Problem Solving Strategies

@psa hi :-)

hey

Yes, what's the problem?

let me upload the question and my work, one sec

I know that I'm really close because I know what the answer should be

Sorry, it's a bit long, but hopefully you can just scan through it.

I'm pretty sure what I should be getting in the end is $\ddot\alpha = -\frac{m+M}{M}\frac{g}{\ell}\alpha$ (this is what I got when I used Lagrangian mechanics)

which is very close to that final answer, but obviously it's not quite that

The x position of the centre of mass cannot change because none of the forces acting have a non-zero x component. Yes?

8:32 AM

yeah

So I would put the COM at x = 0. Then at any time the position of the top end will be $x(t)$ and the bottom end $-x(t)$.

The angle $\alpha$ is then given by $\sin\alpha = \frac{x(t)}{\ell/2}$

The torque is $\tau = x(t)mg$, or $\tau = \tfrac12 \ell \sin\alpha ~ mg$

Oh wait ...

Sorry I hadn't realised it was two masses on a weightless rod. I assumed it was a uniform rod of mass $m$.

ah no worries, yeah it's just a mass $M$ at B and $m$ at A

That just means the position of the COM isn't at the centre of the rod. The basic idea is the same.

Let me draw a diagram ...

@psa that's what I had in mind.

The COM always lies on the y axis so $a$ is the distance from the top end of the rod to the COM.

The distance $a$ remains constant as the rod oscilaltes.

alright I'll give that setup a try and see if it's not as much of a mess

all that should be necessary to solve is conservation of momentum in $\mathbf{\hat x}$ and conservation of energy

There will be a tension in the rod as well. So the total normal force on $m$ won't be $mg$.

8:47 AM

I'm pretty sure my energy and momentum previously were correct with the origin at O, but for some reason I just wasn't getting the right answer in the end.

or at least I assume it's incorrect

@JohnRennie Hi :-)

@Wolgwang hi :-)

in JEE Chemistry Club, 7 mins ago, by Wolgwang

For example, yesterday my coach completed the "vectors" chapter but due to some reasons I have not tried all the questions of the module. This coming Sunday, he would start the next chapter (id est Kinematics ), and I don't think I would be able to solve all the questions till then.
After the inception of that chapter, I would be doing the questions of that chapter, so when would I practise the questions of the previous chapter?

Any inputs here

You mean what do you do about completing all the problems from the previous chapter?

@JohnRennie No

@JohnRennie I haven't.

8:58 AM

I'm not sure what you are asking ...

OK lemme make it more clear
I should have completed all the questions of the previous chapter, but I haven't.
And when the next chapter starts, I won't be able to do so.

So what are you asking me? How to complete the questions from the previous chapter?

@JohnRennie Yes.

I don't think there is a magic solution.

Look through the remaining questions and try to prioritise them i.e. see which ones will teach you most. Do those questions first, then any questions you don't have time for you can keep to use for revision nearer the exam.

@JohnRennie I wasn't expecting that :-)

@JohnRennie Thanks :-)

9:10 AM

@JohnRennie hi :)

@AshishAhuja hi :-)

How far have you got with this?

Could you help me with finding the velocity gradient in the above question? I believe I can find the torque using integration myself, but I'm struggling with the gradient.

Incidentally, when a question says the angles are small it usually means you can use the approximation sinθ = θ

9:12 AM

yup

The reason there is a gradient is that when a fluid is in contact with a surface it is stationary with respect to that surface.

ohh I think I got it.

OK :-)

Let me give it a try...

yup it worked. Now I'll try the integration...

Q: how the form factor (stress - torque) is derived?

:-)

I think I answered a question on this. I used to do this sort of thing for a living when I worked as a scientist. Commercial viscometers generally use the cone and plate geometry.

Aha!

0

I am working on an experiment of rheology and I need to calculate shear stress in order to calculate the viscosity. After some research I found that for the type of viscometer I will be using (cone-plate), the stress is calculated by dividing the torque given by the apparatus by a form factor equ...

9:27 AM

@JohnRennie that seems interesting!

@AshishAhuja it was good fun :-)

@JohnRennie Atomic mass = mass of one atom. So , does atomic weight of Na = 23gms

I'm not a chemist, so my advice on this may not be 100% reliable. But I would say the molar mass of Na is 23g.

Ohk. Yes sir. Thank you.

You will often read statements like the atomic weight of sodium is 23, but this means 23 atomic mass units - often abbreviated to AMU.

One AMU is the mass of a hydrogen atom.

9:36 AM

Ok.

Hmm I'm making some mistake in the integration. I'm getting the right answer but the $\theta$ in the denominator is missing. I'll type out what I'm doing, it's probably something dumb I'm missing...

