A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details

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A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details

Mathematics
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This is the attached diagram
1 Attachment
So this thing is just heading towards O which then acts as a point of rotation?
Yes

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Other answers:

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
I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.
You have two regimes - translational (before the ridge), then rotational.
The kinetic energy of the block before hand is \[K_i=\frac{1}{2}Mv^2\]
The kinetic energy of the block just after impact, as it enters the rotational regime, is\[K_f=\frac{1}{2}I \omega^2\]
Is this making sense so far?
Yes
If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)
Now, your problem is now about finding the moment of inertia for a block whose axis of rotation is about one of its edges. A solid, rectangular parallelpiped is given by...
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
The new moment of inertia will be\[I=I_{cm}+Mh^2\] where h is the distance from the CM axis to the new, parallel axis.
\[I_{cm}=\frac{Ma^2}{6}\]
\[Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}\]
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?
I think the moment of inertia should be\[I=\frac{Ma^2}{6}+Mh^2=\frac{Ma^2}{6}+M \left( \frac{a}{\sqrt{2}} \right)^2=\frac{Ma^2}{6}+\frac{Ma^2}{2}\]
\[=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}\]
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.

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