Problem Solving Strategies

@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."

Tanuj

5:54 AM

@JohnRennie Hi

John Rennie

Hi, what's up?

Tanuj

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.

John Rennie

Yes, that seems a good argument

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

Tanuj

@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

John Rennie

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

Tanuj

6:01 AM

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

John Rennie

We need to choose axes to determine vector quantites

Tanuj

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

John Rennie

Remember that a vector like acceleration can be positive or negative

OK, take x to be positive in the direction up the incline

Tanuj

@JohnRennie yea

John Rennie

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

Tanuj

6:04 AM

@JohnRennie cool , acceleration is down the incline for both upward and downward journeys

John Rennie

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

Tanuj

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

John Rennie

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

Tanuj

outwards positive

@JohnRennie sorry , we can take inside as positive

John Rennie

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

Tanuj

6:09 AM

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

John Rennie

Yes

Tanuj

And it still stays outwards and negative for downward journey

John Rennie

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

Tanuj

@JohnRennie cool.

@JohnRennie one more question

John Rennie

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

Tanuj

6:12 AM

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

John Rennie

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

Tanuj

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

John Rennie

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

Tanuj

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 ?

John Rennie

Tanuj

6:19 AM

@JohnRennie cool

John Rennie

Better with labels ...

Tanuj

yea.

John Rennie

$\omega(t) = L_0/I(t)$

So yu just need to write down $I$ as a function of time.

Tanuj

okay. Cool

John Rennie

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

Tanuj

6:24 AM

Yeah.

It can't be linear.

got it . Thanks :)

John Rennie

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

John Rennie

6:36 AM

More? :-)

Tanuj

@JohnRennie sorry , are you busy ?

yea :)

John Rennie

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

Tanuj

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.

John Rennie

Calculate the total momentum in our initial frame ...

Tanuj

0 ?

John Rennie

6:40 AM

Total momentum = m1v1 + m2v2

yea.

So it's ...

hi

Ahh not again

@Akash.B Morning :-)

@Tanuj Focus grasshopper

Tanuj

6:41 AM

@JohnRennie Oh , sorry . yes

John Rennie

Total momentum is ... ?

Akash. B

Yeah I hope it would be good mornig

I have a question

Tanuj

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

Akash. B

Is there a possibilities of another big bang?

John Rennie

@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

Akash. B

6:43 AM

okay

Tanuj

@JohnRennie hmm. Just at that instant.

John Rennie

So the total momentum is ... ?

Tanuj

@JohnRennie $140$ only ?

John Rennie

Correct! :-)

Tanuj

cool.

John Rennie

6:44 AM

Now the COM frame is the one in which the total momentum is zero

Tanuj

@JohnRennie why

John Rennie

That's the definition of the centre of mass frame

Tanuj

okay.

John Rennie

Center-of-momentum frame

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...

Tanuj

oh I got it !

Akash. B

6:46 AM

Is that 30

Tanuj

@Akash.B I think it should be 10

Akash. B

Okay then you may be right

It was a guess

John Rennie

@Tanuj Yes

Tanuj

cool

@JohnRennie you can discuss the big bang now ;)

John Rennie

:-)

@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.

Tanuj

6:50 AM

@JohnRennie I think he is 14.

John Rennie

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 :-)

Tanuj

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

John Rennie

@Tanuj it's simpler than you might think

Tanuj

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

Akash. B

What is this ?

Is this a novel?

Too long

Answers

John Rennie

6:57 AM

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

Avnish Kabaj

@JohnRennie do you have time for another question

John Rennie

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

Avnish Kabaj

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

John Rennie

OK ... ?

Avnish Kabaj

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

John Rennie

7:07 AM

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

Avnish Kabaj

Oh

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

John Rennie

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

Avnish Kabaj

Rectangle

@JohnRennie yes

John Rennie

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

Avnish Kabaj

@JohnRennie yes

John Rennie

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)

Avnish Kabaj

@JohnRennie $\pi r^2 h$

John Rennie

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)

Avnish Kabaj

Yup

John Rennie

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

Avnish Kabaj

@JohnRennie my bad

One third of it

John Rennie

7:14 AM

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

Avnish Kabaj

Yup

John Rennie

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

Avnish Kabaj

@JohnRennie yes

John Rennie

Negative because it's acting downwards ...