Sir , AMU or g / mol.

AMU

Ohk

The reason they give both is that the number *23* is both:
1. the atomic weight in AMU
2. the molar mass in grams

9:38 AM

Ohk sir

Getting it.

I'm considering a ring at a "height" $x$ from the bottom. The radius of this ring is then $\frac{x}{\theta}$ and the thickness ("height") is $dx$. Surface area $$\frac{2\pi \times x dx}{\theta}$$

Do you mean a ring of radius $x$?

Like I used in the answer I linked above?

no.

I saw your answer, and the approach I tried first was different, and that's what I'm typing here.

I mean a ring of height $dx$, and then I'll integrate from $0$ to $R \times \theta$

Since $tan \theta = \theta$

Or does this make no sense at all?

9:43 AM

Oh, OK, so it is basically a ring of radius $r$ but you're using $h = r\theta$ as the variable to integrate.

Yes. To me naturally integrating over height made more sense, but you can express it like that as well.

This seems like doing the calculation the hard way ...

Hmm...

It's just two more steps, I'll type it out, if that's okay?

OK ...

Continuing from

5 mins ago, by Ashish Ahuja

I'm considering a ring at a "height" $x$ from the bottom. The radius of this ring is then $\frac{x}{\theta}$ and the thickness ("height") is $dx$. Surface area $$\frac{2\pi \times x dx}{\theta}$$

$$dF = \frac{\eta 2 \pi x dx \omega}{\theta \times \theta}$$

Now using $dT$ for torque:

$$dT = dF \times \frac{x}{\theta} = \frac{\eta 2 \pi x^2 dx \omega}{\theta ^3}$$

Integrating over the limits of

0 to $R \theta$

we get $$T = \frac{\eta 2 \pi R^3 \omega}{3}$$

now I've made a mistake somewhere since the $\theta$ is missing...

I'll go try the method given in your answer I guess.

9:52 AM

@AshishAhuja Let's start with the relations $r = x/\theta$ so $dr = dx/\theta$. We'll need these in a moment. OK so far?

ok. Here $h$ is what I was using $x$ for right?

Oops, yes.

Ohh then I think I got my mistake.

:-)

I believe I'm going wrong in the surface area calculation.

9:54 AM

Yes. And you're going wrong because you picked such a hard way to do the calculation!

oh well sometimes the hard way is what comes naturally to mind :D

:-)

I had actually thought about integrating in the way you've done in the answer when I first got the question in a test, but what confused me is since we're considering a flat annulus, how do we justify the surface area = $2 \pi r dr$?

When we calculate the torque we will be taking the cross product of $\mathbf r$ with $d\mathbf F$. Yes?

yes

ohh got it

9:59 AM

And this means all that matters is the component of $d\mathbf F$ normal to $r$ i.e. the vertical component.

And that's going to end up being proportional to the vertical component of the vector area.

Yup. That clears it all up, thanks a lot.

:-)

@JohnRennie quick Q

Free sir ?

My confusion is with amount of O2 reacted.

2 mole of Na to react is given. That is the need and that has been done.

So , 2 mole of Na reacts with 1/2 mole of O2

Why does my teacher even think to write this ( red circle ).

@Sarabsrimt I guess they just thought it would make it clearer ...

Ohk. We can’t we say 1 mole of Na react with 1/4th mole of O2 right

Since requirement is 2 mole of Na

10:08 AM

Well the mole ratio is 4:1

So that could be one mole Na and ¹⁄₄ mole O₂

Or 2 moles Na and ¹⁄₂ mole O₂

Ohk ohk.

Or 4 moles Na and 1 mole O₂

Ohk

we can say this as ratio

In this case you have 2 moles Na, so it's 2 moles Na reacting with ¹⁄₂ mole O₂

but asked in Q is 2mole

ohk got it

10:10 AM

@Sarabsrimt are you trying to solve it using equivalents?

Yes.

Ah ok, so first you'd write down the oxidation states of the reactants and products since this is a redox reaction.

Na and O2 both have a zero oxidation state on the reactant side.

On the product side Na has an oxidation state of +1 while oxygen has a state of -2

Ohk

Now we find the n factor for Na and O2. Should I finish the whole thing?

Umm. I know this part actually and have written it too. Eq of O2 = 1/2 mole * 4

My Q was related to amount of O2 would react with Na

Yeah, thanks though.

@JohnRennie Thanks a lot. I got it now.

Pretty interesting chapter.

10:15 AM

@Sarabsrimt ah ok for that either you could write a balanced chemical reaction (which is much easier than equivalence for this case) or you'd equate Equivalence of Na = Equivalence of O2

@AshishAhuja yeah, I got it now.

@Sarabsrimt :-)

For second reaction

Al + ...

