

5:27 PM
Okay.

6:23 PM
@AvnishKabaj here
The images are in reverse order