@PinkAura Like this?

@JohnRennie yes sir

@PinkAura Use your favourite bitmap editor. I used an old program called Paint Shop Pro because I have it on my PC, but most bitmap editors will work. I just used the "invert colours" function.

@JohnRennie idk what it is :(

can you suggest any of them which is free @JohnRennie?

are you there, sir? @JohnRennie

@PinkAura Hi

You can have a copy of Paint Shop Pro

@PinkAura Are you there? If so I'll post a link to it then delete it.

Hello Everyone...

@JohnRennie yes sir

what do i have to do?

okay i clicked on it

it is downloading

So it should be downloading now. Yes?

you can delete that ig

@JohnRennie yes it finished downloading

what should i do now?

It's a self extracting zip. Just run it and let it extract.

it got extracted

what should i do now?

The program is in the PSP folder. Copy that somewhere convenient, then go into the folder and run the psp.exe file inside that folder.

ok

It's a basic bitmap editor but it's quite powerful.

i see, i am new to this, i dont know what bitmap is

The image you posted earlier was a PNG file, which is a type of bitmap. I opened it in Paint Shop Pro and used the Colours/Negative image function.

This

@JohnRennie it shows this, what should i do next?

Ah, are you using a Macbook?

@JohnRennie yes

Ah, sorry, it's a Windows program. I just assumed you were using Windows.

It won't work on a Mac.

ohh damn , that's why i hate apple lol

I can't help with a Mac I'm afraid because I don't use them. Sorry :-(

@JohnRennie any other apps for mac?

@JohnRennie my roommate uses windows , can i run it on his laptop?

One of these possibly

@PinkAura Yes

@JohnRennie so , i just need to share this file to his laptop, right?

I'd copy it onto a USB key and transfer it to your friend's laptop that way.

@JohnRennie umm ... currently he's not here, he would return in the evening...

btw does it convert photos ? can't we convert pdfs using these all at once?

because im afraid there are lot of them

It converts photos. I don't know of a way to convert pdfs.

i see, thank you so much for the help, that's the closest i could get

You're welcome :-)

OK :-)

Hi @JohnRennie :-)

How is it going ?

@Pizza Hi :-)

Sorry for the slow reply but I have lots of students asking me stuff today.

@JohnRennie Ok don't worry, I wanted to know how it was going these days

I noticed that many concepts from physics are also taken into mathematics in vector fields, rotors

For example, calculate the flux of the vector field through a surface

@JohnRennie Hi, are you free?

@Gian'sPizzeria I'm sorry I'm really busy at the moment :-(

If you want to post your question i can ping you as soon as I'm free ...

Okay :⁠-⁠)

A pulley of mass M has the shape of a homogeneous hollow cylinder with external radius R = 8 cm and internal radius r = 6 cm. It is bound to its axis of symmetry, around which it can rotate without friction, by rigid rays of negligible mass. At the end of one of the rays, at a distance r from the axis, a mass m1 = 1 kg of negligible dimensions is fixed. The moment of inertia of the pulley (mass m1 included) with respect to the rotation axis is I = 1.36 × 10^2 kg m^2.

An inextensible rope of negligible mass is rolled around the pulley to which a mass m2 = 100 g is suspended. Determine: 1) equilibrium angle $\theta_0$ 2) The mass $M$

@Gian'sPizzeria How far have you got with this?

I can't find the M area @JohnRennie

M=areaM•density

Are you trying to find an expression for the moment of inertia of the cylinder?

I'm trying to find rho

However I'm making the 2nd point

@JohnRennie yes

And the density is then just ρ = M/V

Or have I misunderstood what you are trying to do?

Mmm but isn't it a surface?

@JohnRennie no

Okay

@JohnRennie I have to find M

So I think this rho should be put into another expression

To then find M

The way you are going to find M is to write down an expression for the total MOI of the cylindrical shell. This will contain M as a variable.

Then you can equate this to the value given and from that you can solve for M.

Yes?

Yes, but I don't know how to do it

In theory: I=1/2M•R^2-1/2m•r^2+m1•r^2

Is it right?

I'm not sure if this is the formula to use

The way we are going to find the MOI of the shell is to start with the MOI of a solid cylinder of radius R and mass m₁ and subtract the MOI of a smaller cylinder of mass m₂. We don't know m₁ and m₂, but we do know that m₁ - m₂ = M.

OK so far?

Yes

I=1/2M•R^2-1/2m•r^2+m1•r^2 , but is this formula correct?

@JohnRennie okay

@Gian'sPizzeria That equation would be correct if M was the mass of the large cylinder (what I've called m₁) and m was the mass of the small cylinder (what I've called m₂).

Yes?

Yes

Oh, wait, the small mass is m₁

Let's use M₁ and M₂ for the cylinders.

Yes?

Yes?

Yes

OK :-)

So we need another equation relating M₁, M₂ and M so we can solve the two equations to get M₁ and M₂. Yes?

Yes

Yes?

Yes

Yes

So V₁/V₂ = R²/r²

Yes?

Yes

Now we have to solve the system with the 3 equations and from there find M?

Yes :-)

Do you want to go through that together or will you have a go on your own?

Mmm I understand how to do it

OK

Can we make the first point?

Yes

Yes

Yes

The torque created by m₁ is the force m₁g times the normal distance to the axis that I have labelled in blue as d

Yes?

Yes

And d = r sinθ

So τ₂ = m₁gr sinθ

OK so far?

Yes

And now we just equate τ₁ and τ₂ since at equilibrium the two torques have to be equal

m₂gR = m₁gr sinθ

Thanks very much

You're welcome :-)

Hello @JohnRennie Sir

@JohnRennie I have a doubt related to discharge tube. If both anode rays and cathode rays are produced simultaneously in opposite direction how does paddle wheel move? And why in opposite direction when considering anode rays and cathode rays?