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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

This is the attached diagram

##### 1 Attachment

- anonymous

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

- anonymous

Yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

*just go to do something*

- anonymous

Okay, I will be waiting

- anonymous

Still waiting?

- anonymous

Yes

- anonymous

I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.

- anonymous

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

- anonymous

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

- anonymous

The kinetic energy of the block just after impact, as it enters the rotational regime, is\[K_f=\frac{1}{2}I \omega^2\]

- anonymous

Is this making sense so far?

- anonymous

Yes

- anonymous

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

- anonymous

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...

- anonymous

I get something close, but different.

- anonymous

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

- anonymous

Which is wrong

- anonymous

Yes

- anonymous

So how do you calculate it around that corner?

- anonymous

Parallel Axis Theorem

- anonymous

Oh yes, you are right

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

You see the corner wise distance will be sqrt 2

- anonymous

I mean sqrt 2 times a

- anonymous

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

- anonymous

Yes, you're right there.

- anonymous

So change h to a.sqrt(2)

- anonymous

a/sqrt(2)

- anonymous

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

- anonymous

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

- anonymous

am i right?

- anonymous

I think the moment of inertia should be\[I=\frac{Ma^2}{6}+Mh^2=\frac{Ma^2}{6}+M \left( \frac{a}{\sqrt{2}} \right)^2=\frac{Ma^2}{6}+\frac{Ma^2}{2}\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Ok I understand that

- anonymous

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

- anonymous

Any how I have got the thing

- anonymous

Did your answer agree?

- anonymous

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

- anonymous

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

- anonymous

Okay. Happy physics.

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.