Mole ratio is 4:3

Same thing. In any reaction the number of equivalence is the same for all reactants and products.

That is because of mole ratio right

10:20 AM

No.

What is Al reacted was 4 equivalents

Then equivalents of O2 will also be 4. You can try it out by actually calculating the moles.

Hmm ok.

But answer changes here. @AshishAhuja

if I took eq of Al as 4

Give me a second, let me try it out on paper myself...

Ohk. Sure

10:24 AM

@Sarabsrimt uhh you're getting number of moles of O2 = 1, right?

Ohh ohh. My mistake.

yes O2 moles = 1

Ok. And then n factor (or valency factor) = 4, so we get number of equivalents of O2 = 4 * 1 = number of equivalents of Al.

Yes. But I’m 1st Q

no fo eq of O2 = 2

Not 4

5 mins ago, by Sarabsri mt

if I took eq of Al as 4

Aren't we talking about this?

Yes

right

I thought you meant to say that equivalent of O2 would be 2 alwasy

@AshishAhuja here

Did you mean to say equivalence only for 1 reaction.

10:29 AM

Ohh no. I meant to say that the number of equivalents would be the same for all reactants and products for a given reaction, that's the basis of equivalent concept, and that's the reason n factor is defined the way it is.

For example for the reaction A + B -> C + D

Ohk

no. of equiv. of A = no. of equiv. of B = no. of equiv. of C = no. of equiv. of D

Ohk. Got it.

Thanks a lot.

@AshishAhuja ✌️

:)

N factor of Na2O = 2 right

Na2 = 2 and O = -2

10:32 AM

It matters. If you're talking about it in the reaction above, then no. As a standalone salt, yes.

definition of N factor is reaction dependent.

Ohk.I ask for the reaction since that is what I need for so living equivalents

For that reaction, first we write the oxidation states.

(reactant) Na = 0, O = 0

(product) Na = +1, O = -2

Now, we need to equate equivalence of O2 = equivalence of Na

Ohk

for Na, number of electrons involved when one molecule of Na underwent reaction would be 1 (change in oxidation state)

similarly for O2 it would be 4 (two for each oxygen atom)

so nfactor (Na) = 1, nfactor(o2) = 4

Yes

10:36 AM

now n factor(Na) * moles of Na = n factor(O2) * moles of O2 = number of equivalence; and now you'd get the answer.

One thing before this.

$Na_2$ = 2 Na in terms of no of moles right.

2 moles of both

?

Uhh Na2 does not exist.

Ohk. But if it did . Take H2 for example.

2H and H2

Both have 2 moles

You could say so, yes. But you need to be specific that it's "2 moles of H atoms".

Ohk.

Our reaction is Na2O right

@AshishAhuja here. 2 * 1 for Na

How do we know moles for O ?

10:40 AM

@Sarabsrimt yes, 2 because 2 moles of Na and 1 because of the n factor.

For no of moles of O

you mean O2?

Yes

O2

Ok.

We equate the number of equivalence.

10 mins ago, by Ashish Ahuja

no. of equiv. of A = no. of equiv. of B = no. of equiv. of C = no. of equiv. of D

We have number of equivalence of Na = 2 * 1 as you said above.

2 * 1 = x * 4

10:41 AM

yes

x = 1/2 mole of O2

?

Oh great.

Btw. This is how my teacher told me about n factor

Same way. But I didn’t understand it well enough.

Noe I did more

Ohk. Thanks a lot @AshishAhuja

Only problem was that my teacher said even - 6 we can write

As n factor.

See you later then.

n factor is more or less a trick which used to be used by chemists at a time when chemical equations weren't known. So I guess you should take the mod value of n factor. Saying n factor = -6 and getting negative moles of O2 does not make much physical sense.

Yeah . Right

Bye.

12:46 PM

waiting for 7th March ?

No of electrons = summation of atomic number * no of moles * $N_A$ is the formula in my textbook.

Q is to find no of electrons in 5gm of CaCO3 .

No of electrons = 5/100 * $N_A$ * 50 is the answer.

My quick confusion is that if I ask no of electrons in CaCO3 = 1 mole * 50(atomic number ) as no of electrons . Why $N_A$ was not multiplied.

Also , the derivation of the formula . I am not getting how to solve it.

Any idea how we can derive this formula.

1:34 PM

I solved it.

3:41 PM

Can we say mass of an atom = mass of nucleus ( proton + neutron ) + mass of electron? I’m sure it is not enough but what else do we need ?

Sid

5:50 PM

@Sarabsrimt at your level, I think that's the simplest way to understand it.

In fact, you can even think of it as simply the mass of nucleus itself. The protons and neutrons are massive compared to the electrons.