Problem Solving Strategies

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Harjot Dhillon

09:18

Hello @JohnRennie

How could I invite you in jee math chat room on math SE

John Rennie

@HarjotDhillon Hi :-)

2 hours later…

Kavin Ishwaran

11:18

@JohnRennie Will you be around at 9:30PM IST ?

John Rennie

11:30

@KavinIshwaran Yes :-)

Kavin Ishwaran

11:49

Ok :-)

1 hour later…

Bumblebee

12:55

@JohnRennie As you are a scientist who has been working in the natural sciences for many years, I wanted to ask if you knew the answer to this question: what conditions must something pass in order to be classified as scientific knowledge?

2 hours later…

PandaScientist

15:24

Hello @JohnRennie !

John Rennie

Hi :-)

PandaScientist

I'm sending the question

This is a similar question. Can I go through my solution? So that you could point the mistake

John Rennie

OK ...

PandaScientist

If pulley moves by x, the block moves by 2x

Not at x, let the acceleration of cm of pulley be $$\frac{a+b+Mg-kx}{M}$$

Where a and b are the tensions from left to right

Then since pulley is not slipping:

$$\frac{2(b-a)}{M} + \frac{a+b+Mg-kx}{M}=\frac{mg-b}{m}$$

John Rennie

@PandaScientist I think this is wrong ...

The Mg shouldn't be there.

PandaScientist

15:33

I felt that before, but then I thought if gravity is working on block , shouldn't it be working on the pulley too?

@JohnRennie and...I did it without mg too. Still got no answer :-(

John Rennie

I will have to think about this one. I don't immediately see how to do it.

PandaScientist

Can I proceed without mg and continue?

John Rennie

And I don't want to to do it right now. I will try to look at it tomorrow morning.

PandaScientist

ok sir

John Rennie

Sorry :-(

PandaScientist

15:37

Np sir :-) your assurance is enough !

One quick doubt, in cases where we have torsional SHMs, their actual $\omega$ is different from the $\omega$ in $-{\omega}^2 \theta$ , right?

John Rennie

No, I don't think so.

PandaScientist

Can you pls explain?

John Rennie

Obviously we get ω² = k/I instead of k/m but otherwise it's the same equation.

PandaScientist

I Think I didn't convey my doubt properly ! Let me rephrase ...

John Rennie

OK ... ?

PandaScientist

15:44

Say, there's a circular disc rotating but follows SHM

It's omega is diff. from the omega in ${\omega}^2 \theta$ right?

John Rennie

So when we rotate it an angle θ there is a restoring torque τ = kθ

Yes?

PandaScientist

Yes

John Rennie

And τ = Iα so we end up with:
α = -(k/I) θ

Oh I think I see.

PandaScientist

...and now, I got it :-)

John Rennie

OK :-)

PandaScientist

15:48

They are different.

@JohnRennie thank you sir :-)

Kavin Ishwaran

16:05

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

16:23

@JohnRennie How to do this ?

John Rennie

@KavinIshwaran You need to first find the current at steady state. If you use Kirchoff round the outer loop you will find the current is zero. Yes?

Kavin Ishwaran

@JohnRennie At Steady state, the branch A B is ineffective right ?

John Rennie

Yes. There is zero current through the capacitor at steady state.

Kavin Ishwaran

@JohnRennie Yes

John Rennie

Now consider the top loop. The voltage drop in R is zero because I = 0, so the only voltage changes are the two batteries and the capacitor. Yes?

Kavin Ishwaran

16:29

But we are applying it at steady state knowing no current flows through AB right ?

John Rennie

Yes

Because we know the steady state current through a capacitor is always zero.

Kavin Ishwaran

Yes

John Rennie

So if you go round the top loop since the voltage across R is zero, and we know the voltage across the batteries, we can find the voltage cross the capacitor.

Yes?

Kavin Ishwaran

Yes

John Rennie

And that voltage comes out as ... ?
(actually this surprised me a bit)

Kavin Ishwaran

16:36

@JohnRennie Suppose if we take a point next to the middle loop's battery's positive terminal as C (lets take it as zero) and going form C to B through A gives the potential drop to be zero

So PD = zero ?

John Rennie

@KavinIshwaran Yes :-) Option (1)

Kavin Ishwaran

I get it :-)

John Rennie

I must admit I wouldn't have guessed that without doing the analysis.

Kavin Ishwaran

Oh !

But this question troubled me a lot :-)

John Rennie

But you are happy you know how to approach it now?