Problem Solving Strategies

We known that the potential generated by a charge pointwise $q$ is $V(r) = kq/r $ and the equipotential surfaces (in 3D) are spheres centered in the charge with $r\geq 0$ where $r=d(O,P)$, i.e. the distance between the origin and a generic point $P$. In fact if we are in space, where an orthono...

Abcd

@JohnRennie Hi. Can you see it please?

12 hours ago, by Abcd

5 hours ago, by Abcd

12 hours ago, by Abcd

5 mins ago, by Abcd

I felt that the motion can cross any level of complexity...

Sebastiano

@Abcd, @JohnRennie Good morning

Abcd

@Sebastiano Morning.

Sebastiano

@Abcd I wanted to know if it is possible to ask the same question that I placed on the Mathematics website on the Physics website.

Abcd

@Sebastiano No, an exact same question isn't allowed, AFAIK.

Sebastiano

11:39 AM

@Abcd My question has not yet been taken into consideration.

John Rennie

@Abcd well (a) is certainly true because the velocity has to increase then decrease

Abcd

@Sebastiano so delete it from maths and post it on physics.SE (and start a bounty ;))

@JohnRennie Agreed.

Sebastiano

@Abcd OK!

John Rennie

(b) is untrue. The acceleration can be arbitrarily large provided it remains large for only a very short time.

Abcd

Okay, then?

John Rennie

11:42 AM

The minimum value of $\alpha$ will be the constant value that gets the particle from $x=0$ to $x = 1/2$ in 0.5 seconds.

Abcd

Didn't get you.

John Rennie

@Abcd We know the particle accelerates from rest and slows to a stop again.

Abcd

yes

John Rennie

Suppose the particle has a constant acceleration $a$ for half a second and reaches $x=0.5$ at $t=0.5$. Then it has a constant deceleration $-a$ for another half second, which brings it to rest at $x=1$ at $t=1$.

Abcd

But we aren't allowed to suppose anything because nothing is mentioned in the question...

John Rennie

11:46 AM

If I use $s = \tfrac{1}{2}at^2$ that gives me $a=4$. Yes?

Abcd

yes

Sebastiano

@Abcd Ok I have deleted my question. Now it is on Physics website.

@JohnRennie Good morning and good work also @Abcd

John Rennie

Now my claim is that if $a$ varies with time the maximum value of $|a|$ cannot be less than 4.

Abcd

how?

@Sebastiano Which good work :P ?

John Rennie

Suppose the motion is asymmetrical i.e. at $t=0.5$ the particle is not at $x=0$. It could for example be at $x=0.25$ or $x=0.75$. OK so far?

Abcd

11:50 AM

yes

John Rennie

Suppose it's at $x=0.75$ at $t=0.5$. Since it got farther than $x=0.5$ it must have accelerated harder than $a=4$.

Akash. B

Why sun is red in evening?

John Rennie

Now suppose it's at $x=0.25$. That means it has to decelerate harder than $a=4$ to come to a stop again.

Akash. B

hi

Abcd

@Akash.B It has to do with scattering ...You'll get your answer in one google search though...

John Rennie

11:52 AM

@Akash.B we expect you to put at least some effort into researching a question

4

Akash. B

OKAY can I ask you another question?

John Rennie

@Abcd before we got distracted I was attempting to persuade you that the minimum acceleration must be 4 ...

That means (c) is true

Abcd

3 mins ago, by John Rennie

Now suppose it's at $x=0.25$. That means it has to decelerate harder than $a=4$ to come to a stop again.

This part please^

Akash. B

Why light split into different colors when it is passed through a prism?

Abcd

@Akash.B Research first please!

All these answers are available on first google search, and they are even taught in middle school and high school.

Akash. B

11:55 AM

How

John Rennie

@Abcd If the particle is at $x=0.25$ at $t=0.5$ that means it has to cover 0.75 in 0.5 seconds.

While if it's at $x=0.5$ it only has to cover 0.5 in 0.5 seconds.

(it feels really weird to be giving distances without units, but that's the way the question is)

Akash. B

Which force is causing the earth to rotate in its axis?

Abcd

@Akash.B Agh!!! Every answer is on google.

John Rennie

@Akash.B these aren't really "problem solving" questions. Can you ask them in the main chat?

Akash. B

Main chat?

Abcd

11:59 AM

@Akash.B $\bar{h}$

John Rennie

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

@JohnRennie yes, so?

I asked,they are telling that they does not discuss physics ?

Abcd

@Akash.B lol, they must be kidding xD. But don't ask questions that way...Please share your specific problem.

John Rennie

@Abcd that means it has to move faster to cover 0.75, so it has to decelerate harder than $a = -4$ to shed the extra speed.

