Question

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

Answer 1

This is the attached diagram

Answer 2

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

Answer 3

Yes

Answer 4

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

Answer 5

*just go to do something*

Answer 6

Okay, I will be waiting

Answer 7

Still waiting?

Answer 8

Yes

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Is this making sense so far?

Answer 14

Yes

Answer 15

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

Answer 16

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...

Answer 17

I get something close, but different.

Answer 18

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

Answer 19

Which is wrong

Answer 20

Yes

Answer 21

So how do you calculate it around that corner?

Answer 22

Parallel Axis Theorem

Answer 23

Oh yes, you are right

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

You see the corner wise distance will be sqrt 2

Answer 30

I mean sqrt 2 times a

Answer 31

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

Answer 32

Yes, you're right there.

Answer 33

So change h to a.sqrt(2)

Answer 34

a/sqrt(2)

Answer 35

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

Answer 36

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

Answer 37

am i right?

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

Ok I understand that

Answer 43

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

Answer 44

Any how I have got the thing

Answer 45

Did your answer agree?

Answer 46

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

Answer 47

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

Answer 48

Okay. Happy physics.

Answer 49

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