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This is the attached diagram

So this thing is just heading towards O which then acts as a point of rotation?

Yes

Okay...I just to do something, then I'll take a look.

*just go to do something*

Okay, I will be waiting

Still waiting?

Yes

You have two regimes - translational (before the ridge), then rotational.

The kinetic energy of the block before hand is \[K_i=\frac{1}{2}Mv^2\]

Is this making sense so far?

Yes

If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)

I get something close, but different.

Oh I am sorry, I calculated the I around the Cm

Which is wrong

Yes

So how do you calculate it around that corner?

Parallel Axis Theorem

Oh yes, you are right

\[I_{cm}=\frac{Ma^2}{6}\]

\[Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}\]

I think it is \[1/6(Ma^2)+2M\]

So \[I=\frac{5Ma^2}{12}\]

You see the corner wise distance will be sqrt 2

I mean sqrt 2 times a

\[w^2=\frac{12v^2}{5a^2}\]

Yes, you're right there.

So change h to a.sqrt(2)

a/sqrt(2)

\[h=\frac{a}{\sqrt{2}}\]

So I am supposed to solve the equation \[0.5 Mv^2=0.5[(1/6)Ma^2+2M]\omega\]
Am I right

am i right?

\[=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}\]

Then\[\frac{1}{2}Mv^2=\frac{1}{2}\frac{2Ma^2}{3} \omega ^2\]

i.e.\[v^2=\frac{2a^2}{3} \omega ^2 \rightarrow \omega ^2 = \frac{3v^2}{2a^2}\]

Ok I understand that

Are you agreeing? I'm doing this on-the-fly, so if you pick up anything, interrupt.

Any how I have got the thing

Did your answer agree?

Lokisan, I have got what I expected,
I can handle the rest

There was some problem with the site, and some of our converstation got split up

Okay. Happy physics.

lol, thanks for the genius comment. we'll just see how this question pans out.