Abcd

12:02 PM

@JohnRennie How are you getting the value 4?

John Rennie

17 mins ago, by John Rennie

Suppose the particle has a constant acceleration $a$ for half a second and reaches $x=0.5$ at $t=0.5$. Then it has a constant deceleration $-a$ for another half second, which brings it to rest at $x=1$ at $t=1$.

16 mins ago, by John Rennie

If I use $s = \tfrac{1}{2}at^2$ that gives me $a=4$. Yes?

Abcd

$0.75 = 0.5 a (0.5)^2$

John Rennie

I'm talking about the position of the particle at $t=0.5$

Abcd

Oh yes

It gives $a=6$

John Rennie

Which is greater than 4 :-)

Abcd

12:07 PM

Got it @JohnRennie. Though this approach wouldn't have struck me ever...

Is there any other way to do it?

John Rennie

If you make the assumption that the acceleration and deceleration are constant values then yes you can do it using $s=\tfrac{1}{2}at^2$.

Abcd

@JohnRennie And do you have any idea about the second question?

9 hours ago, by Abcd

John Rennie

What does degree of association mean? The equilibrium constant?

Abcd

@JohnRennie You wouldn't believe that till now I read it as "dissociation".

Strange!

@JohnRennie Well degree of dissociation is the number of moles of substance dissociated per mole of it.

degree of association would be opposite of it.

John Rennie

So it means half of the A originally present has associated? And half of the A2 created by the association of $A$ has dissociated?

Abcd

12:16 PM

@JohnRennie Yes, I guess.

So if you took x moles of A ...$(0.5)x$ remained

John Rennie

So if $A_0$ is the total number of moles of $A$ present that means we have $A_0/2$ moles of $A$. So there must have been $A/4$ moles of $A2$ before the $A2$ started to react.

Abcd

@JohnRennie yes, then?

John Rennie

And half the $A_2$ has reacted so we now have $A_0/2$ moles of $A$, $A_0/8$ moles of $A_2$ and $A_0/8$ moles of $B$ and $C$.

Abcd

@JohnRennie yup

John Rennie

So we just have to figure out how many moles of $A$ are present at equilibrium.

Abcd

12:23 PM

$\text{vapour density of A =} D$, $\text{vapour density of mixture of}~ \ce{A_2,B, A}= d$

John Rennie

And I must admit I can't see how to do that ...

Abcd

@JohnRennie $VD \propto \dfrac 1 {\text{number of moles}}$

King Tut

is this chemistry

Abcd

@KingTut Yes :-) , physical chemistry.

John Rennie

I have to say I don't understand what the question is asking ...

King Tut

12:25 PM

@Abcd vapor density is ratio of molar masses of gas to the hydrogen

Abcd

I know that...

John Rennie

What are $M_A$ and $M_B$? Mass? No. of moles?

Molecular weight?

Abcd

@JohnRennie they haven't given anything :/ . But I think its molecular weight...because capital M is used for that only,

King Tut

@Abcd use the point of john rennei and the average molecular weight at equilibrium

also dont include C because it is solid

Abcd

@KingTut but we don't know the masses of each individual...

King Tut

12:30 PM

ok

but in option its given in terms of that

Abcd

And $<M>= \dfrac {m}{\sum n_i}$ where m is mass of mixture and n_i is number of moles of ith component and M_i is its molar mass

King Tut

@Abcd you can use total mass upon total moles

@Abcd No its wong

it should be \frac{ \sum n_i M_i }{ \sum n_i}

@Abcd yes and m = \sum n_i M_i

Abcd

@KingTut DO you have mathajx in chat enabled?

King Tut

no

Abcd

@KingTut Wait a second.

@KingTut Please see how to enable it:

7

A: How can I enable MathJax in chat?

Update 2017-05-01 The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.) So here is my take. I modified the ...

King Tut

12:36 PM

They have made a good website and forgot to put latex in the chatting room

Abcd

@KingTut You can enable it...

I can see latex

But you can't

because you haven't enabled

King Tut

@abcd paste in the console

?

Abcd

@KingTut you have to make a bookmark...

The instructions are there

King Tut

@Abcd if i put dollar sign here, you can read it in latex? $\nabla$

Abcd

yes

I can see an inverted triangle

King Tut

12:41 PM

so answer is D

Abcd

@KingTut how??

I get $\dfrac{8M_a}{6M_a + M_b}$

King Tut

@Abcd use hint of john rennei

@abcd use that average molecular weight $2d = \frac{ M_A\frac{a}{2} + 2 M_A\frac{a}{8}+M_B\frac{a}{8}}{\frac{a}{2}+\frac{a}{8}+\frac{a}{8}}$

and also $M_A = 2D$

Abcd

@KingTut thanks, I'll retry.

