6:14 AM
Sir do you know a good theory based book on astronomy with less or no math?

@PrateekMourya hi :-)
I don't know much about astronomy I'm afraid.
There is an astronomy SE and they may have a chat room.

@JohnRennie Hi

@YouKnowMe hi :-)

Are you free?

In questions like this remember that Newton's second law always applies.

6:26 AM
Yes
First we will find acceleration

So if we know the acceleration of B then we know the force on B because it's just F = ma

of system

@YouKnowMe yes. And the acceleration of the system is ... ?

3/2

Yes, so the acceleration of B is ... ?

6:28 AM
3/2

So the force on B is ... ?

BOOM! :-)

But why did force get distributed?

Suppose A is a spring. Actually it is kind of since all solids are slightly elastic. Then A is slightly compressed by the force being applied to its left side. Yes?

6:30 AM
Yes

And because it is compressed it exerts a force against whatever is at its ends.
One of those ends is at the left where the force is being applied, and the other is at the right side where it pushes against B.

@JohnRennie What is the force exerted by A on B?

3N
The only thing exerting a force on B is A, so the whole 3N force on B comes from the force exerted on it by A.
@PrateekMourya this book looks interesting:
6

I'm looking for a modern astrophysics text. I've had Barbara Ryden's book Foundations of Astrophysics recommended to me and it sounds okay, but the only review of it I've seen slagged off the problem sets as "more complex than they would initially seem, and often don't actually derive from the ch...

It looks as if the Astronomy SE allows book recommendation questions, so if you can pin down exactly what sort of book you want you could post a question there.

@JohnRennie Thanks :-p

7:18 AM
$\vec{F_B}=q\vec v \times \vec{B}$
@JohnRennie If the force acting on a charge is given by the above equation. Will the velocity of charge change due to the this force?

The force is always at right angles to the velocity because of the cross product. Yes?

Yes

And what sort of motion has a constant acceleration at right angles to the velocity?

Circular

Yes :-) What the Lorentz force does is make the charge move in a circle at constant speed.
The electron will move in a circular path in the xz plane (z axis coming out of the page towards us). Yes?

7:29 AM
NVM I got it ;-)

:-)

Hello @JohnRennie sir

@PrateekMourya hi :-)

Sir suppose a ring is rotating
And a particle in its inerior
Not on ring
How can one apply rotation forces(pseudo here?
I mean revolving around its axis

What are you asking? How to model the motion of the particle in the rest frame of the ring?

7:37 AM
I am so super confused with pseudo forces of rotation frame
The derivation they take
Is to consider a frame rotating such that z axis is fixed

As a general rule avoid working in non-inertial frames. They are always confusing and it's easy to make mistakes.
Even simple things become surprisingly complicated.

@JohnRennie hi

@JackRod hi :-)

sir suppose I have a chalk

7:40 AM
@PrateekMourya in the ground frame nothing will happen to P.

and I dropped it from a certain height and I ask u what are the possible degree of freedom what would be ur answer @JohnRennie

In the rest frame of the ring, i.e. in a rotating frame, there will be coriolis pseudoforce on P.

What stops p then in that frame?
At 5:20

Give me a moment to look at the video ...
@PrateekMourya there is nothing happening at 5:20. Just Verma talking. I don't understand what you are asking. Also that is a disk not a ring. Are you asking why a particle on a rotating disk slides outwards?

Yes
I am asking both in case of ring and disk
Sorry for wrong time stops
I was referring why particle goes away

7:52 AM
@JohnRennie hi sir
11 mins ago, by Jack Rod
and I dropped it from a certain height and I ask u what are the possible degree of freedom what would be ur answer @JohnRennie

@PrateekMourya this is a big topic and would be a long discussion. I don't want to do this at th moment.
@JackRod I have no idea because I'm not sure what the question means.
A piece of chalk isn't a statistical system so it isn't clear to me what degrees of freedom would mean.

no sir in genreal
if I say we dropped a chalk one can say it has only one degree of freedom
which is along the forces

The motion would be one dimensional in the sense that all the motion is along a straight line, but I wouldn't use the term degree of freedom to describe this.

I was studying the core classical physics
they have used termonolgy there

I guess you could say the chalk has only one degree of freedom because it only moves along a straight line.

8:02 AM
naa sir he said it has n degree of freedom

I would have to see the lecture to understand what the lecturer meant.

i feel like i forgot newtonina mech when we deal it with 4d vector
@JohnRennie

What time?

17.49
@JohnRennie

What he is saying is that you cannot predict the trajectory of the chalk if you only know its position. You have to know both its position and its velocity.

8:11 AM
why there is a need of a degree of freedom in this

He doesn't mention "degrees of freedom" in that part of the video.

he does sir just roll back some 5-6 min

What he is saying is that the equation of motion is a second order equation i.e. Newton's second law.

he is talking about degree of freedom in initial phase

When he is talking about the cube?
Or when he is talking about a system of point particles?

8:16 AM
system of point particles

All he is saying is that for a system of point particles each particle has a position vector $\mathbf r = (x, y, z)$.
So each particle needs three numbers to specify its position.
So if we have N particles we need 3N numbers to specify the positions of all the particles in the system.

Sir if you get time please review this question if there is something missing please tell me or correct it before anyone closes this
0

https://youtu.be/FzfOZn3xnMM In this video sir took a disk and demonstrated that the particles kept on it flies away Dur to centrifugal force Now suppose i took a ring in free space and a particle in its interior (Not on circumference) If we observe it from a frame attached to ring axis such that...

A particle on a disk has a force exerted on it by the disk - specifically the frictional force between the particle and the surface of the disk.
But for a ring the particle and the ring are not touching so the ring exerts no force on the particle.
That's the difference between the two systems.

9:04 AM
So in order to apply pseudo force on it from a rotating frame it must somehow interact with the fram?
@JohnRennie sir?

7 hours later…
4:30 PM
1

If we observe it from a frame of reference of the ring , since ring is rotating the particle must experience a centrifugal force and hence it will fly off . But from space frame, it doesn't. Why does this happens ? The centrifugal force is not the only inertial force in a rotating reference fra...

@JohnRennie is this seems fine to you too
I am confirming
Since it will play an important role in building my concept

@PrateekMourya hi :-)
@PrateekMourya it looks OK, though I'm not sure how helpful it is ...

@JohnRennie Respected sir I am friend of Knight..
Sir, I’m friend of Knight. He told me to talk to you in my need. Please allow me to converse with you sir.”

@KnightadmiresChappo hi friend of Knight :-)

Sir, when can I ask my questions from you?

4:50 PM
I'm usually around from about 05:30 to 12:30 UK time - that's about 11:00 to 18:00 Indian time, and I'm sometimes around at this time.

okay sir thanku so much sir

@KnightadmiresChappo I'm not sure to be honest. I'll have to think about it.
Is that a JEE question? It looks hard for a JEE question.

@JohnRennie Sir, it’s on 2nd year UG course
no sir..its university question

What you need to do is figure out the acceleration of the bead, then the force is just F = ma. The force will be normal to the wire since the wire is frictionless.
The acceleration comes from the bead's motion outwards along the wire, but also because the direction of the bead's velocity changes as the wire rotates.
I think I have done the equations of motion for this before but it was a long time ago and I can't remember how the problem is solved.

5:09 PM
okay..sir tomorrow I will try and let you know..

OK :-)

Thankyou sir

@KnightadmiresChappo I found a description of the problem here:
There's also a solution using Lagrangian mechanics. I don't know if you've studied Hamiltonian and Lagrangian mechanics yet.

@JohnRennie Okay sir, I will try to do it using Lagrangian mechanics.