Problem Solving Strategies

RR.

03:17

@Wolgwang It's given that both of them are Vrms... So, ig use the formula V^2 = VR^2 + (VL-VC)^2

So, VC = 100 √3 V

Wolgwang

04:10

Yeah. Thanks :)

Prateek Mourya

Here whats does state means? What's the relation between state and system (on which observation is made)

John Rennie

04:35

@NaveenV This is one of the key things about QM that students find difficult. Quantum particles are always delocalised and are best thought of as fuzzy clouds not little balls. An electron in, for example, a 1s orbital is like a cloud that fills the whole orbital and exists everywhere in the orbital simultaneously.

The Bohr radius is not the distance of the electron from the nucleus. Instead it is a parameter that sets the size of the cloud. Specifically for a 1s electron the orbitals is:
ψ₁ₛ = A exp(-r/a₀)
where A is a constant.

@NaveenV The conductivity is proportional to the number of conduction electrons and the mobility of the electrons.

In all conductors the mobility decreases with temperature i.e. this happens in both metals and semiconductors. But in semiconductors the number of conduction electrons increases with temperature and this outweighs the decrease in mobility.

@PrateekMourya What book is that from?

Prateek Mourya

06:02

That's

@JohnRennie Thermodynamics and stat mech by richard Fitzpatrick

John Rennie

OK, what he's saying is that we have a collection of many systems S e.g. S could be a hydrogen atom and the ensemble would be a collection of many hydrogen atoms.

And each hydrogen atom is in some state ψ that is in general a superposition of the eigenstates.

So what he's saying is you pick one of the hydrogen atoms at random and do a measurement on it, then pick a second hydrogen atom at random and do a separate measurement on the second atom.

So the system is the atom (or whatever) that you choose from the ensemble at random, and the state is the wavefunction of that atom.

Does this make sense?

cOnnectOrTR 12

06:41

Hi @JohnRennie

John Rennie

Hi :-)

cOnnectOrTR 12

When two bodies move towards each other the distance between them decreases at a rate a+b m/s where a and b are their speeds.

John Rennie

Yes

cOnnectOrTR 12

The time taken to cover that distance by the two bodies will be distance/(a+b). Can you prove?

John Rennie

Suppose the initial position of the blue ball is x = 0 and the initial position of the red ball is x = 𝓁

At a time t the blue ball has moved right a distance s = at so its position at time t is:

x₁ = at

OK so far?

cOnnectOrTR 12

06:50

Okay. Yes

John Rennie

Now, the red ball starts at x = 𝓁 and moves left at speed b so at time t the position of the red ball is:
x₂ = 𝓁 - bt
Yes?

cOnnectOrTR 12

Ok

John Rennie

And when the time that the balls have covered the distance 𝓁 is the time that they meet i.e. we are looking for the time when x₁ = x₂.

Does this make sense so far?

cOnnectOrTR 12

@JohnRennie are you saying time taken by one body to cover the distance l is same as the time when they meet ?

John Rennie

The two balls start a distance 𝓁 apart i.e. the distance between them starts at 𝓁.
Then when the meet the distance between them is zero.
So if we find the time when they meet that is the time for the distance between the balls to decrease from 𝓁 to zero.

Yes?

cOnnectOrTR 12

07:01

Yes

John Rennie

And that's what the question is asking.

So we just have to find the time that the balls meet.

cOnnectOrTR 12

I asked how is time =l/(a+b) ?

John Rennie

Well x₁ = at
and x₂ = 𝓁 - bt
so if we set x₁ = x₂ then we get:
at = 𝓁 - bt
Yes?

cOnnectOrTR 12

Ok

John Rennie

And if we rearrange this we get:
at + bt = 𝓁
t(a + b) = 𝓁
t = 𝓁/(a + b)

That's how we show the time is 𝓁/(a + b)

cOnnectOrTR 12

07:13

👏👏👏

John Rennie

:-)

TonyPhysicslover

08:56

Hey

Hey @JohnRennie Could you please help me this question ?

John Rennie

@TonyPhysicslover Hi :-)

TonyPhysicslover

Hey, sir !

John Rennie

Suppose the ball is stationary. If it's stationary the net vertical force on it must be zero. Yes?

TonyPhysicslover

Yes

John Rennie

There are two forces acting on the ball. The normal force N between the ball and the hemisphere, and the tension in the string, T.

And both of these forces change as the ball moves up the sphere.

TonyPhysicslover

09:08

But sir what about the downwards force due to gravity ?

John Rennie

Yes, sorry, we have N, T and the downwards force mg.

TonyPhysicslover

yes

John Rennie

So if we know the angles of the two forces N and T we can find their horizontal and vertical components, and we know the horizontal components must sum to zero and the vertical components must sum to mg.
Yes?

TonyPhysicslover

Yes sir

John Rennie

We are not given the height of the nail above the hemisphere or the radius of the hemisphere so we can't do a calculation with the angles, but that's OK as we can answer the question without having to do the full calculation.

Suppose we have pulled the ball right to the top so the ball is resting on to op the hemisphere.