King Tut

are you from school

Abcd

12:57 PM

@KingTut Yes, what about you?

King Tut

yes me also

Abcd

1:50 PM

@KingTut grade 12? How do you know so high level maths dude?

King Tut

No i am in grad 10, i asked my cousin who is in higher classes, I think he is in college

@Abcd I am trying to learn higher physics because in school we dont have much physic

Abcd

2:06 PM

@KingTut how old are you?

King Tut

why? 15

Abcd

@KingTut Oh, you must be a brilliant student :) . You know a lot at this young age.

King Tut

:o really My cousin mostly teaches me, in school I dont have good performance

Abcd

3:29 PM

(@GaurangTandon (sorry for pinging) is coriolis force in syllabus? )

@KingTut Which country are you from?

King Tut

@Abcd first you ping and then be sorry for it? Such behavior is worthy of honor. :D

@Abcd i am from south asia

Abcd

@KingTut lol what do you mean?

I didn't want to disturb him ... But it was urgent because there are many problems on this force

King Tut

ok

Abcd

@KingTut malaysia

King Tut

No

@abcd do you know where hanoi is

Abcd

3:35 PM

@JohnRennie how to calculate coriolis force?

John Rennie

@Abcd that's a somewhat vague question ...

Abcd

@KingTut I can always google :) ... Its in vietnam.

@JohnRennie Just a second, I'll share the question.

King Tut

@John is coriolis force a term in the acceleration in polar coordinate

i mean when we differentiate velocity in polar coordinate we get acceleration, in r hat and theta hat direction

Abcd

A horizontal disc rotates with a constant angular velocity $\omega = \pu{6 rad s^{-1}}$ about a vertical axis passing through its centre. A small body of mass $m= \pu {0.5 kg}$ moves along a diameter of disc with velocity $v' = \pu{50 cm/s}$ which is constant relative to the other disk. Find the force that the disc exerts on the body at the moment when it is located at the distance $r= \pu{30 cm}$ from the rotation axis.

@JohnRennie here^

John Rennie

@KingTut it's kind of a vaguely defined term. The coriolis force is strictly speaking the force required to keep the object moving straight,

Abcd

3:40 PM

I calculated the frictional force and the normal force.

Then I looked up the solution and there was a new force which I had heard in geography class 2 years ago...coriolis force...Never been taught this...How to calculate it?

John Rennie

What does \pu do? It's giving a MathJax error here.

Abcd

3 hours ago, by Abcd

7

A: How can I enable MathJax in chat?

Update 2017-05-01 The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.) So here is my take. I modified the ...

@JohnRennie What mathrm does ...

it is for units which chemistry people developed I think.

John Rennie

That's what I get.

Abcd

@JohnRennie Because your mathjax lacks the mhchem package.

2 mins ago, by Abcd

3 hours ago, by Abcd

7

A: How can I enable MathJax in chat?

Update 2017-05-01 The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.) So here is my take. I modified the ...

Please upgrade :-)

Then you'll be able to read : $\ce{2H2 + O2 -> 2H2O}$

Gaurang Tandon

@Abcd no; but rotational mechanics has some advanced questions that were asked last year; not exactly "coriolis-force" but "coriolis-force" like

actually only one was asked IIRC

some disc rotating in a circle about a pivot (like buffalos pumping water out of well)

John Rennie

3:47 PM

$\require{mhchem}$

Gaurang Tandon

i guess but i don't recall exacty

John Rennie

Nope still doesn't render

Abcd

@JohnRennie Change your bookmark link through that ^

Manishearth has posted a link

@GaurangTandon You did it?

John Rennie

It works! :-)

@Abcd isn't the force just $\mu m g$?

Abcd

14 mins ago, by Abcd

A horizontal disc rotates with a constant angular velocity $\omega = \pu{6 rad s^{-1}}$ about a vertical axis passing through its centre. A small body of mass $m= \pu {0.5 kg}$ moves along a diameter of disc with velocity $v' = \pu{50 cm/s}$ which is constant relative to the other disk. Find the force that the disc exerts on the body at the moment when it is located at the distance $r= \pu{30 cm}$ from the rotation axis.

@JohnRennie No...

John Rennie

3:55 PM

@Abcd How can the disk exert a force greater than the frictional force?

Gaurang Tandon

@Abcd well not so soon...

for advanced only

Abcd

@JohnRennie there's no "other" before disk in the question.

That was a mistype

@JohnRennie my point is that the net force on the body in disk frame is 0 right?

SO something balances the weight (aka Normal force)

Something balances the centrifugal force (aka frictional force)

But see the solution:

John Rennie

Ah, OK, I had misunderstood the question. You mean the small body moves in a straight line relative to the disk, so it's actually moving in a spiral.

