@UtkarshJha sorry to be so slow replying - yesterday turned out to be a really busy day :-(

@UtkarshJha yes you are correct. We define a frame centred on our reference point, then we can define position vectors r and velocities v for other points. Then the angular velocity is defined by ω = r x v.

It may seem odd to take a point that isn't the axis of rotation, but mathematically it still works and the conservation laws still apply e.g. angular momentum is still conserved unless an external torque is acting.

physics.stackexchange.com/questions/635263/… @JohnRennie can I know what edits should I do to the question to reopen it?

@IITM That question isn't going to be reopened, but we can discuss it here if you want.

@JohnRennie sure,I would like to get it answered here

@IITM Can I use MathJax? Do you have the MathJax addon for chat?

Incidentally there is a solution here if you want to read through it.

@JohnRennie yes you can use mathjax

@JohnRennie the solution is actually in the book itself but I did not understand that

OK. There's a really quick and easy way to solve this.

As you say in your question the force on the mass is:

$$ F =mg\sin\theta - mgbx\cos\theta $$

And this has the form:

$$ F = A - kx $$

where $k = mgb\cos\theta$

OK so far?

@JohnRennie Thanks for replying :). Can you tell what is the answer to question (2)? I think that the frame fixed to the body should be one that is translating all the time (and which may be accelerating w.r.t. an inertial frame), because it is only then that we can see the position vector rotating about the point and measure omega.

@UtkarshJha well the frame is the rest frame of our reference point, so by definition it has the same velocity as the reference point. If the reference point is not in inertial motion then the frame is the inertial frame that has the same instantaneous velocity as the reference point.

Yes. This is what we mean by the instantaneous centre of rotation.

@JohnRennie ok understood

@JohnRennie sorry for the late reply I was eating then

so couldn't message

@IITM and that's just the equation for simple harmonic motion with force constant $k = mgb\cos\theta$.

Then the period of the motion is $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{1/mgb\cos\theta}$. Yes?

@JohnRennie yes

got the problem

but can you say what is the mistake in my process

And the time for the particle to come to rest is half the period. That's how you get $$t=\frac{\pi}{\sqrt{bgcos\theta}}$$

any mistake in my process? why didn't I get the answer?

@IITM I'm answering another question at the moment. I'll look through your working as soon as I'm free.

@JohnRennie Ok,Thank you so much for the explanation,I'll be waiting for the reply

@JohnRennie. Thanks for helping :) I think I understand things better now. Thanks.

@UtkarshJha :-)

@IITM hi, sorry to take so long.

Your working looks fine to me, but you ended up calculating the value of x when the particle stopped, not the time at which it stopped.

@JohnRennie but from last equation $(\frac{dx}{dt})^2=2(gxsin\theta-b\frac{x^2}{2}gcos\theta)$,I could find relation for $x$ and $t$ so as I got $x$ I could get $t$?

but this didn't work...:(