Avnish Kabaj

I've solved the question using this method

I was actually enquiing about another method

John Rennie

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.

Avnish Kabaj

Just a second

This was mainly taught to us by some examples

I'll upload them

It's very useful for curved surfaces

John Rennie

@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

Avnish Kabaj

@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.

John Rennie

@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.

Avnish Kabaj

7:34 AM

Ok

John Rennie

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

Avnish Kabaj

@JohnRennie that makes sense

Thanks!

A lot

John Rennie

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

Avnish Kabaj

@JohnRennie yup

Tanuj

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 ?

John Rennie

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

Tanuj

yea

John Rennie

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

Tanuj

yea

John Rennie

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

Tanuj

7:55 AM

its clockwise

John Rennie

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?

Tanuj

@JohnRennie yes

John Rennie

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

Tanuj

@JohnRennie I see what you did there ;)

John Rennie

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

Tanuj

8:00 AM

yea , got it.

John Rennie

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

Tanuj

hmm why not !

Akash. B

8:37 AM

why the infrared rays are not visible to naked eyes?

Avnish Kabaj

@Akash.B

Hello

Akash. B

hi

Avnish Kabaj

Did you Google that question?

Before asking

Akash. B

Do you have an answer?

Avnish Kabaj

Please google once

Akash. B

8:41 AM

Yeah I got It google.co.in/…

John Rennie

@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.

Akash. B

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

John Rennie

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.

Madhuchhanda Mandal

Akash. B

Look at this picture

I Am having a confusion

Madhuchhanda Mandal

8:53 AM

@JohnRennie @Abcd @Tanuj @GaurangTandon Please look at the Fission Problem posted above

(why option 1 is not correct?)

John Rennie

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

Akash. B

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 ?

John Rennie

@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.

Akash. B

@JohnRennie Isn't earth round?

John Rennie

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

Akash. B

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?

John Rennie

@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.

Akash. B

@JohnRennie well I have googled it

It is due to earth lining

hmm

Anyone?

Madhuchhanda Mandal

9:38 AM

@JohnRennie Yeah I was asking the Physics behind this.. Anyway I will remember the fact

John Rennie

@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.

Madhuchhanda Mandal

@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).

John Rennie

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.

Madhuchhanda Mandal

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

John Rennie

@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.

Madhuchhanda Mandal

@JohnRennie Thanks a lot !!! :-)😊

Abcd

10:36 AM

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

John Rennie

@Abcd for gases. Specifically for ideal gases.

Abcd

But why?

How does it work?

John Rennie

Because $P = nRT/V$

For constant T and V we get $P \propto n$

Abcd

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

But we use directly $P=n$

John Rennie

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.

Gaurang Tandon

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

Madhuchhanda Mandal

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

Abcd

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)

John Rennie

@Abcd gamma?

Oh, as in the adiabatic coefficient?

Abcd

@JohnRennie Gamma= Poisson's ratio

@JohnRennie Yes

31 secs ago, by Abcd

(assume its a mixture of nitrogen and oxygen only)

John Rennie

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

Abcd

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

John Rennie

So they'll have very close to the ideal diatomic specific heats

Figures here

Basically 1.4 at regular temperatures

Abcd

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 $C_{v_{eq}}$

John Rennie

$1.05^{-1.4}P_0$ presumably

Abcd

@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

John Rennie

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

Abcd

Agh!!!

Okay :/

John Rennie

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

Abcd

2:01 PM

@JohnRennie so?

John Rennie

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

Abcd

@JohnRennie Can we prove it.

$\gamma = \dfrac R{C_{Veq}}+1$

John Rennie

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.

Abcd

Okay, thanks!!

Tanuj

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

John Rennie

Dispersion means optical dispersion

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

Tanuj

3:01 PM

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

John Rennie

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?

Tanuj

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

John Rennie

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

Tanuj

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

John Rennie

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

Tanuj

3:06 PM

@JohnRennie all good

John Rennie

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)

Tanuj

Red refracts the least .

John Rennie

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.

Tanuj

@JohnRennie yeah

John Rennie

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.

Tanuj

3:09 PM

@JohnRennie so does it become white on the other side ?

John Rennie

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.

Tanuj

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

John Rennie

Bingo! :-)

Tanuj