TonyPhysicslover

09:12

Okay

T + N = mg ?

John Rennie

In that case the tension in the string is zero and the normal force is mg. This is because the ball doesn't need to be pulled upwards by the string. It will just rest on top of the hemisphere.
Yes?

TonyPhysicslover

Yes but in the original question, It is mentioned that a Force F is pulling the string

So can we still assume that ?

John Rennie

The force F is equal to the tension in the string. So we can just consider the tension in the string.

TonyPhysicslover

okay

John Rennie

So we know that at the top of ball T = 0 and N = mg.

TonyPhysicslover

09:17

yes

John Rennie

10 mins ago, by John Rennie

If we look at this diagram both N and T point upwards. If θ is the angle of N to the vertical and φ is the angle of T to the vertical then taking the vertical components of the forces gives us:
N cosθ + T cosφ = mg
Yes?

TonyPhysicslover

Yes

John Rennie

And by looking at the horizontal forces we also get:
N sinθ = T sinφ

Yes

TonyPhysicslover

Yes okay

John Rennie

From that second equation we see that T > 0

TonyPhysicslover

09:24

Yess

John Rennie

But we know that when the ball was at the top T = 0. Yes?

TonyPhysicslover

@JohnRennie but sir, if the Force pulling the string is equal to the tension and F is not equal to 0 , then the tension shouldn't be 0 at the top ?

John Rennie

At the top the ball would stay motionless on top of the sphere even if the string wasn't there. Yes?

TonyPhysicslover

Yes

John Rennie

I guess what the question doesn't make clear is that F is the minimum force needed to keep the ball stationary.

At the top we could be lifting the ball slightly e.g. set F to ¹⁄₂mg and n that case N would be ¹⁄₂mg as well.

But the point is that we don't need any eternal force i.e. F can be zero and N = mg.

TonyPhysicslover

09:29

So we're assuming that The pulling Force F varies with position of the ball and not constant overall ?

John Rennie

Yes

It's the force needed to keep the ball stationary

TonyPhysicslover

okay

RR.

@JohnRennie Hi sir!

Could you help me out with this question?

John Rennie

I'm busy answering another question at the moment. Sorry :-(

RR.

Ohh... Okay sir... Whenever you find the time =)

Wolgwang

10:09

@RR. Have you tried to use that when it starts rolling normal at the point of contact on the left side becomes zero? ( I am trying clarify my doubt)

RR.

Meaning? I'm confused on what you are referring to?

John Rennie

@RR. Hi

RR.

Hi sir!

Wolgwang

This N=0

Force such be such that.

What do you think about it?

John Rennie

For the sphere to roll it has to pivot about this point. Yes?

RR.

10:14

Yes sir...

John Rennie

If the centre of the sphere is to the left of that point the torque on the sphere is anticlockwise and it will stay in the hole. If the centre of the sphere is to the right of the hole the torque is clockwise and the sphere will roll out of the hole.
Yes?

RR.

Yes sir

So, when centre of the sphere is just above the pivot, it would just begin to roll?

John Rennie

Yes :-)

RR.

Ahh... Okay... Hadn't thought of that...

John Rennie

So now it's just geometry to work out what θ is when the centre of the sphere is exactly above the right edge of the hole.

I'll leave you to do that :-)

Wolgwang

10:19

@JohnRennie Hi! It should give the same result if I find the nomal at left and make it equal to zero?

RR.

Sure sir... I'll get on that =)

John Rennie

@Wolgwang Yes

@Wolgwang Though that seems harder to me ...

Wolgwang

@JohnRennie It doesn't :(

I had tried that the first time

Then I thought that there can be an instant when it just loses contact and after some angle rotation it gets back.

It was kinda a less stricter inequality.

John Rennie

@Wolgwang Hmm, OK, I'd have to work through the problem to see what is going on.

Naveen V

11:10

@JohnRennie can you tell me more about mobility also thank you very much for qm clarification I was very confused thinking that but I forgot about fuzzy clouds model when reasoning

TonyPhysicslover

11:42

@JohnRennie Sir, I tried to solve the question that @RR. posted above, I was able to solve the question using the fact that the normal forces at each point of contact acts towards the centre of football, but why is it necessary that the normal contact forces at each point acts towards the centre.

For Example, consider the Figure F2 The plank is in the horizontal position (not tilted) , Where the normal contact forces at each point of contact acts prependicular to the plank, Using this arrangement too, Net N in vertical direction = mg ( balances out)

RR.

@Wolgwang The solution given followed your method... Though, I preferred sir's method as it was easier... You could check if it solves your doubt

Wolgwang

12:10

@RR. Oh! Thanks.

I was doing something slightly different. That why I was getting sonething elde.

@RR. Can you please send that COM approach solution?

RR.

15:41

@Wolgwang Sure... Give me a min

RR.

16:04

6 hours ago, by RR.

So, when centre of the sphere is just above the pivot, it would just begin to roll?

@Wolgwang Because of this

Wolgwang

Cool. Thanks

Transcript for

Problem Solving Strategies