Debanjan Biswas

Someone can get easily confused in some cases

It is really hard to tell which approximation is valid or not valid beforehand

@JohnRennie Sir, I think I got it. The process you mentioned is only applicable when the contact is fixed. The method also implies that it doesn't matter if the surface is frictionless or with friction. The reason is that the contact is fixed somehow

@JohnRennie why are we getting different expressions for omega?

And the expression for ω_com can be verified from energy conservation

Because ω_com=dθ/dt and simply ω_con.=d(π/2-θ)/dt=-ω_com

Sir, I doubt if the expression for omega here is correct assuming the former angular velocity respect to com is correct

And it results in a different value of time

The angular velocity square respect to the contact is equal to ω²=3g(sinθ_0-sinθ)/L

But sir now I have got another question. If I follow the later process you mentioned which is very simple and elegant, results in a different value of time. Here

@JohnRennie Sir, my apologies. Omega has real value. But, as I mentioned the first integral also gives finite results for theta not not equal to zero. Because the value of the function that is being integrated including the term (cos theta not - cos theta) gets bounded and the area under curve approaches a finite value

@JohnRennie In the method you mentioned what I got is more terrible. Omega becomes imaginary for theta less than theta not

Sir, what is going on?

I don't know if wolfram alpha got it wrong but I checked for many values of theta not and the integral doesn't converge only if theta not is zero

But how does it converge?

But sir I checked with theta not equal to 1 in wolpharm alpha and it showed a finite value. Does it mean that integral also converges for theta not not equal to π/2?

Hmmm sir, I understand it now

@JohnRennie sir, what is the reason behind the rod not start rotating even after it's set free?

Sir I am here

@JohnRennie Sir I am here now, could you please pin a time for the discussion? I live in India

Sir I am trying to say that if theta not was zero, there would be no torque on the rod and hence it is understandable why it takes infinite time. But suppose we have theta not π/6, why wouldn't it fall? Doesn't taking infinite time to fall means for every value of theta not between 0 and π/2, the rod is in a unstable equilibrium?

It may not be verified experimentally because I suppose even on a plane with very very low coefficient of friction, even due to the slightest disturbance it would fall in a short time

What makes the rod fall so slow?

It takes infinite time to fall even though the torque on the rod relative to its centre always has a finite value

Yes sir, but why is it like that? I have seen similar cases before but they were intuitive. But it seems non intuitive and unrealistic

But the problem is the integral doesn't converge. It diverges and results in an infinite time. It would be no problem if θ_0 was zero but it's not the only case

Now expressing ω as dθ/dt and taking the integral from θ_0 to π/2, the time can be calculated

i. e ω²=12g(cosθ_0-cosθ)/L(1+3sin²θ)

Suppose there's a rod tilted at an angle θ_0 with vertical with one of its ends on a frictionless plane. Now I wanted to calculate the time how long will it take for the rod to land on the plane horizontally. First I derived the equation for the square of it's angular velocity respect to its centre of mass

Sir, I have a question that is bothering me for quite a time

If someone induces circular motion, there will be a stretching and compression of the string periodically happening in molecular level. Is there a way to say in general that every change in macroscopic motion will result in a change in oscillatory and rotational motion of molecules? Please pardon me if anything is unclear. I will try to express it correctly

@JohnRennie As it seems to me every macroscopic motion no matter how simple it is always contain some oscillating, rotational motion at molecular level. Just assume a small ball tied to a string in a horizontal position resting on a frictionless plane with one of its end fixed at some point

And its not equal to the work done+initial EPE

I am just seeing the rise in GPE which is m²g²/2k

@JohnRennie Sir, you have explained everything in the h Bar but I am just too dumb to understand. I couldn't figure out how is the energy stored in the spring after you rise it to its natural length slowly? Please help me with this

Why isn't there a possible shape of conductor in which charges can never reach electrostatic equilibrium if an external field is present or additional charges are given?

Can you recommend me a book where the derivation of resistance of a sphere, hemisphere or any arbitrary 3d shape is shown?

Isn't that a different scenario? In my experiment, it's always attached to the ceiling

:-)

@JohnRennie Yes, I really like this thing about physics

@naturallyInconsistent Thanks for pointing out my mistake in the question

@JohnRennie Thanks for your assistance

Oh my, I ignored the change in gravitational potential in this whole process

Ah, yes

Sir, wait so now elastic energy+work done = increase in gravitational potential energy?

Yes that is the increase in gravitational potential energy

@JohnRennie yes

@PM2Ring oh I see! Please pardon me for the confusion

prove that √(a²-1) is irrational if a is rational and ≠1,-1