Abcd

ok

@JohnRennie How did they calculate coriolis force?

John Rennie

@Abcd I'm not sure to be honest, but I probably wouldn't do the calculation that way.

Abcd

4:10 PM

@JohnRennie They have just added perpendicular vectors it seems.

@JohnRennie Would you get the same answer with your calculation?

John Rennie

Well, let's see ...

The radial velocity is 0.5 m/sec, and the tangential velocity is $r\omega$, so the total velocity is $$ v^2 = (0.5)^2 + (r\omega)^2 $$

And the radial distance is just $r=0.5t$

So $$ v = 0.5 \sqrt{1 + \omega^2 t^2} $$

OK so far?

Nehal Samee

@JohnRennie .

spring is hanging at 20m height from the ground . A ball of mass 0.2kg is shot towards the spring with velocity 49m/s such that extension of spring is 3m . What will be the rebound velocity of ball on the ground ?
The problem I mentioned ...

John Rennie

@NehalSamee I can only do one problem at a time :-)

Nehal Samee

@JohnRennie .

Keep it in your memory ... Answer at your free will ...

Abcd

@JohnRennie yes

Nehal Samee

4:15 PM

I'm next on the serial 😂😂😂

John Rennie

@Abcd and $a = dv/dt$

So if we differentiate that expression wrt time and multiply by $m$ we should get the (horizontal) force ...

Abcd

@JohnRennie just a second it should be $R- 0.5t$

Where R is radius of disk

John Rennie

@Abcd Doesn't it start at the middle and move outwards?

Abcd

@JohnRennie He doesn't tell anything about that

John Rennie

@Abcd Well that seems a reasonable assumption ...

Abcd

4:18 PM

Okay

John Rennie

Oh hang on, no that won't work. Rats.

I hate working in non-inertial frames!

OK I guess taking the vector sum of the three forces is the only way to do it.

Abcd

But how to find that third force..

John Rennie

But I can't remember the equation for the Coriolis force. I'd have to Google it.

King Tut

@John we can differentiate velocity in polar coordinate is that there is a term with theta cap

John Rennie

@KingTut I can't remember how to do differentiation in polar coordinates! :-)

But yes, given we have expressions for the radial and tangential velocity we could differentiate using the correct expressions for polar coordinates and we should get the correct answer.

In fact it's a form of covariant differentiation, which I do all the time in GR so I should be able to remember it. But I suspect that's beyond what JEE would expect.

Coriolis force

In physics, the Coriolis force is an inertial force that acts on objects that are in motion relative to a rotating reference frame. In a reference frame with clockwise rotation, the force acts to the left of the motion of the object. In one with anticlockwise (or counterclockwise) rotation, the force acts to the right. Deflection of an object due to the Coriolis force is called the Coriolis effect. Though recognized previously by others, the mathematical expression for the Coriolis force appeared in an 1835 paper by French scientist Gaspard-Gustave de Coriolis, in connection with the theory of...

$$ a_C = -2\omega \times \mathbf v $$

Abcd

4:27 PM

@JohnRennie Is coriolis force $-2m(\vec{\omega}\times \vec{v})$

John Rennie

Snap :-)

Abcd

I was seeing a video on youtube that derives it! It's cool!

John Rennie

And since $\omega$ and $v$ are at right angles it's just $-2m\omega v$

I guess you just have to remember that equation for the Coriolis force ...

Abcd

I saw the derivation...so I can remember.

Deflects things sideways:Coriolis Force by prof. Hc Verma

►

John Rennie

@Abcd Cool :-) Shall I ping Nehal now to look at his question?

Abcd

4:30 PM

@JohnRennie Just 3 minutes more.

John Rennie

OK ...

Abcd

@JohnRennie What is the direction?

John Rennie

Well it's - omega cross v ... erm ... that's the right hand rule isn't it ...

Abcd

@JohnRennie okay, so $(-\hat{i}\times \hat{k})= \hat{j}$

John Rennie

I get it pointing in the opposite direction to the tangential velocity ...

Abcd

4:33 PM

Is $\hat{j}$ correct?

John Rennie

I'm not sure about using Cartesian coordinates for a rotating system ...

Abcd

35 mins ago, by Abcd

SO something balances the weight (aka Normal force)

34 mins ago, by Abcd

Something balances the centrifugal force (aka frictional force)

Is my method correct?

And then we'll just add the coriolis force coz we are in rotating frame

So direction of normal force is $\hat{k} $

John Rennie

Yes. The acceleration towards the centre is just $r\omega^2$. We aren't told what causes this acceleration, whether it's a frictional force or whatever.

Abcd

@JohnRennie Okay, now you can help @NehalSamee with his problem. thanks :) !

John Rennie

@NehalSamee are you around?

... it would appear not.

The h Bar

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